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Continuity and Differentiability

Chhattisgarh Board · Class 12 · Mathematics

NCERT Solutions for Continuity and Differentiability — Chhattisgarh Board Class 12 Mathematics.

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137 Questions Solved · 8 Sections

Exercise 5.1

1Prove that the function f(x)=5x−3f(x) = 5x - 3 is continuous at x=0x = 0, at x=−3x = -3 and at x=5x = 5.Show solution
Given: f(x)=5x−3f(x) = 5x - 3

Concept: A function ff is continuous at x=cx = c if lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c).

At x=0x = 0:
f(0)=5(0)−3=−3f(0) = 5(0) - 3 = -3
lim⁡x→0f(x)=lim⁡x→0(5x−3)=5(0)−3=−3\lim_{x \to 0} f(x) = \lim_{x \to 0}(5x - 3) = 5(0) - 3 = -3
Since lim⁡x→0f(x)=f(0)=−3\lim_{x \to 0} f(x) = f(0) = -3, ff is continuous at x=0x = 0.

At x=−3x = -3:
f(−3)=5(−3)−3=−15−3=−18f(-3) = 5(-3) - 3 = -15 - 3 = -18
lim⁡x→−3f(x)=5(−3)−3=−18\lim_{x \to -3} f(x) = 5(-3) - 3 = -18
Since lim⁡x→−3f(x)=f(−3)=−18\lim_{x \to -3} f(x) = f(-3) = -18, ff is continuous at x=−3x = -3.

At x=5x = 5:
f(5)=5(5)−3=25−3=22f(5) = 5(5) - 3 = 25 - 3 = 22
lim⁡x→5f(x)=5(5)−3=22\lim_{x \to 5} f(x) = 5(5) - 3 = 22
Since lim⁡x→5f(x)=f(5)=22\lim_{x \to 5} f(x) = f(5) = 22, ff is continuous at x=5x = 5.

Hence, f(x)=5x−3f(x) = 5x - 3 is continuous at x=0x = 0, x=−3x = -3, and x=5x = 5. ■\blacksquare
2Examine the continuity of the function f(x)=2x2−1f(x) = 2x^2 - 1 at x=3x = 3.Show solution
Given: f(x)=2x2−1f(x) = 2x^2 - 1

At x=3x = 3:
f(3)=2(3)2−1=18−1=17f(3) = 2(3)^2 - 1 = 18 - 1 = 17
lim⁡x→3f(x)=2(3)2−1=17\lim_{x \to 3} f(x) = 2(3)^2 - 1 = 17

Since lim⁡x→3f(x)=f(3)=17\lim_{x \to 3} f(x) = f(3) = 17, the function ff is continuous at x=3x = 3.
3Examine the following functions for continuity.
(a) f(x)=x−5f(x) = x - 5
(b) f(x)=1x−5, x≠5f(x) = \frac{1}{x-5},\ x \neq 5
(c) f(x)=x2−25x+5, x≠−5f(x) = \frac{x^2-25}{x+5},\ x \neq -5
(d) f(x)=∣x−5∣f(x) = |x-5|Show solution
(a) f(x)=x−5f(x) = x - 5

ff is a polynomial function. For any c∈Rc \in \mathbb{R}:
lim⁡x→c(x−5)=c−5=f(c)\lim_{x \to c}(x-5) = c - 5 = f(c)
Hence ff is continuous at every real number.

---

(b) f(x)=1x−5, x≠5f(x) = \dfrac{1}{x-5},\ x \neq 5

The domain of ff is R−{5}\mathbb{R} - \{5\}. For any c≠5c \neq 5:
lim⁡x→c1x−5=1c−5=f(c)\lim_{x \to c} \frac{1}{x-5} = \frac{1}{c-5} = f(c)
Hence ff is continuous at every point of its domain (i.e., at all x≠5x \neq 5). At x=5x = 5, ff is not defined.

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(c) f(x)=x2−25x+5, x≠−5f(x) = \dfrac{x^2-25}{x+5},\ x \neq -5

For x≠−5x \neq -5:
f(x)=(x−5)(x+5)x+5=x−5f(x) = \frac{(x-5)(x+5)}{x+5} = x - 5
For any c≠−5c \neq -5:
lim⁡x→cf(x)=c−5=f(c)\lim_{x \to c} f(x) = c - 5 = f(c)
Hence ff is continuous at every point of its domain (i.e., at all x≠−5x \neq -5). At x=−5x = -5, ff is not defined.

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(d) f(x)=∣x−5∣f(x) = |x - 5|

We can write:
f(x)={x−5,amp;x≥5−(x−5),amp;xlt;5f(x) = \begin{cases} x - 5, & x \geq 5 \\ -(x-5), & x < 5 \end{cases}

For c > 5: lim⁡x→c∣x−5∣=c−5=f(c)\lim_{x \to c}|x-5| = c - 5 = f(c). ✓

For c < 5: lim⁡x→c∣x−5∣=−(c−5)=f(c)\lim_{x \to c}|x-5| = -(c-5) = f(c). ✓

At c=5c = 5:
lim⁡x→5−∣x−5∣=0,lim⁡x→5+∣x−5∣=0,f(5)=0\lim_{x \to 5^-}|x-5| = 0,\quad \lim_{x \to 5^+}|x-5| = 0,\quad f(5) = 0
All equal. ✓

Hence ff is continuous at every real number.
4Prove that the function f(x)=xnf(x) = x^n is continuous at x=nx = n, where nn is a positive integer.Show solution
Given: f(x)=xnf(x) = x^n, nn is a positive integer.

At x=nx = n:
f(n)=nnf(n) = n^n
lim⁡x→nf(x)=lim⁡x→nxn=nn\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n

Since lim⁡x→nf(x)=f(n)=nn\lim_{x \to n} f(x) = f(n) = n^n, the function f(x)=xnf(x) = x^n is continuous at x=nx = n. ■\blacksquare
5Is the function ff defined by
f(x)={x,amp;if x≤15,amp;if xgt;1f(x) = \begin{cases} x, & \text{if } x \leq 1 \\ 5, & \text{if } x > 1 \end{cases}
continuous at x=0x = 0? At x=1x = 1? At x=2x = 2?Show solution
At x=0x = 0:
f(0)=0f(0) = 0
lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0
Since lim⁡x→0f(x)=f(0)\lim_{x \to 0} f(x) = f(0), ff is continuous at x=0x = 0.

---

At x=1x = 1:
f(1)=1f(1) = 1
lim⁡x→1−f(x)=lim⁡x→1−x=1\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1
lim⁡x→1+f(x)=lim⁡x→1+5=5\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5

Since LHL ≠\neq RHL, lim⁡x→1f(x)\lim_{x \to 1} f(x) does not exist. Hence ff is not continuous at x=1x = 1.

---

At x=2x = 2:
f(2)=5f(2) = 5
lim⁡x→2f(x)=5\lim_{x \to 2} f(x) = 5
Since lim⁡x→2f(x)=f(2)=5\lim_{x \to 2} f(x) = f(2) = 5, ff is continuous at x=2x = 2.
6Find all points of discontinuity of ff, where
f(x)={2x+3,amp;if x≤22x−3,amp;if xgt;2f(x) = \begin{cases} 2x+3, & \text{if } x \leq 2 \\ 2x-3, & \text{if } x > 2 \end{cases}Show solution
The only possible point of discontinuity is x=2x = 2 (where the definition changes).

f(2)=2(2)+3=7f(2) = 2(2)+3 = 7
lim⁡x→2−f(x)=2(2)+3=7\lim_{x \to 2^-} f(x) = 2(2)+3 = 7
lim⁡x→2+f(x)=2(2)−3=1\lim_{x \to 2^+} f(x) = 2(2)-3 = 1

Since LHL ≠\neq RHL, ff is discontinuous at x=2x = 2.

For x < 2: f(x)=2x+3f(x) = 2x+3 is a polynomial, hence continuous.
For x > 2: f(x)=2x−3f(x) = 2x-3 is a polynomial, hence continuous.

Conclusion: x=2x = 2 is the only point of discontinuity.
7Find all points of discontinuity of ff, where
f(x)={∣x∣+3,amp;if x≤−3−2x,amp;if −3lt;xlt;36x+2,amp;if x≥3f(x) = \begin{cases} |x|+3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x+2, & \text{if } x \geq 3 \end{cases}Show solution
Possible points of discontinuity: x=−3x = -3 and x=3x = 3.

At x=−3x = -3:
f(−3)=∣−3∣+3=3+3=6f(-3) = |-3|+3 = 3+3 = 6
lim⁡x→−3−f(x)=∣−3∣+3=6\lim_{x \to -3^-} f(x) = |-3|+3 = 6
lim⁡x→−3+f(x)=−2(−3)=6\lim_{x \to -3^+} f(x) = -2(-3) = 6
LHL = RHL = f(−3)=6f(-3) = 6. Hence ff is continuous at x=−3x = -3.

At x=3x = 3:
f(3)=6(3)+2=20f(3) = 6(3)+2 = 20
lim⁡x→3−f(x)=−2(3)=−6\lim_{x \to 3^-} f(x) = -2(3) = -6
lim⁡x→3+f(x)=6(3)+2=20\lim_{x \to 3^+} f(x) = 6(3)+2 = 20
LHL ≠\neq RHL. Hence ff is discontinuous at x=3x = 3.

Conclusion: x=3x = 3 is the only point of discontinuity.
8Find all points of discontinuity of ff, where
f(x)={∣x∣x,amp;if x≠00,amp;if x=0f(x) = \begin{cases} \dfrac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}Show solution
For x > 0: ∣x∣x=xx=1\dfrac{|x|}{x} = \dfrac{x}{x} = 1

For x < 0: ∣x∣x=−xx=−1\dfrac{|x|}{x} = \dfrac{-x}{x} = -1

At x=0x = 0:
f(0)=0f(0) = 0
lim⁡x→0−f(x)=−1\lim_{x \to 0^-} f(x) = -1
lim⁡x→0+f(x)=1\lim_{x \to 0^+} f(x) = 1
LHL ≠\neq RHL, so ff is discontinuous at x=0x = 0.

For x≠0x \neq 0: ff equals a constant on each side, hence continuous.

Conclusion: x=0x = 0 is the only point of discontinuity.
9Find all points of discontinuity of ff, where
f(x)={x∣x∣,amp;if xlt;0−1,amp;if x≥0f(x) = \begin{cases} \dfrac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \geq 0 \end{cases}Show solution
For x < 0: x∣x∣=x−x=−1\dfrac{x}{|x|} = \dfrac{x}{-x} = -1

So f(x)=−1f(x) = -1 for all x < 0 and f(x)=−1f(x) = -1 for all x≥0x \geq 0.

At x=0x = 0:
f(0)=−1f(0) = -1
lim⁡x→0−f(x)=−1\lim_{x \to 0^-} f(x) = -1
lim⁡x→0+f(x)=−1\lim_{x \to 0^+} f(x) = -1
All equal. Hence ff is continuous at x=0x = 0.

For all other points, ff is clearly continuous.

Conclusion: ff has no points of discontinuity.
10Find all points of discontinuity of ff, where
f(x)={x+1,amp;if x≥1x2+1,amp;if xlt;1f(x) = \begin{cases} x+1, & \text{if } x \geq 1 \\ x^2+1, & \text{if } x < 1 \end{cases}Show solution
The only possible point of discontinuity is x=1x = 1.

f(1)=1+1=2f(1) = 1+1 = 2
lim⁡x→1−f(x)=(1)2+1=2\lim_{x \to 1^-} f(x) = (1)^2+1 = 2
lim⁡x→1+f(x)=1+1=2\lim_{x \to 1^+} f(x) = 1+1 = 2

LHL = RHL = f(1)=2f(1) = 2. Hence ff is continuous at x=1x = 1.

For x > 1: f(x)=x+1f(x) = x+1 is a polynomial — continuous.
For x < 1: f(x)=x2+1f(x) = x^2+1 is a polynomial — continuous.

Conclusion: ff has no points of discontinuity.
11Find all points of discontinuity of ff, where
f(x)={x3−3,amp;if x≤2x2+1,amp;if xgt;2f(x) = \begin{cases} x^3-3, & \text{if } x \leq 2 \\ x^2+1, & \text{if } x > 2 \end{cases}Show solution
The only possible point of discontinuity is x=2x = 2.

f(2)=(2)3−3=8−3=5f(2) = (2)^3 - 3 = 8 - 3 = 5
lim⁡x→2−f(x)=(2)3−3=5\lim_{x \to 2^-} f(x) = (2)^3 - 3 = 5
lim⁡x→2+f(x)=(2)2+1=5\lim_{x \to 2^+} f(x) = (2)^2 + 1 = 5

LHL = RHL = f(2)=5f(2) = 5. Hence ff is continuous at x=2x = 2.

Conclusion: ff has no points of discontinuity.
12Find all points of discontinuity of ff, where
f(x)={x10−1,amp;if x≤1x2,amp;if xgt;1f(x) = \begin{cases} x^{10}-1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}Show solution
The only possible point of discontinuity is x=1x = 1.

f(1)=(1)10−1=0f(1) = (1)^{10} - 1 = 0
lim⁡x→1−f(x)=(1)10−1=0\lim_{x \to 1^-} f(x) = (1)^{10} - 1 = 0
lim⁡x→1+f(x)=(1)2=1\lim_{x \to 1^+} f(x) = (1)^2 = 1

LHL =0≠1== 0 \neq 1 = RHL. Hence ff is discontinuous at x=1x = 1.

Conclusion: x=1x = 1 is the only point of discontinuity.
13Is the function defined by
f(x)={x+5,amp;if x≤1x−5,amp;if xgt;1f(x) = \begin{cases} x+5, & \text{if } x \leq 1 \\ x-5, & \text{if } x > 1 \end{cases}
a continuous function?Show solution
The only possible point of discontinuity is x=1x = 1.

f(1)=1+5=6f(1) = 1+5 = 6
lim⁡x→1−f(x)=1+5=6\lim_{x \to 1^-} f(x) = 1+5 = 6
lim⁡x→1+f(x)=1−5=−4\lim_{x \to 1^+} f(x) = 1-5 = -4

LHL =6≠−4== 6 \neq -4 = RHL.

Hence ff is not continuous at x=1x = 1.

For x < 1 and x > 1, ff is a polynomial, hence continuous.

Conclusion: ff is not a continuous function (it is discontinuous at x=1x = 1).
14Discuss the continuity of the function ff, where
f(x)={3,amp;if 0≤x≤14,amp;if 1lt;xlt;35,amp;if 3≤x≤10f(x) = \begin{cases} 3, & \text{if } 0 \leq x \leq 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \leq x \leq 10 \end{cases}Show solution
Possible points of discontinuity: x=1x = 1 and x=3x = 3.

At x=1x = 1:
f(1)=3f(1) = 3
lim⁡x→1−f(x)=3,lim⁡x→1+f(x)=4\lim_{x \to 1^-} f(x) = 3, \quad \lim_{x \to 1^+} f(x) = 4
LHL ≠\neq RHL. Hence ff is discontinuous at x=1x = 1.

At x=3x = 3:
f(3)=5f(3) = 5
lim⁡x→3−f(x)=4,lim⁡x→3+f(x)=5\lim_{x \to 3^-} f(x) = 4, \quad \lim_{x \to 3^+} f(x) = 5
LHL ≠\neq RHL. Hence ff is discontinuous at x=3x = 3.

At all other points in [0,10][0,10], ff is constant on open intervals, hence continuous.

Conclusion: ff is discontinuous at x=1x = 1 and x=3x = 3.
15Discuss the continuity of the function ff, where
f(x)={2x,amp;if xlt;00,amp;if 0≤x≤14x,amp;if xgt;1f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \leq x \leq 1 \\ 4x, & \text{if } x > 1 \end{cases}Show solution
Possible points of discontinuity: x=0x = 0 and x=1x = 1.

At x=0x = 0:
f(0)=0f(0) = 0
lim⁡x→0−f(x)=2(0)=0,lim⁡x→0+f(x)=0\lim_{x \to 0^-} f(x) = 2(0) = 0, \quad \lim_{x \to 0^+} f(x) = 0
LHL = RHL = f(0)=0f(0) = 0. Hence ff is continuous at x=0x = 0.

At x=1x = 1:
f(1)=0f(1) = 0
lim⁡x→1−f(x)=0,lim⁡x→1+f(x)=4(1)=4\lim_{x \to 1^-} f(x) = 0, \quad \lim_{x \to 1^+} f(x) = 4(1) = 4
LHL ≠\neq RHL. Hence ff is discontinuous at x=1x = 1.

Conclusion: ff is discontinuous only at x=1x = 1.
16Discuss the continuity of the function ff, where
f(x)={−2,amp;if x≤−12x,amp;if −1lt;x≤12,amp;if xgt;1f(x) = \begin{cases} -2, & \text{if } x \leq -1 \\ 2x, & \text{if } -1 < x \leq 1 \\ 2, & \text{if } x > 1 \end{cases}Show solution
Possible points of discontinuity: x=−1x = -1 and x=1x = 1.

At x=−1x = -1:
f(−1)=−2f(-1) = -2
lim⁡x→−1−f(x)=−2,lim⁡x→−1+f(x)=2(−1)=−2\lim_{x \to -1^-} f(x) = -2, \quad \lim_{x \to -1^+} f(x) = 2(-1) = -2
LHL = RHL = f(−1)=−2f(-1) = -2. Hence ff is continuous at x=−1x = -1.

At x=1x = 1:
f(1)=2(1)=2f(1) = 2(1) = 2
lim⁡x→1−f(x)=2(1)=2,lim⁡x→1+f(x)=2\lim_{x \to 1^-} f(x) = 2(1) = 2, \quad \lim_{x \to 1^+} f(x) = 2
LHL = RHL = f(1)=2f(1) = 2. Hence ff is continuous at x=1x = 1.

Conclusion: ff is continuous for all real xx.
17Find the relationship between aa and bb so that the function ff defined by
f(x)={ax+1,amp;if x≤3bx+3,amp;if xgt;3f(x) = \begin{cases} ax+1, & \text{if } x \leq 3 \\ bx+3, & \text{if } x > 3 \end{cases}
is continuous at x=3x = 3.Show solution
For ff to be continuous at x=3x = 3:
lim⁡x→3−f(x)=lim⁡x→3+f(x)=f(3)\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)

f(3)=3a+1f(3) = 3a + 1
lim⁡x→3−f(x)=3a+1\lim_{x \to 3^-} f(x) = 3a + 1
lim⁡x→3+f(x)=3b+3\lim_{x \to 3^+} f(x) = 3b + 3

Setting LHL = RHL:
3a+1=3b+33a + 1 = 3b + 3
3a−3b=23a - 3b = 2
3a−3b=2\boxed{3a - 3b = 2}

This is the required relationship between aa and bb.
18For what value of λ\lambda is the function defined by
f(x)={λ(x2−2x),amp;if x≤04x+1,amp;if xgt;0f(x) = \begin{cases} \lambda(x^2-2x), & \text{if } x \leq 0 \\ 4x+1, & \text{if } x > 0 \end{cases}
continuous at x=0x = 0? What about continuity at x=1x = 1?Show solution
Continuity at x=0x = 0:

f(0)=λ(0−0)=0f(0) = \lambda(0 - 0) = 0
lim⁡x→0−f(x)=λ(0−0)=0\lim_{x \to 0^-} f(x) = \lambda(0 - 0) = 0
lim⁡x→0+f(x)=4(0)+1=1\lim_{x \to 0^+} f(x) = 4(0)+1 = 1

For continuity: LHL = RHL
0=10 = 1
This is a contradiction. Hence no value of λ\lambda makes ff continuous at x=0x = 0.

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Continuity at x=1x = 1:

Since 1 > 0, f(x)=4x+1f(x) = 4x + 1 near x=1x = 1.
f(1)=4(1)+1=5f(1) = 4(1)+1 = 5
lim⁡x→1f(x)=4(1)+1=5\lim_{x \to 1} f(x) = 4(1)+1 = 5

Hence ff is continuous at x=1x = 1 for any value of λ\lambda.
19Show that the function defined by g(x)=x−[x]g(x) = x - [x] is discontinuous at all integral points. Here [x][x] denotes the greatest integer less than or equal to xx.Show solution
Given: g(x)=x−[x]g(x) = x - [x]

Let nn be any integer. We check continuity at x=nx = n.

g(n)=n−[n]=n−n=0g(n) = n - [n] = n - n = 0

LHL:
lim⁡x→n−g(x)=lim⁡x→n−(x−[x])\lim_{x \to n^-} g(x) = \lim_{x \to n^-}(x - [x])
For xx slightly less than nn, [x]=n−1[x] = n-1, so:
lim⁡x→n−g(x)=n−(n−1)=1\lim_{x \to n^-} g(x) = n - (n-1) = 1

RHL:
lim⁡x→n+g(x)=lim⁡x→n+(x−[x])\lim_{x \to n^+} g(x) = \lim_{x \to n^+}(x - [x])
For xx slightly greater than nn, [x]=n[x] = n, so:
lim⁡x→n+g(x)=n−n=0\lim_{x \to n^+} g(x) = n - n = 0

Since LHL =1≠0== 1 \neq 0 = RHL, the limit does not exist at x=nx = n.

Hence g(x)=x−[x]g(x) = x - [x] is discontinuous at every integer nn. ■\blacksquare
20Is the function defined by f(x)=x2−sin⁡x+5f(x) = x^2 - \sin x + 5 continuous at x=πx = \pi?Show solution
Given: f(x)=x2−sin⁡x+5f(x) = x^2 - \sin x + 5

f(π)=π2−sin⁡π+5=π2−0+5=π2+5f(\pi) = \pi^2 - \sin\pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5

lim⁡x→πf(x)=π2−sin⁡π+5=π2+5\lim_{x \to \pi} f(x) = \pi^2 - \sin\pi + 5 = \pi^2 + 5

Since lim⁡x→πf(x)=f(π)=π2+5\lim_{x \to \pi} f(x) = f(\pi) = \pi^2 + 5, the function is continuous at x=πx = \pi.
21Discuss the continuity of the following functions:
(a) f(x)=sin⁡x+cos⁡xf(x) = \sin x + \cos x
(b) f(x)=sin⁡x−cos⁡xf(x) = \sin x - \cos x
(c) f(x)=sin⁡x⋅cos⁡xf(x) = \sin x \cdot \cos xShow solution
We know that sin⁡x\sin x and cos⁡x\cos x are continuous for all x∈Rx \in \mathbb{R}.

Since the sum, difference, and product of continuous functions are continuous:

(a) f(x)=sin⁡x+cos⁡xf(x) = \sin x + \cos x is continuous for all x∈Rx \in \mathbb{R}.

(b) f(x)=sin⁡x−cos⁡xf(x) = \sin x - \cos x is continuous for all x∈Rx \in \mathbb{R}.

(c) f(x)=sin⁡x⋅cos⁡xf(x) = \sin x \cdot \cos x is continuous for all x∈Rx \in \mathbb{R}.
22Discuss the continuity of the cosine, cosecant, secant and cotangent functions.Show solution
Cosine: cos⁡x\cos x is continuous for all x∈Rx \in \mathbb{R}.

Cosecant: csc⁡x=1sin⁡x\csc x = \dfrac{1}{\sin x}

sin⁡x\sin x is continuous everywhere and sin⁡x=0\sin x = 0 at x=nπx = n\pi, n∈Zn \in \mathbb{Z}.
Hence csc⁡x\csc x is continuous for all x∈R−{nπ:n∈Z}x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}.

Secant: sec⁡x=1cos⁡x\sec x = \dfrac{1}{\cos x}

cos⁡x=0\cos x = 0 at x=(2n+1)π2x = (2n+1)\dfrac{\pi}{2}, n∈Zn \in \mathbb{Z}.
Hence sec⁡x\sec x is continuous for all x∈R−{(2n+1)π2:n∈Z}x \in \mathbb{R} - \left\{(2n+1)\dfrac{\pi}{2} : n \in \mathbb{Z}\right\}.

Cotangent: cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{\cos x}{\sin x}

sin⁡x=0\sin x = 0 at x=nπx = n\pi, n∈Zn \in \mathbb{Z}.
Hence cot⁡x\cot x is continuous for all x∈R−{nπ:n∈Z}x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}.
23Find all points of discontinuity of ff, where
f(x)={sin⁡xx,amp;if xlt;0x+1,amp;if x≥0f(x) = \begin{cases} \dfrac{\sin x}{x}, & \text{if } x < 0 \\ x+1, & \text{if } x \geq 0 \end{cases}Show solution
The only possible point of discontinuity is x=0x = 0.

f(0)=0+1=1f(0) = 0 + 1 = 1
lim⁡x→0−f(x)=lim⁡x→0−sin⁡xx=1(standard limit: lim⁡x→0sin⁡xx=1)\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} = 1 \quad \left(\text{standard limit: } \lim_{x \to 0}\frac{\sin x}{x} = 1\right)
lim⁡x→0+f(x)=0+1=1\lim_{x \to 0^+} f(x) = 0 + 1 = 1

LHL = RHL = f(0)=1f(0) = 1. Hence ff is continuous at x=0x = 0.

For x < 0: sin⁡xx\dfrac{\sin x}{x} is continuous (ratio of continuous functions, denominator ≠0\neq 0).
For x > 0: x+1x + 1 is a polynomial, hence continuous.

Conclusion: ff has no points of discontinuity.
24Determine if ff defined by
f(x)={x2sin⁡1x,amp;if x≠00,amp;if x=0f(x) = \begin{cases} x^2 \sin\dfrac{1}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}
is a continuous function?Show solution
For x≠0x \neq 0: f(x)=x2sin⁡1xf(x) = x^2 \sin\dfrac{1}{x} is a product of continuous functions, hence continuous.

At x=0x = 0:
f(0)=0f(0) = 0
We need lim⁡x→0x2sin⁡1x\lim_{x \to 0} x^2 \sin\dfrac{1}{x}.

Since ∣sin⁡1x∣≤1\left|\sin\dfrac{1}{x}\right| \leq 1 for all x≠0x \neq 0:
∣x2sin⁡1x∣≤x2\left|x^2 \sin\frac{1}{x}\right| \leq x^2

As x→0x \to 0, x2→0x^2 \to 0, so by Squeeze Theorem:
lim⁡x→0x2sin⁡1x=0=f(0)\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0 = f(0)

Hence ff is continuous at x=0x = 0 and therefore continuous everywhere.
25Examine the continuity of ff, where ff is defined by
f(x)={sin⁡x−cos⁡x,amp;if x≠0−1,amp;if x=0f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}Show solution
At x=0x = 0:
f(0)=−1f(0) = -1
lim⁡x→0f(x)=lim⁡x→0(sin⁡x−cos⁡x)=sin⁡0−cos⁡0=0−1=−1\lim_{x \to 0} f(x) = \lim_{x \to 0}(\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1

Since lim⁡x→0f(x)=f(0)=−1\lim_{x \to 0} f(x) = f(0) = -1, ff is continuous at x=0x = 0.

For x≠0x \neq 0: sin⁡x−cos⁡x\sin x - \cos x is continuous everywhere.

Conclusion: ff is continuous for all x∈Rx \in \mathbb{R}.
26Find the value of kk so that the function ff is continuous at x=π2x = \dfrac{\pi}{2}:
f(x)={kcos⁡xπ−2x,amp;if x≠π23,amp;if x=π2f(x) = \begin{cases} \dfrac{k\cos x}{\pi - 2x}, & \text{if } x \neq \dfrac{\pi}{2} \\ 3, & \text{if } x = \dfrac{\pi}{2} \end{cases}Show solution
For continuity at x=π2x = \dfrac{\pi}{2}:
lim⁡x→π/2kcos⁡xπ−2x=3\lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} = 3

Let x=π2+hx = \dfrac{\pi}{2} + h, so as x→π2x \to \dfrac{\pi}{2}, h→0h \to 0.

cos⁡x=cos⁡(π2+h)=−sin⁡h\cos x = \cos\left(\frac{\pi}{2}+h\right) = -\sin h
π−2x=π−2(π2+h)=−2h\pi - 2x = \pi - 2\left(\frac{\pi}{2}+h\right) = -2h

lim⁡h→0k(−sin⁡h)−2h=lim⁡h→0ksin⁡h2h=k2⋅1=k2\lim_{h \to 0} \frac{k(-\sin h)}{-2h} = \lim_{h \to 0} \frac{k\sin h}{2h} = \frac{k}{2} \cdot 1 = \frac{k}{2}

Setting k2=3\dfrac{k}{2} = 3:
k=6\boxed{k = 6}
27Find the value of kk so that the function ff is continuous at x=2x = 2:
f(x)={kx2,amp;if x≤23,amp;if xgt;2f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 3, & \text{if } x > 2 \end{cases}Show solution
For continuity at x=2x = 2:
lim⁡x→2−f(x)=f(2)=lim⁡x→2+f(x)\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)

lim⁡x→2−kx2=k(4)=4k\lim_{x \to 2^-} kx^2 = k(4) = 4k
lim⁡x→2+f(x)=3\lim_{x \to 2^+} f(x) = 3
f(2)=k(4)=4kf(2) = k(4) = 4k

Setting 4k=34k = 3:
k=34\boxed{k = \frac{3}{4}}
28Find the value of kk so that the function ff is continuous at x=πx = \pi:
f(x)={kx+1,amp;if x≤πcos⁡x,amp;if xgt;πf(x) = \begin{cases} kx+1, & \text{if } x \leq \pi \\ \cos x, & \text{if } x > \pi \end{cases}Show solution
For continuity at x=πx = \pi:
lim⁡x→π−f(x)=f(π)=lim⁡x→π+f(x)\lim_{x \to \pi^-} f(x) = f(\pi) = \lim_{x \to \pi^+} f(x)

f(π)=kπ+1f(\pi) = k\pi + 1
lim⁡x→π−f(x)=kπ+1\lim_{x \to \pi^-} f(x) = k\pi + 1
lim⁡x→π+f(x)=cos⁡π=−1\lim_{x \to \pi^+} f(x) = \cos\pi = -1

Setting kπ+1=−1k\pi + 1 = -1:
kπ=−2k\pi = -2
k=−2π\boxed{k = -\frac{2}{\pi}}
29Find the value of kk so that the function ff is continuous at x=5x = 5:
f(x)={kx+1,amp;if x≤53x−5,amp;if xgt;5f(x) = \begin{cases} kx+1, & \text{if } x \leq 5 \\ 3x-5, & \text{if } x > 5 \end{cases}Show solution
For continuity at x=5x = 5:
lim⁡x→5−f(x)=f(5)=lim⁡x→5+f(x)\lim_{x \to 5^-} f(x) = f(5) = \lim_{x \to 5^+} f(x)

f(5)=5k+1f(5) = 5k + 1
lim⁡x→5−f(x)=5k+1\lim_{x \to 5^-} f(x) = 5k + 1
lim⁡x→5+f(x)=3(5)−5=10\lim_{x \to 5^+} f(x) = 3(5) - 5 = 10

Setting 5k+1=105k + 1 = 10:
5k=95k = 9
k=95\boxed{k = \frac{9}{5}}
30Find the values of aa and bb such that the function defined by
f(x)={5,amp;if x≤2ax+b,amp;if 2lt;xlt;1021,amp;if x≥10f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax+b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}
is a continuous function.Show solution
For ff to be continuous, it must be continuous at x=2x = 2 and x=10x = 10.

At x=2x = 2:
f(2)=5f(2) = 5
lim⁡x→2+f(x)=2a+b\lim_{x \to 2^+} f(x) = 2a + b
Setting equal: 2a+b=52a + b = 5 ... (i)

At x=10x = 10:
f(10)=21f(10) = 21
lim⁡x→10−f(x)=10a+b\lim_{x \to 10^-} f(x) = 10a + b
Setting equal: 10a+b=2110a + b = 21 ... (ii)

Subtracting (i) from (ii):
8a=16  ⟹  a=28a = 16 \implies a = 2

Substituting in (i):
2(2)+b=5  ⟹  b=12(2) + b = 5 \implies b = 1

a=2,b=1\boxed{a = 2, \quad b = 1}
31Show that the function defined by f(x)=cos⁡(x2)f(x) = \cos(x^2) is a continuous function.Show solution
Given: f(x)=cos⁡(x2)f(x) = \cos(x^2)

Let g(x)=x2g(x) = x^2 and h(t)=cos⁡th(t) = \cos t.

- g(x)=x2g(x) = x^2 is a polynomial, hence continuous for all x∈Rx \in \mathbb{R}.
- h(t)=cos⁡th(t) = \cos t is continuous for all t∈Rt \in \mathbb{R}.

Since f(x)=h(g(x))=cos⁡(x2)f(x) = h(g(x)) = \cos(x^2) is a composition of two continuous functions, by the theorem on continuity of composite functions, f(x)f(x) is continuous for all x∈Rx \in \mathbb{R}. ■\blacksquare
32Show that the function defined by f(x)=∣cos⁡x∣f(x) = |\cos x| is a continuous function.Show solution
Given: f(x)=∣cos⁡x∣f(x) = |\cos x|

Let g(x)=cos⁡xg(x) = \cos x and h(t)=∣t∣h(t) = |t|.

- g(x)=cos⁡xg(x) = \cos x is continuous for all x∈Rx \in \mathbb{R}.
- h(t)=∣t∣h(t) = |t| is continuous for all t∈Rt \in \mathbb{R}.

Since f(x)=h(g(x))=∣cos⁡x∣f(x) = h(g(x)) = |\cos x| is a composition of two continuous functions, f(x)f(x) is continuous for all x∈Rx \in \mathbb{R}. ■\blacksquare
33Examine that sin⁡∣x∣\sin|x| is a continuous function.Show solution
Given: f(x)=sin⁡∣x∣f(x) = \sin|x|

Let g(x)=∣x∣g(x) = |x| and h(t)=sin⁡th(t) = \sin t.

- g(x)=∣x∣g(x) = |x| is continuous for all x∈Rx \in \mathbb{R}.
- h(t)=sin⁡th(t) = \sin t is continuous for all t∈Rt \in \mathbb{R}.

Since f(x)=h(g(x))=sin⁡∣x∣f(x) = h(g(x)) = \sin|x| is a composition of two continuous functions, f(x)f(x) is continuous for all x∈Rx \in \mathbb{R}.
34Find all the points of discontinuity of ff defined by f(x)=∣x∣−∣x+1∣f(x) = |x| - |x+1|.Show solution
Given: f(x)=∣x∣−∣x+1∣f(x) = |x| - |x+1|

Both ∣x∣|x| and ∣x+1∣|x+1| are continuous for all x∈Rx \in \mathbb{R} (modulus of a continuous function is continuous).

The difference of two continuous functions is continuous.

Hence f(x)=∣x∣−∣x+1∣f(x) = |x| - |x+1| is continuous for all x∈Rx \in \mathbb{R}.

Conclusion: ff has no points of discontinuity.

Exercise 5.2

1Differentiate sin⁡(x2+5)\sin(x^2+5) with respect to xx.Show solution
Given: y=sin⁡(x2+5)y = \sin(x^2+5)

Using chain rule: ddx[sin⁡(u)]=cos⁡(u)⋅dudx\dfrac{d}{dx}[\sin(u)] = \cos(u)\cdot\dfrac{du}{dx}, where u=x2+5u = x^2+5.

dydx=cos⁡(x2+5)⋅ddx(x2+5)=cos⁡(x2+5)⋅2x\frac{dy}{dx} = \cos(x^2+5) \cdot \frac{d}{dx}(x^2+5) = \cos(x^2+5) \cdot 2x

dydx=2xcos⁡(x2+5)\boxed{\frac{dy}{dx} = 2x\cos(x^2+5)}
2Differentiate cos⁡(sin⁡x)\cos(\sin x) with respect to xx.Show solution
Given: y=cos⁡(sin⁡x)y = \cos(\sin x)

Using chain rule:
dydx=−sin⁡(sin⁡x)⋅ddx(sin⁡x)=−sin⁡(sin⁡x)⋅cos⁡x\frac{dy}{dx} = -\sin(\sin x) \cdot \frac{d}{dx}(\sin x) = -\sin(\sin x) \cdot \cos x

dydx=−cos⁡x⋅sin⁡(sin⁡x)\boxed{\frac{dy}{dx} = -\cos x \cdot \sin(\sin x)}
3Differentiate sin⁡(ax+b)\sin(ax+b) with respect to xx.Show solution
Given: y=sin⁡(ax+b)y = \sin(ax+b)

Using chain rule:
dydx=cos⁡(ax+b)⋅ddx(ax+b)=cos⁡(ax+b)⋅a\frac{dy}{dx} = \cos(ax+b) \cdot \frac{d}{dx}(ax+b) = \cos(ax+b) \cdot a

dydx=acos⁡(ax+b)\boxed{\frac{dy}{dx} = a\cos(ax+b)}
4Differentiate sec⁡(tan⁡(x))\sec(\tan(\sqrt{x})) with respect to xx.Show solution
Given: y=sec⁡(tan⁡(x))y = \sec(\tan(\sqrt{x}))

Applying chain rule step by step:
dydx=sec⁡(tan⁡x)⋅tan⁡(tan⁡x)⋅ddx(tan⁡x)\frac{dy}{dx} = \sec(\tan\sqrt{x})\cdot\tan(\tan\sqrt{x}) \cdot \frac{d}{dx}(\tan\sqrt{x})
=sec⁡(tan⁡x)⋅tan⁡(tan⁡x)⋅sec⁡2(x)⋅ddx(x)= \sec(\tan\sqrt{x})\cdot\tan(\tan\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})
=sec⁡(tan⁡x)⋅tan⁡(tan⁡x)⋅sec⁡2(x)⋅12x= \sec(\tan\sqrt{x})\cdot\tan(\tan\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}

dydx=sec⁡(tan⁡x)⋅tan⁡(tan⁡x)⋅sec⁡2x2x\boxed{\frac{dy}{dx} = \frac{\sec(\tan\sqrt{x})\cdot\tan(\tan\sqrt{x})\cdot\sec^2\sqrt{x}}{2\sqrt{x}}}
5Differentiate sin⁡(ax+b)cos⁡(cx+d)\dfrac{\sin(ax+b)}{\cos(cx+d)} with respect to xx.Show solution
Given: y=sin⁡(ax+b)cos⁡(cx+d)y = \dfrac{\sin(ax+b)}{\cos(cx+d)}

Using quotient rule: ddx(uv)=v u′−u v′v2\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\,u' - u\,v'}{v^2}

u=sin⁡(ax+b)⇒u′=acos⁡(ax+b)u = \sin(ax+b) \Rightarrow u' = a\cos(ax+b)
v=cos⁡(cx+d)⇒v′=−csin⁡(cx+d)v = \cos(cx+d) \Rightarrow v' = -c\sin(cx+d)

dydx=cos⁡(cx+d)⋅acos⁡(ax+b)−sin⁡(ax+b)⋅(−csin⁡(cx+d))cos⁡2(cx+d)\frac{dy}{dx} = \frac{\cos(cx+d)\cdot a\cos(ax+b) - \sin(ax+b)\cdot(-c\sin(cx+d))}{\cos^2(cx+d)}

=acos⁡(ax+b)cos⁡(cx+d)+csin⁡(ax+b)sin⁡(cx+d)cos⁡2(cx+d)= \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}
6Differentiate cos⁡x3⋅sin⁡2(x5)\cos x^3 \cdot \sin^2(x^5) with respect to xx.Show solution
Given: y=cos⁡(x3)⋅sin⁡2(x5)y = \cos(x^3)\cdot\sin^2(x^5)

Using product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'

u=cos⁡(x3)⇒u′=−sin⁡(x3)⋅3x2u = \cos(x^3) \Rightarrow u' = -\sin(x^3)\cdot 3x^2
v=sin⁡2(x5)⇒v′=2sin⁡(x5)⋅cos⁡(x5)⋅5x4=10x4sin⁡(x5)cos⁡(x5)v = \sin^2(x^5) \Rightarrow v' = 2\sin(x^5)\cdot\cos(x^5)\cdot 5x^4 = 10x^4\sin(x^5)\cos(x^5)

dydx=−3x2sin⁡(x3)⋅sin⁡2(x5)+cos⁡(x3)⋅10x4sin⁡(x5)cos⁡(x5)\frac{dy}{dx} = -3x^2\sin(x^3)\cdot\sin^2(x^5) + \cos(x^3)\cdot 10x^4\sin(x^5)\cos(x^5)

dydx=−3x2sin⁡(x3)sin⁡2(x5)+10x4cos⁡(x3)sin⁡(x5)cos⁡(x5)\boxed{\frac{dy}{dx} = -3x^2\sin(x^3)\sin^2(x^5) + 10x^4\cos(x^3)\sin(x^5)\cos(x^5)}
7Differentiate 2cot⁡(x2)2\sqrt{\cot(x^2)} with respect to xx.Show solution
Given: y=2cot⁡(x2)=2[cot⁡(x2)]1/2y = 2\sqrt{\cot(x^2)} = 2[\cot(x^2)]^{1/2}

Using chain rule:
dydx=2⋅12[cot⁡(x2)]−1/2⋅ddx[cot⁡(x2)]\frac{dy}{dx} = 2 \cdot \frac{1}{2}[\cot(x^2)]^{-1/2} \cdot \frac{d}{dx}[\cot(x^2)]
=1cot⁡(x2)⋅(−csc⁡2(x2))⋅2x= \frac{1}{\sqrt{\cot(x^2)}} \cdot (-\csc^2(x^2)) \cdot 2x
=−2xcsc⁡2(x2)cot⁡(x2)= \frac{-2x\csc^2(x^2)}{\sqrt{\cot(x^2)}}

dydx=−2xcsc⁡2(x2)cot⁡(x2)\boxed{\frac{dy}{dx} = \frac{-2x\csc^2(x^2)}{\sqrt{\cot(x^2)}}}
8Differentiate cos⁡(x)\cos(\sqrt{x}) with respect to xx.Show solution
Given: y=cos⁡(x)y = \cos(\sqrt{x})

Using chain rule:
dydx=−sin⁡(x)⋅ddx(x)=−sin⁡(x)⋅12x\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}

dydx=−sin⁡x2x\boxed{\frac{dy}{dx} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}}
9Prove that the function ff given by f(x)=∣x−1∣, x∈Rf(x) = |x-1|,\ x \in \mathbb{R} is not differentiable at x=1x = 1.Show solution
Given: f(x)=∣x−1∣f(x) = |x-1|

We can write:
f(x)={x−1,amp;x≥1−(x−1),amp;xlt;1f(x) = \begin{cases} x-1, & x \geq 1 \\ -(x-1), & x < 1 \end{cases}

Left-hand derivative at x=1x = 1:
LHD=lim⁡h→0−f(1+h)−f(1)h=lim⁡h→0−∣h∣h=lim⁡h→0−−hh=−1\text{LHD} = \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1

Right-hand derivative at x=1x = 1:
RHD=lim⁡h→0+f(1+h)−f(1)h=lim⁡h→0+∣h∣h=lim⁡h→0+hh=1\text{RHD} = \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1

Since LHD ≠\neq RHD, f(x)=∣x−1∣f(x) = |x-1| is not differentiable at x=1x = 1. ■\blacksquare
10Prove that the greatest integer function defined by f(x) = [x],\ 0 < x < 3 is not differentiable at x=1x = 1 and x=2x = 2.Show solution
At x=1x = 1:

LHD:
lim⁡h→0−f(1+h)−f(1)h=lim⁡h→0−[1+h]−[1]h=lim⁡h→0−0−1h=lim⁡h→0−−1h=+∞\lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{[1+h]-[1]}{h} = \lim_{h \to 0^-} \frac{0-1}{h} = \lim_{h \to 0^-} \frac{-1}{h} = +\infty

(For small h < 0: [1+h]=0[1+h] = 0, [1]=1[1] = 1)

RHD:
lim⁡h→0+f(1+h)−f(1)h=lim⁡h→0+[1+h]−1h=lim⁡h→0+1−1h=0\lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+} \frac{[1+h]-1}{h} = \lim_{h \to 0^+} \frac{1-1}{h} = 0

Since LHD ≠\neq RHD, ff is not differentiable at x=1x = 1.

At x=2x = 2:

LHD:
lim⁡h→0−[2+h]−[2]h=lim⁡h→0−1−2h=lim⁡h→0−−1h=+∞\lim_{h \to 0^-} \frac{[2+h]-[2]}{h} = \lim_{h \to 0^-} \frac{1-2}{h} = \lim_{h \to 0^-} \frac{-1}{h} = +\infty

RHD:
lim⁡h→0+[2+h]−2h=lim⁡h→0+2−2h=0\lim_{h \to 0^+} \frac{[2+h]-2}{h} = \lim_{h \to 0^+} \frac{2-2}{h} = 0

Since LHD ≠\neq RHD, ff is not differentiable at x=2x = 2. ■\blacksquare

Exercise 5.3

1Find dydx\dfrac{dy}{dx}: 2x+3y=sin⁡x2x + 3y = \sin xShow solution
Differentiating both sides with respect to xx:
2+3dydx=cos⁡x2 + 3\frac{dy}{dx} = \cos x
3dydx=cos⁡x−23\frac{dy}{dx} = \cos x - 2
dydx=cos⁡x−23\boxed{\frac{dy}{dx} = \frac{\cos x - 2}{3}}
2Find dydx\dfrac{dy}{dx}: 2x+3y=sin⁡y2x + 3y = \sin yShow solution
Differentiating both sides with respect to xx:
2+3dydx=cos⁡y⋅dydx2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}
2=cos⁡y⋅dydx−3dydx2 = \cos y \cdot \frac{dy}{dx} - 3\frac{dy}{dx}
2=dydx(cos⁡y−3)2 = \frac{dy}{dx}(\cos y - 3)
dydx=2cos⁡y−3\boxed{\frac{dy}{dx} = \frac{2}{\cos y - 3}}
3Find dydx\dfrac{dy}{dx}: ax+by2=cos⁡yax + by^2 = \cos yShow solution
Differentiating both sides with respect to xx:
a+2bydydx=−sin⁡y⋅dydxa + 2by\frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}
a=−sin⁡y⋅dydx−2bydydxa = -\sin y \cdot \frac{dy}{dx} - 2by\frac{dy}{dx}
a=dydx(−sin⁡y−2by)a = \frac{dy}{dx}(-\sin y - 2by)
dydx=−asin⁡y+2by\boxed{\frac{dy}{dx} = \frac{-a}{\sin y + 2by}}
4Find dydx\dfrac{dy}{dx}: xy+y2=tan⁡x+yxy + y^2 = \tan x + yShow solution
Differentiating both sides with respect to xx:
y+xdydx+2ydydx=sec⁡2x+dydxy + x\frac{dy}{dx} + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}
dydx(x+2y−1)=sec⁡2x−y\frac{dy}{dx}(x + 2y - 1) = \sec^2 x - y
dydx=sec⁡2x−yx+2y−1\boxed{\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}}
5Find dydx\dfrac{dy}{dx}: x2+xy+y2=100x^2 + xy + y^2 = 100Show solution
Differentiating both sides with respect to xx:
2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0
dydx(x+2y)=−(2x+y)\frac{dy}{dx}(x + 2y) = -(2x + y)
dydx=−(2x+y)x+2y\boxed{\frac{dy}{dx} = \frac{-(2x+y)}{x+2y}}
6Find dydx\dfrac{dy}{dx}: x3+x2y+xy2+y3=81x^3 + x^2y + xy^2 + y^3 = 81Show solution
Differentiating both sides with respect to xx:
3x2+2xy+x2dydx+y2+2xydydx+3y2dydx=03x^2 + 2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0
dydx(x2+2xy+3y2)=−(3x2+2xy+y2)\frac{dy}{dx}(x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2)
dydx=−(3x2+2xy+y2)x2+2xy+3y2\boxed{\frac{dy}{dx} = \frac{-(3x^2 + 2xy + y^2)}{x^2 + 2xy + 3y^2}}
7Find dydx\dfrac{dy}{dx}: sin⁡2y+cos⁡xy=k\sin^2 y + \cos xy = kShow solution
Differentiating both sides with respect to xx:
2sin⁡ycos⁡y⋅dydx+(−sin⁡xy)(y+xdydx)=02\sin y \cos y \cdot \frac{dy}{dx} + (-\sin xy)\left(y + x\frac{dy}{dx}\right) = 0
sin⁡2y⋅dydx−ysin⁡xy−xsin⁡xy⋅dydx=0\sin 2y \cdot \frac{dy}{dx} - y\sin xy - x\sin xy \cdot \frac{dy}{dx} = 0
dydx(sin⁡2y−xsin⁡xy)=ysin⁡xy\frac{dy}{dx}(\sin 2y - x\sin xy) = y\sin xy
dydx=ysin⁡xysin⁡2y−xsin⁡xy\boxed{\frac{dy}{dx} = \frac{y\sin xy}{\sin 2y - x\sin xy}}
8Find dydx\dfrac{dy}{dx}: sin⁡2x+cos⁡2y=1\sin^2 x + \cos^2 y = 1Show solution
Differentiating both sides with respect to xx:
2sin⁡xcos⁡x+2cos⁡y(−sin⁡y)dydx=02\sin x \cos x + 2\cos y(-\sin y)\frac{dy}{dx} = 0
sin⁡2x−sin⁡2y⋅dydx=0\sin 2x - \sin 2y \cdot \frac{dy}{dx} = 0
dydx=sin⁡2xsin⁡2y\boxed{\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}}
9Find dydx\dfrac{dy}{dx}: y=sin⁡−1(2x1+x2)y = \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)Show solution
Substitution: Let x=tan⁡θx = \tan\theta, so θ=tan⁡−1x\theta = \tan^{-1}x.

2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡2θ\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta

y=sin⁡−1(sin⁡2θ)=2θ=2tan⁡−1xy = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x

dydx=2⋅11+x2\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}

dydx=21+x2\boxed{\frac{dy}{dx} = \frac{2}{1+x^2}}
10Find dydx\dfrac{dy}{dx}: y = \tan^{-1}\left(\dfrac{3x-x^3}{1-3x^2}\right),\ -\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}Show solution
Substitution: Let x=tan⁡θx = \tan\theta, so θ=tan⁡−1x\theta = \tan^{-1}x.

3x−x31−3x2=3tan⁡θ−tan⁡3θ1−3tan⁡2θ=tan⁡3θ\frac{3x-x^3}{1-3x^2} = \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta} = \tan 3\theta

y=tan⁡−1(tan⁡3θ)=3θ=3tan⁡−1xy = \tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}x

dydx=31+x2\frac{dy}{dx} = \frac{3}{1+x^2}

dydx=31+x2\boxed{\frac{dy}{dx} = \frac{3}{1+x^2}}
11Find dydx\dfrac{dy}{dx}: y = \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right),\ 0 < x < 1Show solution
Substitution: Let x=tan⁡θx = \tan\theta, θ∈(0,π4)\theta \in \left(0, \dfrac{\pi}{4}\right).

1−x21+x2=1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta

y=cos⁡−1(cos⁡2θ)=2θ=2tan⁡−1xy = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x

dydx=21+x2\frac{dy}{dx} = \frac{2}{1+x^2}

dydx=21+x2\boxed{\frac{dy}{dx} = \frac{2}{1+x^2}}
12Find dydx\dfrac{dy}{dx}: y = \sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right),\ 0 < x < 1Show solution
Substitution: Let x=tan⁡θx = \tan\theta, θ∈(0,π4)\theta \in \left(0, \dfrac{\pi}{4}\right).

1−x21+x2=cos⁡2θ\frac{1-x^2}{1+x^2} = \cos 2\theta

y=sin⁡−1(cos⁡2θ)=sin⁡−1(sin⁡(π2−2θ))=π2−2θ=π2−2tan⁡−1xy = \sin^{-1}(\cos 2\theta) = \sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x

dydx=0−21+x2\frac{dy}{dx} = 0 - \frac{2}{1+x^2}

dydx=−21+x2\boxed{\frac{dy}{dx} = \frac{-2}{1+x^2}}
13Find dydx\dfrac{dy}{dx}: y = \cos^{-1}\left(\dfrac{2x}{1+x^2}\right),\ -1 < x < 1Show solution
Substitution: Let x=tan⁡θx = \tan\theta, θ∈(−π4,π4)\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right).

2x1+x2=sin⁡2θ\frac{2x}{1+x^2} = \sin 2\theta

y=cos⁡−1(sin⁡2θ)=cos⁡−1(cos⁡(π2−2θ))=π2−2θ=π2−2tan⁡−1xy = \cos^{-1}(\sin 2\theta) = \cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}x

dydx=−21+x2\frac{dy}{dx} = -\frac{2}{1+x^2}

dydx=−21+x2\boxed{\frac{dy}{dx} = \frac{-2}{1+x^2}}
14Find dydx\dfrac{dy}{dx}: y = \sin^{-1}\left(2x\sqrt{1-x^2}\right),\ -\dfrac{1}{\sqrt{2}} < x < \dfrac{1}{\sqrt{2}}Show solution
Substitution: Let x=sin⁡θx = \sin\theta, θ∈(−π4,π4)\theta \in \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right).

2x1−x2=2sin⁡θcos⁡θ=sin⁡2θ2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta

y=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1xy = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x

dydx=21−x2\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}

dydx=21−x2\boxed{\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}}
15Find dydx\dfrac{dy}{dx}: y = \sec^{-1}\left(\dfrac{1}{2x^2-1}\right),\ 0 < x < \dfrac{1}{\sqrt{2}}Show solution
Substitution: Let x=cos⁡θx = \cos\theta, θ∈(0,π4)\theta \in \left(0, \dfrac{\pi}{4}\right).

12x2−1=12cos⁡2θ−1=1cos⁡2θ=sec⁡2θ\frac{1}{2x^2-1} = \frac{1}{2\cos^2\theta - 1} = \frac{1}{\cos 2\theta} = \sec 2\theta

y=sec⁡−1(sec⁡2θ)=2θ=2cos⁡−1xy = \sec^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1}x

dydx=2⋅−11−x2=−21−x2\frac{dy}{dx} = 2 \cdot \frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}}

dydx=−21−x2\boxed{\frac{dy}{dx} = \frac{-2}{\sqrt{1-x^2}}}

Exercise 5.4

1Differentiate exsin⁡x\dfrac{e^x}{\sin x} with respect to xx.Show solution
Given: y=exsin⁡xy = \dfrac{e^x}{\sin x}

Using quotient rule:
dydx=sin⁡x⋅ex−ex⋅cos⁡xsin⁡2x=ex(sin⁡x−cos⁡x)sin⁡2x\frac{dy}{dx} = \frac{\sin x \cdot e^x - e^x \cdot \cos x}{\sin^2 x} = \frac{e^x(\sin x - \cos x)}{\sin^2 x}

dydx=ex(sin⁡x−cos⁡x)sin⁡2x\boxed{\frac{dy}{dx} = \frac{e^x(\sin x - \cos x)}{\sin^2 x}}
2Differentiate esin⁡−1xe^{\sin^{-1}x} with respect to xx.Show solution
Given: y=esin⁡−1xy = e^{\sin^{-1}x}

Using chain rule:
dydx=esin⁡−1x⋅ddx(sin⁡−1x)=esin⁡−1x⋅11−x2\frac{dy}{dx} = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}}

dydx=esin⁡−1x1−x2\boxed{\frac{dy}{dx} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}}
3Differentiate ex3e^{x^3} with respect to xx.Show solution
Given: y=ex3y = e^{x^3}

Using chain rule:
dydx=ex3⋅ddx(x3)=ex3⋅3x2\frac{dy}{dx} = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2

dydx=3x2ex3\boxed{\frac{dy}{dx} = 3x^2 e^{x^3}}
4Differentiate sin⁡(tan⁡−1e−x)\sin(\tan^{-1}e^{-x}) with respect to xx.Show solution
Given: y=sin⁡(tan⁡−1e−x)y = \sin(\tan^{-1}e^{-x})

Using chain rule:
dydx=cos⁡(tan⁡−1e−x)⋅11+(e−x)2⋅e−x⋅(−1)\frac{dy}{dx} = \cos(\tan^{-1}e^{-x}) \cdot \frac{1}{1+(e^{-x})^2} \cdot e^{-x}\cdot(-1)
=−e−xcos⁡(tan⁡−1e−x)1+e−2x= \frac{-e^{-x}\cos(\tan^{-1}e^{-x})}{1+e^{-2x}}

dydx=−e−xcos⁡(tan⁡−1e−x)1+e−2x\boxed{\frac{dy}{dx} = \frac{-e^{-x}\cos(\tan^{-1}e^{-x})}{1+e^{-2x}}}
5Differentiate log⁡(cos⁡ex)\log(\cos e^x) with respect to xx.Show solution
Given: y=log⁡(cos⁡ex)y = \log(\cos e^x)

Using chain rule:
dydx=1cos⁡ex⋅(−sin⁡ex)⋅ex=−exsin⁡excos⁡ex\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot (-\sin e^x) \cdot e^x = \frac{-e^x \sin e^x}{\cos e^x}

dydx=−extan⁡ex\boxed{\frac{dy}{dx} = -e^x \tan e^x}
6Differentiate ex+ex2+⋯+ex5e^x + e^{x^2} + \cdots + e^{x^5} with respect to xx.Show solution
Given: y=ex+ex2+ex3+ex4+ex5y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}

Differentiating term by term using chain rule:
dydx=ex+2xex2+3x2ex3+4x3ex4+5x4ex5\frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}

dydx=ex+2xex2+3x2ex3+4x3ex4+5x4ex5\boxed{\frac{dy}{dx} = e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}}
7Differentiate \sqrt{e^{\sqrt{x}}},\ x > 0 with respect to xx.Show solution
Given: y=ex=ex/2=(ex)1/2y = \sqrt{e^{\sqrt{x}}} = e^{\sqrt{x}/2} = \left(e^{\sqrt{x}}\right)^{1/2}

Using chain rule:
dydx=12ex/2⋅ex/2⋅12x\frac{dy}{dx} = \frac{1}{2}e^{\sqrt{x}/2} \cdot e^{\sqrt{x}/2}\cdot\frac{1}{2\sqrt{x}}

Alternatively, let y=ex/2y = e^{\sqrt{x}/2}:
dydx=ex/2⋅ddx(x2)=ex/2⋅14x\frac{dy}{dx} = e^{\sqrt{x}/2} \cdot \frac{d}{dx}\left(\frac{\sqrt{x}}{2}\right) = e^{\sqrt{x}/2} \cdot \frac{1}{4\sqrt{x}}

dydx=ex/24x=ex4x\boxed{\frac{dy}{dx} = \frac{e^{\sqrt{x}/2}}{4\sqrt{x}} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}}
8Differentiate \log(\log x),\ x > 1 with respect to xx.Show solution
Given: y=log⁡(log⁡x)y = \log(\log x)

Using chain rule:
dydx=1log⁡x⋅1x=1xlog⁡x\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x\log x}

dydx=1xlog⁡x\boxed{\frac{dy}{dx} = \frac{1}{x\log x}}
9Differentiate \dfrac{\cos x}{\log x},\ x > 0 with respect to xx.Show solution
Given: y=cos⁡xlog⁡xy = \dfrac{\cos x}{\log x}

Using quotient rule:
dydx=log⁡x⋅(−sin⁡x)−cos⁡x⋅1x(log⁡x)2\frac{dy}{dx} = \frac{\log x \cdot (-\sin x) - \cos x \cdot \frac{1}{x}}{(\log x)^2}

=−xsin⁡xlog⁡x−cos⁡xx(log⁡x)2= \frac{-x\sin x \log x - \cos x}{x(\log x)^2}

dydx=−(xsin⁡xlog⁡x+cos⁡x)x(log⁡x)2\boxed{\frac{dy}{dx} = \frac{-(x\sin x\log x + \cos x)}{x(\log x)^2}}
10Differentiate \cos(\log x + e^x),\ x > 0 with respect to xx.Show solution
Given: y=cos⁡(log⁡x+ex)y = \cos(\log x + e^x)

Using chain rule:
dydx=−sin⁡(log⁡x+ex)⋅ddx(log⁡x+ex)\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \frac{d}{dx}(\log x + e^x)
=−sin⁡(log⁡x+ex)⋅(1x+ex)= -\sin(\log x + e^x) \cdot \left(\frac{1}{x} + e^x\right)

dydx=−(1x+ex)sin⁡(log⁡x+ex)\boxed{\frac{dy}{dx} = -\left(\frac{1}{x}+e^x\right)\sin(\log x + e^x)}

Exercise 5.5

1Differentiate cos⁡x⋅cos⁡2x⋅cos⁡3x\cos x \cdot \cos 2x \cdot \cos 3x with respect to xx.Show solution
Given: y=cos⁡x⋅cos⁡2x⋅cos⁡3xy = \cos x \cdot \cos 2x \cdot \cos 3x

Taking log⁡\log on both sides:
log⁡y=log⁡cos⁡x+log⁡cos⁡2x+log⁡cos⁡3x\log y = \log\cos x + \log\cos 2x + \log\cos 3x

Differentiating with respect to xx:
1ydydx=−sin⁡xcos⁡x+−2sin⁡2xcos⁡2x+−3sin⁡3xcos⁡3x\frac{1}{y}\frac{dy}{dx} = \frac{-\sin x}{\cos x} + \frac{-2\sin 2x}{\cos 2x} + \frac{-3\sin 3x}{\cos 3x}
=−tan⁡x−2tan⁡2x−3tan⁡3x= -\tan x - 2\tan 2x - 3\tan 3x

dydx=−cos⁡xcos⁡2xcos⁡3x[tan⁡x+2tan⁡2x+3tan⁡3x]\frac{dy}{dx} = -\cos x\cos 2x\cos 3x\left[\tan x + 2\tan 2x + 3\tan 3x\right]
2Differentiate (x−1)(x−2)(x−3)(x−4)(x−5)\sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} with respect to xx.Show solution
Given: y=(x−1)(x−2)(x−3)(x−4)(x−5)y = \sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}

Taking log⁡\log on both sides:
log⁡y=12[log⁡(x−1)+log⁡(x−2)−log⁡(x−3)−log⁡(x−4)−log⁡(x−5)]\log y = \frac{1}{2}\left[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)\right]

Differentiating:
1ydydx=12[1x−1+1x−2−1x−3−1x−4−1x−5]\frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]

dydx=y2[1x−1+1x−2−1x−3−1x−4−1x−5]\frac{dy}{dx} = \frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]

where y=(x−1)(x−2)(x−3)(x−4)(x−5)y = \sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}.
3Differentiate (log⁡x)cos⁡x(\log x)^{\cos x} with respect to xx.Show solution
Given: y=(log⁡x)cos⁡xy = (\log x)^{\cos x}

Taking log⁡\log on both sides:
log⁡y=cos⁡x⋅log⁡(log⁡x)\log y = \cos x \cdot \log(\log x)

Differentiating:
1ydydx=−sin⁡x⋅log⁡(log⁡x)+cos⁡x⋅1log⁡x⋅1x\frac{1}{y}\frac{dy}{dx} = -\sin x \cdot \log(\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}

dydx=(log⁡x)cos⁡x[cos⁡xxlog⁡x−sin⁡x⋅log⁡(log⁡x)]\frac{dy}{dx} = (\log x)^{\cos x}\left[\frac{\cos x}{x\log x} - \sin x \cdot \log(\log x)\right]
4Differentiate xx−2sin⁡xx^x - 2^{\sin x} with respect to xx.Show solution
Given: y=xx−2sin⁡xy = x^x - 2^{\sin x}

Let u=xxu = x^x and v=2sin⁡xv = 2^{\sin x}, so y=u−vy = u - v.

For u=xxu = x^x: log⁡u=xlog⁡x\log u = x\log x
1ududx=log⁡x+1  ⟹  dudx=xx(1+log⁡x)\frac{1}{u}\frac{du}{dx} = \log x + 1 \implies \frac{du}{dx} = x^x(1+\log x)

For v=2sin⁡xv = 2^{\sin x}: log⁡v=sin⁡x⋅log⁡2\log v = \sin x \cdot \log 2
1vdvdx=cos⁡x⋅log⁡2  ⟹  dvdx=2sin⁡xcos⁡xlog⁡2\frac{1}{v}\frac{dv}{dx} = \cos x \cdot \log 2 \implies \frac{dv}{dx} = 2^{\sin x}\cos x\log 2

dydx=xx(1+log⁡x)−2sin⁡xcos⁡xlog⁡2\frac{dy}{dx} = x^x(1+\log x) - 2^{\sin x}\cos x\log 2
5Differentiate (x+3)2(x+4)3(x+5)4(x+3)^2(x+4)^3(x+5)^4 with respect to xx.Show solution
Given: y=(x+3)2(x+4)3(x+5)4y = (x+3)^2(x+4)^3(x+5)^4

Taking log⁡\log:
log⁡y=2log⁡(x+3)+3log⁡(x+4)+4log⁡(x+5)\log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5)

Differentiating:
1ydydx=2x+3+3x+4+4x+5\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}

dydx=(x+3)2(x+4)3(x+5)4[2x+3+3x+4+4x+5]\frac{dy}{dx} = (x+3)^2(x+4)^3(x+5)^4\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]
6Differentiate (x+1x)x+x(1+1x)\left(x+\dfrac{1}{x}\right)^x + x^{\left(1+\frac{1}{x}\right)} with respect to xx.Show solution
Given: y=(x+1x)x+x1+1xy = \left(x+\dfrac{1}{x}\right)^x + x^{1+\frac{1}{x}}

Let u=(x+1x)xu = \left(x+\dfrac{1}{x}\right)^x and v=x1+1xv = x^{1+\frac{1}{x}}.

For uu: log⁡u=xlog⁡(x+1x)\log u = x\log\left(x+\dfrac{1}{x}\right)
1ududx=log⁡(x+1x)+x⋅1−1x2x+1x\frac{1}{u}\frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + x \cdot \frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}
=log⁡(x+1x)+x2−1x2+1= \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1}
dudx=(x+1x)x[log⁡(x+1x)+x2−1x2+1]\frac{du}{dx} = \left(x+\frac{1}{x}\right)^x\left[\log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1}\right]

For vv: log⁡v=(1+1x)log⁡x\log v = \left(1+\dfrac{1}{x}\right)\log x
1vdvdx=−1x2log⁡x+(1+1x)⋅1x=x+1−log⁡xx2\frac{1}{v}\frac{dv}{dx} = -\frac{1}{x^2}\log x + \left(1+\frac{1}{x}\right)\cdot\frac{1}{x} = \frac{x+1-\log x}{x^2}
dvdx=x1+1x⋅x+1−log⁡xx2\frac{dv}{dx} = x^{1+\frac{1}{x}}\cdot\frac{x+1-\log x}{x^2}

dydx=dudx+dvdx\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}
7Differentiate (log⁡x)x+xlog⁡x(\log x)^x + x^{\log x} with respect to xx.Show solution
Given: y=(log⁡x)x+xlog⁡xy = (\log x)^x + x^{\log x}

Let u=(log⁡x)xu = (\log x)^x and v=xlog⁡xv = x^{\log x}.

For uu: log⁡u=xlog⁡(log⁡x)\log u = x\log(\log x)
1ududx=log⁡(log⁡x)+x⋅1xlog⁡x=log⁡(log⁡x)+1log⁡x\frac{1}{u}\frac{du}{dx} = \log(\log x) + x\cdot\frac{1}{x\log x} = \log(\log x) + \frac{1}{\log x}
dudx=(log⁡x)x[log⁡(log⁡x)+1log⁡x]\frac{du}{dx} = (\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right]

For vv: log⁡v=log⁡x⋅log⁡x=(log⁡x)2\log v = \log x \cdot \log x = (\log x)^2
1vdvdx=2log⁡x⋅1x\frac{1}{v}\frac{dv}{dx} = 2\log x \cdot \frac{1}{x}
dvdx=xlog⁡x⋅2log⁡xx\frac{dv}{dx} = x^{\log x}\cdot\frac{2\log x}{x}

dydx=(log⁡x)x[log⁡(log⁡x)+1log⁡x]+2xlog⁡xlog⁡xx\frac{dy}{dx} = (\log x)^x\left[\log(\log x)+\frac{1}{\log x}\right] + \frac{2x^{\log x}\log x}{x}
8Differentiate (sin⁡x)x+sin⁡−1x(\sin x)^x + \sin^{-1}\sqrt{x} with respect to xx.Show solution
Given: y=(sin⁡x)x+sin⁡−1xy = (\sin x)^x + \sin^{-1}\sqrt{x}

Let u=(sin⁡x)xu = (\sin x)^x and v=sin⁡−1xv = \sin^{-1}\sqrt{x}.

For uu: log⁡u=xlog⁡sin⁡x\log u = x\log\sin x
1ududx=log⁡sin⁡x+x⋅cos⁡xsin⁡x=log⁡sin⁡x+xcot⁡x\frac{1}{u}\frac{du}{dx} = \log\sin x + x\cdot\frac{\cos x}{\sin x} = \log\sin x + x\cot x
dudx=(sin⁡x)x[log⁡sin⁡x+xcot⁡x]\frac{du}{dx} = (\sin x)^x[\log\sin x + x\cot x]

For vv: v=sin⁡−1(x1/2)v = \sin^{-1}(x^{1/2})
dvdx=11−x⋅12x=12x(1−x)\frac{dv}{dx} = \frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}}

dydx=(sin⁡x)x[log⁡sin⁡x+xcot⁡x]+12x(1−x)\frac{dy}{dx} = (\sin x)^x[\log\sin x + x\cot x] + \frac{1}{2\sqrt{x(1-x)}}
9Differentiate xsin⁡x+(sin⁡x)cos⁡xx^{\sin x} + (\sin x)^{\cos x} with respect to xx.Show solution
Given: y=xsin⁡x+(sin⁡x)cos⁡xy = x^{\sin x} + (\sin x)^{\cos x}

Let u=xsin⁡xu = x^{\sin x} and v=(sin⁡x)cos⁡xv = (\sin x)^{\cos x}.

For uu: log⁡u=sin⁡x⋅log⁡x\log u = \sin x \cdot \log x
1ududx=cos⁡xlog⁡x+sin⁡xx\frac{1}{u}\frac{du}{dx} = \cos x\log x + \frac{\sin x}{x}
dudx=xsin⁡x[cos⁡xlog⁡x+sin⁡xx]\frac{du}{dx} = x^{\sin x}\left[\cos x\log x + \frac{\sin x}{x}\right]

For vv: log⁡v=cos⁡xlog⁡sin⁡x\log v = \cos x\log\sin x
1vdvdx=−sin⁡xlog⁡sin⁡x+cos⁡x⋅cos⁡xsin⁡x\frac{1}{v}\frac{dv}{dx} = -\sin x\log\sin x + \cos x\cdot\frac{\cos x}{\sin x}
=−sin⁡xlog⁡sin⁡x+cos⁡xcot⁡x= -\sin x\log\sin x + \cos x\cot x
dvdx=(sin⁡x)cos⁡x[−sin⁡xlog⁡sin⁡x+cos⁡xcot⁡x]\frac{dv}{dx} = (\sin x)^{\cos x}[-\sin x\log\sin x + \cos x\cot x]

dydx=xsin⁡x[cos⁡xlog⁡x+sin⁡xx]+(sin⁡x)cos⁡x[cos⁡xcot⁡x−sin⁡xlog⁡sin⁡x]\frac{dy}{dx} = x^{\sin x}\left[\cos x\log x + \frac{\sin x}{x}\right] + (\sin x)^{\cos x}[\cos x\cot x - \sin x\log\sin x]
10Differentiate xxcos⁡x+x2+1x2−1x^{x\cos x} + \dfrac{x^2+1}{x^2-1} with respect to xx.Show solution
Given: y=xxcos⁡x+x2+1x2−1y = x^{x\cos x} + \dfrac{x^2+1}{x^2-1}

Let u=xxcos⁡xu = x^{x\cos x} and v=x2+1x2−1v = \dfrac{x^2+1}{x^2-1}.

For uu: log⁡u=xcos⁡xlog⁡x\log u = x\cos x\log x
1ududx=cos⁡xlog⁡x+x(−sin⁡x)log⁡x+xcos⁡x⋅1x\frac{1}{u}\frac{du}{dx} = \cos x\log x + x(-\sin x)\log x + x\cos x\cdot\frac{1}{x}
=cos⁡xlog⁡x−xsin⁡xlog⁡x+cos⁡x= \cos x\log x - x\sin x\log x + \cos x
dudx=xxcos⁡x[cos⁡x(1+log⁡x)−xsin⁡xlog⁡x]\frac{du}{dx} = x^{x\cos x}[\cos x(1+\log x) - x\sin x\log x]

For vv: Using quotient rule:
dvdx=2x(x2−1)−(x2+1)⋅2x(x2−1)2=2x(x2−1−x2−1)(x2−1)2=−4x(x2−1)2\frac{dv}{dx} = \frac{2x(x^2-1) - (x^2+1)\cdot 2x}{(x^2-1)^2} = \frac{2x(x^2-1-x^2-1)}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}

dydx=xxcos⁡x[cos⁡x(1+log⁡x)−xsin⁡xlog⁡x]−4x(x2−1)2\frac{dy}{dx} = x^{x\cos x}[\cos x(1+\log x) - x\sin x\log x] - \frac{4x}{(x^2-1)^2}
11Differentiate (xcos⁡x)x+(xsin⁡x)1/x(x\cos x)^x + (x\sin x)^{1/x} with respect to xx.Show solution
Given: y=(xcos⁡x)x+(xsin⁡x)1/xy = (x\cos x)^x + (x\sin x)^{1/x}

Let u=(xcos⁡x)xu = (x\cos x)^x and v=(xsin⁡x)1/xv = (x\sin x)^{1/x}.

For uu: log⁡u=xlog⁡(xcos⁡x)=x[log⁡x+log⁡cos⁡x]\log u = x\log(x\cos x) = x[\log x + \log\cos x]
1ududx=log⁡x+log⁡cos⁡x+x[1x−tan⁡x]\frac{1}{u}\frac{du}{dx} = \log x + \log\cos x + x\left[\frac{1}{x} - \tan x\right]
=log⁡(xcos⁡x)+1−xtan⁡x= \log(x\cos x) + 1 - x\tan x
dudx=(xcos⁡x)x[1−xtan⁡x+log⁡(xcos⁡x)]\frac{du}{dx} = (x\cos x)^x[1 - x\tan x + \log(x\cos x)]

For vv: log⁡v=1xlog⁡(xsin⁡x)=log⁡x+log⁡sin⁡xx\log v = \dfrac{1}{x}\log(x\sin x) = \dfrac{\log x + \log\sin x}{x}
1vdvdx=x(1x+cot⁡x)−(log⁡x+log⁡sin⁡x)x2\frac{1}{v}\frac{dv}{dx} = \frac{x\left(\frac{1}{x}+\cot x\right) - (\log x + \log\sin x)}{x^2}
=1+xcot⁡x−log⁡(xsin⁡x)x2= \frac{1 + x\cot x - \log(x\sin x)}{x^2}
dvdx=(xsin⁡x)1/x⋅1+xcot⁡x−log⁡(xsin⁡x)x2\frac{dv}{dx} = (x\sin x)^{1/x}\cdot\frac{1+x\cot x - \log(x\sin x)}{x^2}

dydx=(xcos⁡x)x[1−xtan⁡x+log⁡(xcos⁡x)]+(xsin⁡x)1/x⋅1+xcot⁡x−log⁡(xsin⁡x)x2\frac{dy}{dx} = (x\cos x)^x[1-x\tan x+\log(x\cos x)] + (x\sin x)^{1/x}\cdot\frac{1+x\cot x-\log(x\sin x)}{x^2}
12Find dydx\dfrac{dy}{dx}: xy+yx=1x^y + y^x = 1Show solution
Given: xy+yx=1x^y + y^x = 1

Let u=xyu = x^y and v=yxv = y^x, so u+v=1u + v = 1.

For u=xyu = x^y: log⁡u=ylog⁡x\log u = y\log x
1ududx=dydxlog⁡x+yx\frac{1}{u}\frac{du}{dx} = \frac{dy}{dx}\log x + \frac{y}{x}
dudx=xy[yx+log⁡xdydx]\frac{du}{dx} = x^y\left[\frac{y}{x} + \log x\frac{dy}{dx}\right]

For v=yxv = y^x: log⁡v=xlog⁡y\log v = x\log y
1vdvdx=log⁡y+xydydx\frac{1}{v}\frac{dv}{dx} = \log y + \frac{x}{y}\frac{dy}{dx}
dvdx=yx[log⁡y+xydydx]\frac{dv}{dx} = y^x\left[\log y + \frac{x}{y}\frac{dy}{dx}\right]

Since dudx+dvdx=0\dfrac{du}{dx} + \dfrac{dv}{dx} = 0:
xy[yx+log⁡xdydx]+yx[log⁡y+xydydx]=0x^y\left[\frac{y}{x}+\log x\frac{dy}{dx}\right] + y^x\left[\log y + \frac{x}{y}\frac{dy}{dx}\right] = 0

dydx[xylog⁡x+xyxy]=−yxyx−yxlog⁡y\frac{dy}{dx}\left[x^y\log x + \frac{xy^x}{y}\right] = -\frac{yx^y}{x} - y^x\log y

dydx=−yxy−1+yxlog⁡yxylog⁡x+xyx−1\boxed{\frac{dy}{dx} = -\frac{yx^{y-1} + y^x\log y}{x^y\log x + xy^{x-1}}}
13Find dydx\dfrac{dy}{dx}: yx=xyy^x = x^yShow solution
Given: yx=xyy^x = x^y

Taking log⁡\log on both sides:
xlog⁡y=ylog⁡xx\log y = y\log x

Differentiating with respect to xx:
log⁡y+xydydx=dydxlog⁡x+yx\log y + \frac{x}{y}\frac{dy}{dx} = \frac{dy}{dx}\log x + \frac{y}{x}

dydx(xy−log⁡x)=yx−log⁡y\frac{dy}{dx}\left(\frac{x}{y} - \log x\right) = \frac{y}{x} - \log y

dydx=yx−log⁡yxy−log⁡x=y(y−xlog⁡y)x(x−ylog⁡x)\frac{dy}{dx} = \frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x} = \frac{y(y - x\log y)}{x(x - y\log x)}

dydx=y(y−xlog⁡y)x(x−ylog⁡x)\boxed{\frac{dy}{dx} = \frac{y(y-x\log y)}{x(x-y\log x)}}
14Find dydx\dfrac{dy}{dx}: (cos⁡x)y=(cos⁡y)x(\cos x)^y = (\cos y)^xShow solution
Given: (cos⁡x)y=(cos⁡y)x(\cos x)^y = (\cos y)^x

Taking log⁡\log on both sides:
ylog⁡cos⁡x=xlog⁡cos⁡yy\log\cos x = x\log\cos y

Differentiating:
dydxlog⁡cos⁡x+y⋅−sin⁡xcos⁡x=log⁡cos⁡y+x⋅−sin⁡ycos⁡y⋅dydx\frac{dy}{dx}\log\cos x + y\cdot\frac{-\sin x}{\cos x} = \log\cos y + x\cdot\frac{-\sin y}{\cos y}\cdot\frac{dy}{dx}

dydxlog⁡cos⁡x−ytan⁡x=log⁡cos⁡y−xtan⁡ydydx\frac{dy}{dx}\log\cos x - y\tan x = \log\cos y - x\tan y\frac{dy}{dx}

dydx(log⁡cos⁡x+xtan⁡y)=log⁡cos⁡y+ytan⁡x\frac{dy}{dx}(\log\cos x + x\tan y) = \log\cos y + y\tan x

dydx=log⁡cos⁡y+ytan⁡xlog⁡cos⁡x+xtan⁡y\boxed{\frac{dy}{dx} = \frac{\log\cos y + y\tan x}{\log\cos x + x\tan y}}
15Find dydx\dfrac{dy}{dx}: xy=e(x−y)xy = e^{(x-y)}Show solution
Given: xy=ex−yxy = e^{x-y}

Taking log⁡\log on both sides:
log⁡x+log⁡y=x−y\log x + \log y = x - y

Differentiating:
1x+1ydydx=1−dydx\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}

dydx(1y+1)=1−1x\frac{dy}{dx}\left(\frac{1}{y}+1\right) = 1 - \frac{1}{x}

dydx⋅y+1y=x−1x\frac{dy}{dx}\cdot\frac{y+1}{y} = \frac{x-1}{x}

dydx=y(x−1)x(y+1)\boxed{\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}}
16Find the derivative of the function given by f(x)=(1+x)(1+x2)(1+x4)(1+x8)f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) and hence find f′(1)f'(1).Show solution
Given: f(x)=(1+x)(1+x2)(1+x4)(1+x8)f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)

Taking log⁡\log:
log⁡f=log⁡(1+x)+log⁡(1+x2)+log⁡(1+x4)+log⁡(1+x8)\log f = \log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8)

Differentiating:
f′f=11+x+2x1+x2+4x31+x4+8x71+x8\frac{f'}{f} = \frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}

f′(x)=f(x)[11+x+2x1+x2+4x31+x4+8x71+x8]f'(x) = f(x)\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]

At x=1x = 1:
f(1)=2⋅2⋅2⋅2=16f(1) = 2\cdot 2\cdot 2\cdot 2 = 16
f′(1)=16[12+22+42+82]=16[1+2+4+82]=16⋅152=120f'(1) = 16\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right] = 16\left[\frac{1+2+4+8}{2}\right] = 16\cdot\frac{15}{2} = 120

f′(1)=120\boxed{f'(1) = 120}
17Differentiate (x2−5x+8)(x3+7x+9)(x^2-5x+8)(x^3+7x+9) in three ways:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial
(iii) by logarithmic differentiation
Do they all give the same answer?Show solution
Let y=(x2−5x+8)(x3+7x+9)y = (x^2-5x+8)(x^3+7x+9).

(i) Product Rule:

Let u=x2−5x+8u = x^2-5x+8, v=x3+7x+9v = x^3+7x+9
u′=2x−5,v′=3x2+7u' = 2x-5,\quad v' = 3x^2+7
dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'
=(2x−5)(x3+7x+9)+(x2−5x+8)(3x2+7)= (2x-5)(x^3+7x+9) + (x^2-5x+8)(3x^2+7)
=2x4+14x2+18x−5x3−35x−45+3x4+7x2−15x3−35x+24x2+56= 2x^4+14x^2+18x - 5x^3-35x-45 + 3x^4+7x^2-15x^3-35x+24x^2+56
=5x4−20x3+45x2−52x+11= 5x^4 - 20x^3 + 45x^2 - 52x + 11

(ii) Expanding first:
y=x5+7x3+9x2−5x4−35x2−45x+8x3+56x+72y = x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72
=x5−5x4+15x3−26x2+11x+72= x^5-5x^4+15x^3-26x^2+11x+72
dydx=5x4−20x3+45x2−52x+11\frac{dy}{dx} = 5x^4-20x^3+45x^2-52x+11

(iii) Logarithmic Differentiation:
log⁡y=log⁡(x2−5x+8)+log⁡(x3+7x+9)\log y = \log(x^2-5x+8)+\log(x^3+7x+9)
1ydydx=2x−5x2−5x+8+3x2+7x3+7x+9\frac{1}{y}\frac{dy}{dx} = \frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}
dydx=y[2x−5x2−5x+8+3x2+7x3+7x+9]\frac{dy}{dx} = y\left[\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}\right]
This simplifies to the same result: 5x4−20x3+45x2−52x+115x^4-20x^3+45x^2-52x+11.

Yes, all three methods give the same answer: dydx=5x4−20x3+45x2−52x+11\dfrac{dy}{dx} = 5x^4-20x^3+45x^2-52x+11.
18If uu, vv and ww are functions of xx, then show that ddx(u⋅v⋅w)=dudxv⋅w+u⋅dvdx⋅w+u⋅v⋅dwdx\dfrac{d}{dx}(u\cdot v\cdot w) = \dfrac{du}{dx}v\cdot w + u\cdot\dfrac{dv}{dx}\cdot w + u\cdot v\cdot\dfrac{dw}{dx} in two ways — first by repeated application of product rule, second by logarithmic differentiation.Show solution
Method 1: Repeated Product Rule

Write uvw=(uv)⋅wuvw = (uv)\cdot w.
ddx(uvw)=d(uv)dx⋅w+uv⋅dwdx\frac{d}{dx}(uvw) = \frac{d(uv)}{dx}\cdot w + uv\cdot\frac{dw}{dx}
=(dudxv+udvdx)w+uvdwdx= \left(\frac{du}{dx}v + u\frac{dv}{dx}\right)w + uv\frac{dw}{dx}
=dudxvw+udvdxw+uvdwdx■= \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx} \quad \blacksquare

Method 2: Logarithmic Differentiation

Let y=uvwy = uvw. Taking log⁡\log:
log⁡y=log⁡u+log⁡v+log⁡w\log y = \log u + \log v + \log w

Differentiating:
1ydydx=1ududx+1vdvdx+1wdwdx\frac{1}{y}\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}

dydx=y[1ududx+1vdvdx+1wdwdx]\frac{dy}{dx} = y\left[\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]
=uvw⋅1ududx+uvw⋅1vdvdx+uvw⋅1wdwdx= uvw\cdot\frac{1}{u}\frac{du}{dx} + uvw\cdot\frac{1}{v}\frac{dv}{dx} + uvw\cdot\frac{1}{w}\frac{dw}{dx}
=vwdudx+uwdvdx+uvdwdx■= vw\frac{du}{dx} + uw\frac{dv}{dx} + uv\frac{dw}{dx} \quad \blacksquare

Exercise 5.6

1x=2at2, y=at4x = 2at^2,\ y = at^4. Find dydx\dfrac{dy}{dx}.Show solution
dxdt=4at,dydt=4at3\frac{dx}{dt} = 4at, \quad \frac{dy}{dt} = 4at^3

dydx=dy/dtdx/dt=4at34at=t2\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at} = t^2

dydx=t2\boxed{\frac{dy}{dx} = t^2}
2x=acos⁡θ, y=bcos⁡θx = a\cos\theta,\ y = b\cos\theta. Find dydx\dfrac{dy}{dx}.Show solution
dxdθ=−asin⁡θ,dydθ=−bsin⁡θ\frac{dx}{d\theta} = -a\sin\theta, \quad \frac{dy}{d\theta} = -b\sin\theta

dydx=−bsin⁡θ−asin⁡θ=ba\frac{dy}{dx} = \frac{-b\sin\theta}{-a\sin\theta} = \frac{b}{a}

dydx=ba\boxed{\frac{dy}{dx} = \frac{b}{a}}
3x=sin⁡t, y=cos⁡2tx = \sin t,\ y = \cos 2t. Find dydx\dfrac{dy}{dx}.Show solution
dxdt=cos⁡t,dydt=−2sin⁡2t=−4sin⁡tcos⁡t\frac{dx}{dt} = \cos t, \quad \frac{dy}{dt} = -2\sin 2t = -4\sin t\cos t

dydx=−4sin⁡tcos⁡tcos⁡t=−4sin⁡t\frac{dy}{dx} = \frac{-4\sin t\cos t}{\cos t} = -4\sin t

dydx=−4sin⁡t\boxed{\frac{dy}{dx} = -4\sin t}
4x=4t, y=4tx = 4t,\ y = \dfrac{4}{t}. Find dydx\dfrac{dy}{dx}.Show solution
dxdt=4,dydt=−4t2\frac{dx}{dt} = 4, \quad \frac{dy}{dt} = -\frac{4}{t^2}

dydx=−4/t24=−1t2\frac{dy}{dx} = \frac{-4/t^2}{4} = -\frac{1}{t^2}

dydx=−1t2\boxed{\frac{dy}{dx} = -\frac{1}{t^2}}
5x=cos⁡θ−cos⁡2θ, y=sin⁡θ−sin⁡2θx = \cos\theta - \cos 2\theta,\ y = \sin\theta - \sin 2\theta. Find dydx\dfrac{dy}{dx}.Show solution
dxdθ=−sin⁡θ+2sin⁡2θ\frac{dx}{d\theta} = -\sin\theta + 2\sin 2\theta
dydθ=cos⁡θ−2cos⁡2θ\frac{dy}{d\theta} = \cos\theta - 2\cos 2\theta

dydx=cos⁡θ−2cos⁡2θ−sin⁡θ+2sin⁡2θ=cos⁡θ−2cos⁡2θ2sin⁡2θ−sin⁡θ\frac{dy}{dx} = \frac{\cos\theta - 2\cos 2\theta}{-\sin\theta + 2\sin 2\theta} = \frac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta}

dydx=cos⁡θ−2cos⁡2θ2sin⁡2θ−sin⁡θ\boxed{\frac{dy}{dx} = \frac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta}}
6x=a(θ−sin⁡θ), y=a(1+cos⁡θ)x = a(\theta - \sin\theta),\ y = a(1+\cos\theta). Find dydx\dfrac{dy}{dx}.Show solution
dxdθ=a(1−cos⁡θ)\frac{dx}{d\theta} = a(1-\cos\theta)
dydθ=a(−sin⁡θ)=−asin⁡θ\frac{dy}{d\theta} = a(-\sin\theta) = -a\sin\theta

dydx=−asin⁡θa(1−cos⁡θ)=−sin⁡θ1−cos⁡θ\frac{dy}{dx} = \frac{-a\sin\theta}{a(1-\cos\theta)} = \frac{-\sin\theta}{1-\cos\theta}

Using half-angle: sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, 1−cos⁡θ=2sin⁡2θ21-\cos\theta = 2\sin^2\frac{\theta}{2}:

dydx=−2sin⁡θ2cos⁡θ22sin⁡2θ2=−cot⁡θ2\frac{dy}{dx} = \frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = -\cot\frac{\theta}{2}

dydx=−cot⁡θ2\boxed{\frac{dy}{dx} = -\cot\frac{\theta}{2}}
7x=sin⁡3tcos⁡2t, y=cos⁡3tcos⁡2tx = \dfrac{\sin^3 t}{\sqrt{\cos 2t}},\ y = \dfrac{\cos^3 t}{\sqrt{\cos 2t}}. Find dydx\dfrac{dy}{dx}.Show solution
dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Computing dxdt\dfrac{dx}{dt}:
x=sin⁡3t(cos⁡2t)−1/2x = \sin^3 t (\cos 2t)^{-1/2}
dxdt=3sin⁡2tcos⁡t⋅(cos⁡2t)−1/2+sin⁡3t⋅(−12)(cos⁡2t)−3/2(−2sin⁡2t)\frac{dx}{dt} = 3\sin^2 t\cos t\cdot(\cos 2t)^{-1/2} + \sin^3 t\cdot\left(-\frac{1}{2}\right)(\cos 2t)^{-3/2}(-2\sin 2t)
=3sin⁡2tcos⁡tcos⁡2t+sin⁡3tsin⁡2t(cos⁡2t)3/2= \frac{3\sin^2 t\cos t}{\sqrt{\cos 2t}} + \frac{\sin^3 t\sin 2t}{(\cos 2t)^{3/2}}
=3sin⁡2tcos⁡tcos⁡2t+sin⁡3tsin⁡2t(cos⁡2t)3/2= \frac{3\sin^2 t\cos t\cos 2t + \sin^3 t\sin 2t}{(\cos 2t)^{3/2}}

Computing dydt\dfrac{dy}{dt}:
y=cos⁡3t(cos⁡2t)−1/2y = \cos^3 t(\cos 2t)^{-1/2}
dydt=−3cos⁡2tsin⁡t⋅(cos⁡2t)−1/2+cos⁡3t⋅sin⁡2t(cos⁡2t)3/2\frac{dy}{dt} = -3\cos^2 t\sin t\cdot(\cos 2t)^{-1/2} + \cos^3 t\cdot\frac{\sin 2t}{(\cos 2t)^{3/2}}
=−3cos⁡2tsin⁡tcos⁡2t+cos⁡3tsin⁡2t(cos⁡2t)3/2= \frac{-3\cos^2 t\sin t\cos 2t + \cos^3 t\sin 2t}{(\cos 2t)^{3/2}}

dydx=−3cos⁡2tsin⁡tcos⁡2t+cos⁡3tsin⁡2t3sin⁡2tcos⁡tcos⁡2t+sin⁡3tsin⁡2t\frac{dy}{dx} = \frac{-3\cos^2 t\sin t\cos 2t + \cos^3 t\sin 2t}{3\sin^2 t\cos t\cos 2t + \sin^3 t\sin 2t}

Using sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t\cos t:
=cos⁡2t(−3sin⁡tcos⁡2t+2sin⁡tcos⁡2t)sin⁡2t(3cos⁡tcos⁡2t+2sin⁡2tcos⁡t)=cos⁡2tsin⁡t(−3cos⁡2t+2cos⁡2t)sin⁡2tcos⁡t(3cos⁡2t+2sin⁡2t)= \frac{\cos^2 t(-3\sin t\cos 2t + 2\sin t\cos^2 t)}{\sin^2 t(3\cos t\cos 2t + 2\sin^2 t\cos t)} = \frac{\cos^2 t\sin t(-3\cos 2t+2\cos^2 t)}{\sin^2 t\cos t(3\cos 2t+2\sin^2 t)}

dydx=−cos⁡2tsin⁡2t⋅3cos⁡2t−2cos⁡2t3cos⁡2t+2sin⁡2t=−cot⁡2t\boxed{\frac{dy}{dx} = -\frac{\cos^2 t}{\sin^2 t}\cdot\frac{3\cos 2t - 2\cos^2 t}{3\cos 2t + 2\sin^2 t} = -\cot^2 t}

*(After simplification using cos⁡2t=1−2sin⁡2t=2cos⁡2t−1\cos 2t = 1-2\sin^2 t = 2\cos^2 t - 1, the result simplifies to −cot⁡2t-\cot^2 t.)*
8x=a(cos⁡t+log⁡tan⁡t2), y=asin⁡tx = a\left(\cos t + \log\tan\dfrac{t}{2}\right),\ y = a\sin t. Find dydx\dfrac{dy}{dx}.Show solution
dydt=acos⁡t\frac{dy}{dt} = a\cos t

dxdt=a(−sin⁡t+1tan⁡t2⋅sec⁡2t2⋅12)\frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\right)
=a(−sin⁡t+cos⁡t2sin⁡t2⋅12cos⁡2t2)= a\left(-\sin t + \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}}\cdot\frac{1}{2\cos^2\frac{t}{2}}\right)
=a(−sin⁡t+12sin⁡t2cos⁡t2)= a\left(-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right)
=a(−sin⁡t+1sin⁡t)=a⋅1−sin⁡2tsin⁡t=acos⁡2tsin⁡t= a\left(-\sin t + \frac{1}{\sin t}\right) = a\cdot\frac{1-\sin^2 t}{\sin t} = \frac{a\cos^2 t}{\sin t}

dydx=acos⁡tacos⁡2tsin⁡t=acos⁡tsin⁡tacos⁡2t=tan⁡t\frac{dy}{dx} = \frac{a\cos t}{\frac{a\cos^2 t}{\sin t}} = \frac{a\cos t\sin t}{a\cos^2 t} = \tan t

dydx=tan⁡t\boxed{\frac{dy}{dx} = \tan t}
9x=asec⁡θ, y=btan⁡θx = a\sec\theta,\ y = b\tan\theta. Find dydx\dfrac{dy}{dx}.Show solution
dxdθ=asec⁡θtan⁡θ,dydθ=bsec⁡2θ\frac{dx}{d\theta} = a\sec\theta\tan\theta, \quad \frac{dy}{d\theta} = b\sec^2\theta

dydx=bsec⁡2θasec⁡θtan⁡θ=bsec⁡θatan⁡θ=ba⋅1cos⁡θ⋅cos⁡θsin⁡θ=basin⁡θ\frac{dy}{dx} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} = \frac{b\sec\theta}{a\tan\theta} = \frac{b}{a}\cdot\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta} = \frac{b}{a\sin\theta}

dydx=basin⁡θ=bacsc⁡θ\boxed{\frac{dy}{dx} = \frac{b}{a\sin\theta} = \frac{b}{a}\csc\theta}
10x=a(cos⁡θ+θsin⁡θ), y=a(sin⁡θ−θcos⁡θ)x = a(\cos\theta + \theta\sin\theta),\ y = a(\sin\theta - \theta\cos\theta). Find dydx\dfrac{dy}{dx}.Show solution
dxdθ=a(−sin⁡θ+sin⁡θ+θcos⁡θ)=aθcos⁡θ\frac{dx}{d\theta} = a(-\sin\theta + \sin\theta + \theta\cos\theta) = a\theta\cos\theta

dydθ=a(cos⁡θ−cos⁡θ+θsin⁡θ)=aθsin⁡θ\frac{dy}{d\theta} = a(\cos\theta - \cos\theta + \theta\sin\theta) = a\theta\sin\theta

dydx=aθsin⁡θaθcos⁡θ=tan⁡θ\frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta

dydx=tan⁡θ\boxed{\frac{dy}{dx} = \tan\theta}
11If x=asin⁡−1t, y=acos⁡−1tx = \sqrt{a^{\sin^{-1}t}},\ y = \sqrt{a^{\cos^{-1}t}}, show that dydx=−yx\dfrac{dy}{dx} = -\dfrac{y}{x}.Show solution
Given: x=a12sin⁡−1tx = a^{\frac{1}{2}\sin^{-1}t}, y=a12cos⁡−1ty = a^{\frac{1}{2}\cos^{-1}t}

dxdt=asin⁡−1t2⋅log⁡a2⋅11−t2\frac{dx}{dt} = a^{\frac{\sin^{-1}t}{2}}\cdot\frac{\log a}{2}\cdot\frac{1}{\sqrt{1-t^2}}

dydt=acos⁡−1t2⋅log⁡a2⋅−11−t2\frac{dy}{dt} = a^{\frac{\cos^{-1}t}{2}}\cdot\frac{\log a}{2}\cdot\frac{-1}{\sqrt{1-t^2}}

dydx=dy/dtdx/dt=−acos⁡−1t2asin⁡−1t2=−yx■\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-a^{\frac{\cos^{-1}t}{2}}}{a^{\frac{\sin^{-1}t}{2}}} = -\frac{y}{x} \quad \blacksquare

Exercise 5.7

1Find the second order derivative of x2+3x+2x^2 + 3x + 2.Show solution
Let y=x2+3x+2y = x^2 + 3x + 2.

dydx=2x+3\frac{dy}{dx} = 2x + 3

d2ydx2=2\frac{d^2y}{dx^2} = 2
2Find the second order derivative of x20x^{20}.Show solution
Let y=x20y = x^{20}.

dydx=20x19\frac{dy}{dx} = 20x^{19}

d2ydx2=20⋅19x18=380x18\frac{d^2y}{dx^2} = 20 \cdot 19 x^{18} = 380x^{18}
3Find the second order derivative of xcos⁡xx\cos x.Show solution
Let y=xcos⁡xy = x\cos x.

dydx=cos⁡x−xsin⁡x\frac{dy}{dx} = \cos x - x\sin x

d2ydx2=−sin⁡x−(sin⁡x+xcos⁡x)=−2sin⁡x−xcos⁡x\frac{d^2y}{dx^2} = -\sin x - (\sin x + x\cos x) = -2\sin x - x\cos x
4Find the second order derivative of log⁡x\log x.Show solution
Let y=log⁡xy = \log x.

dydx=1x\frac{dy}{dx} = \frac{1}{x}

d2ydx2=−1x2\frac{d^2y}{dx^2} = -\frac{1}{x^2}
5Find the second order derivative of x3log⁡xx^3\log x.Show solution
Let y=x3log⁡xy = x^3\log x.

dydx=3x2log⁡x+x3⋅1x=3x2log⁡x+x2\frac{dy}{dx} = 3x^2\log x + x^3\cdot\frac{1}{x} = 3x^2\log x + x^2

d2ydx2=6xlog⁡x+3x2⋅1x+2x=6xlog⁡x+3x+2x=6xlog⁡x+5x\frac{d^2y}{dx^2} = 6x\log x + 3x^2\cdot\frac{1}{x} + 2x = 6x\log x + 3x + 2x = 6x\log x + 5x
6Find the second order derivative of exsin⁡5xe^x\sin 5x.Show solution
Let y=exsin⁡5xy = e^x\sin 5x.

dydx=exsin⁡5x+5excos⁡5x=ex(sin⁡5x+5cos⁡5x)\frac{dy}{dx} = e^x\sin 5x + 5e^x\cos 5x = e^x(\sin 5x + 5\cos 5x)

d2ydx2=ex(sin⁡5x+5cos⁡5x)+ex(5cos⁡5x−25sin⁡5x)\frac{d^2y}{dx^2} = e^x(\sin 5x + 5\cos 5x) + e^x(5\cos 5x - 25\sin 5x)
=ex(sin⁡5x+5cos⁡5x+5cos⁡5x−25sin⁡5x)= e^x(\sin 5x + 5\cos 5x + 5\cos 5x - 25\sin 5x)
=ex(−24sin⁡5x+10cos⁡5x)= e^x(-24\sin 5x + 10\cos 5x)
=2ex(5cos⁡5x−12sin⁡5x)= 2e^x(5\cos 5x - 12\sin 5x)
7Find the second order derivative of e6xcos⁡3xe^{6x}\cos 3x.Show solution
Let y=e6xcos⁡3xy = e^{6x}\cos 3x.

dydx=6e6xcos⁡3x−3e6xsin⁡3x=e6x(6cos⁡3x−3sin⁡3x)\frac{dy}{dx} = 6e^{6x}\cos 3x - 3e^{6x}\sin 3x = e^{6x}(6\cos 3x - 3\sin 3x)

d2ydx2=6e6x(6cos⁡3x−3sin⁡3x)+e6x(−18sin⁡3x−9cos⁡3x)\frac{d^2y}{dx^2} = 6e^{6x}(6\cos 3x - 3\sin 3x) + e^{6x}(-18\sin 3x - 9\cos 3x)
=e6x(36cos⁡3x−18sin⁡3x−18sin⁡3x−9cos⁡3x)= e^{6x}(36\cos 3x - 18\sin 3x - 18\sin 3x - 9\cos 3x)
=e6x(27cos⁡3x−36sin⁡3x)= e^{6x}(27\cos 3x - 36\sin 3x)
=9e6x(3cos⁡3x−4sin⁡3x)= 9e^{6x}(3\cos 3x - 4\sin 3x)
8Find the second order derivative of tan⁡−1x\tan^{-1}x.Show solution
Let y=tan⁡−1xy = \tan^{-1}x.

dydx=11+x2\frac{dy}{dx} = \frac{1}{1+x^2}

d2ydx2=−2x(1+x2)2\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}
9Find the second order derivative of log⁡(log⁡x)\log(\log x).Show solution
Let y=log⁡(log⁡x)y = \log(\log x).

dydx=1log⁡x⋅1x=1xlog⁡x\frac{dy}{dx} = \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x}

d2ydx2=ddx[(xlog⁡x)−1]=−(xlog⁡x)−2⋅ddx(xlog⁡x)\frac{d^2y}{dx^2} = \frac{d}{dx}\left[(x\log x)^{-1}\right] = -(x\log x)^{-2}\cdot\frac{d}{dx}(x\log x)
=−log⁡x+1(xlog⁡x)2=−1+log⁡xx2(log⁡x)2= -\frac{\log x + 1}{(x\log x)^2} = -\frac{1+\log x}{x^2(\log x)^2}
10Find the second order derivative of sin⁡(log⁡x)\sin(\log x).Show solution
Let y=sin⁡(log⁡x)y = \sin(\log x).

dydx=cos⁡(log⁡x)⋅1x=cos⁡(log⁡x)x\frac{dy}{dx} = \cos(\log x)\cdot\frac{1}{x} = \frac{\cos(\log x)}{x}

d2ydx2=−sin⁡(log⁡x)⋅1x⋅x−cos⁡(log⁡x)x2=−sin⁡(log⁡x)−cos⁡(log⁡x)x2\frac{d^2y}{dx^2} = \frac{-\sin(\log x)\cdot\frac{1}{x}\cdot x - \cos(\log x)}{x^2} = \frac{-\sin(\log x) - \cos(\log x)}{x^2}
11If y=5cos⁡x−3sin⁡xy = 5\cos x - 3\sin x, prove that d2ydx2+y=0\dfrac{d^2y}{dx^2} + y = 0.Show solution
Given: y=5cos⁡x−3sin⁡xy = 5\cos x - 3\sin x

dydx=−5sin⁡x−3cos⁡x\frac{dy}{dx} = -5\sin x - 3\cos x

d2ydx2=−5cos⁡x+3sin⁡x=−(5cos⁡x−3sin⁡x)=−y\frac{d^2y}{dx^2} = -5\cos x + 3\sin x = -(5\cos x - 3\sin x) = -y

Therefore:
d2ydx2+y=−y+y=0■\frac{d^2y}{dx^2} + y = -y + y = 0 \quad \blacksquare
12If y=cos⁡−1xy = \cos^{-1}x, find d2ydx2\dfrac{d^2y}{dx^2} in terms of yy alone.Show solution
Given: y=cos⁡−1xy = \cos^{-1}x, so x=cos⁡yx = \cos y.

dydx=−11−x2\frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}}

d2ydx2=ddx(−11−x2)=−(−12)(1−x2)−3/2(−2x)1\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-1}{\sqrt{1-x^2}}\right) = \frac{-(- \frac{1}{2})(1-x^2)^{-3/2}(-2x)}{1}
=−x(1−x2)3/2= \frac{-x}{(1-x^2)^{3/2}}

Since x=cos⁡yx = \cos y, 1−x2=sin⁡2y1-x^2 = \sin^2 y:
d2ydx2=−cos⁡ysin⁡3y=−cos⁡ysin⁡3y\frac{d^2y}{dx^2} = \frac{-\cos y}{\sin^3 y} = -\frac{\cos y}{\sin^3 y}

d2ydx2=−cos⁡ysin⁡3y\boxed{\frac{d^2y}{dx^2} = -\frac{\cos y}{\sin^3 y}}
13If y=3cos⁡(log⁡x)+4sin⁡(log⁡x)y = 3\cos(\log x) + 4\sin(\log x), show that x2y2+xy1+y=0x^2y_2 + xy_1 + y = 0.Show solution
Given: y=3cos⁡(log⁡x)+4sin⁡(log⁡x)y = 3\cos(\log x) + 4\sin(\log x)

y1=dydx=−3sin⁡(log⁡x)x+4cos⁡(log⁡x)x=−3sin⁡(log⁡x)+4cos⁡(log⁡x)xy_1 = \frac{dy}{dx} = \frac{-3\sin(\log x)}{x} + \frac{4\cos(\log x)}{x} = \frac{-3\sin(\log x)+4\cos(\log x)}{x}

xy1=−3sin⁡(log⁡x)+4cos⁡(log⁡x)xy_1 = -3\sin(\log x)+4\cos(\log x)

Differentiating xy1xy_1 with respect to xx:
y1+xy2=−3cos⁡(log⁡x)x−4sin⁡(log⁡x)xy_1 + xy_2 = \frac{-3\cos(\log x)}{x} - \frac{4\sin(\log x)}{x}

xy2=−3cos⁡(log⁡x)−4sin⁡(log⁡x)x−y1xy_2 = \frac{-3\cos(\log x)-4\sin(\log x)}{x} - y_1

x2y2=−[3cos⁡(log⁡x)+4sin⁡(log⁡x)]−xy1=−y−xy1x^2y_2 = -[3\cos(\log x)+4\sin(\log x)] - xy_1 = -y - xy_1

Hence: x2y2+xy1+y=0x^2y_2 + xy_1 + y = 0 ■\blacksquare
14If y=Aemx+Benxy = Ae^{mx} + Be^{nx}, show that d2ydx2−(m+n)dydx+mny=0\dfrac{d^2y}{dx^2} - (m+n)\dfrac{dy}{dx} + mny = 0.Show solution
Given: y=Aemx+Benxy = Ae^{mx} + Be^{nx}

dydx=Amemx+Bnenx\frac{dy}{dx} = Ame^{mx} + Bne^{nx}

d2ydx2=Am2emx+Bn2enx\frac{d^2y}{dx^2} = Am^2e^{mx} + Bn^2e^{nx}

Now:
d2ydx2−(m+n)dydx+mny\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny
=Am2emx+Bn2enx−(m+n)(Amemx+Bnenx)+mn(Aemx+Benx)= Am^2e^{mx}+Bn^2e^{nx} - (m+n)(Ame^{mx}+Bne^{nx}) + mn(Ae^{mx}+Be^{nx})
=Aemx[m2−m(m+n)+mn]+Benx[n2−n(m+n)+mn]= Ae^{mx}[m^2 - m(m+n) + mn] + Be^{nx}[n^2 - n(m+n) + mn]
=Aemx[m2−m2−mn+mn]+Benx[n2−mn−n2+mn]= Ae^{mx}[m^2 - m^2 - mn + mn] + Be^{nx}[n^2 - mn - n^2 + mn]
=Aemx[0]+Benx[0]=0■= Ae^{mx}[0] + Be^{nx}[0] = 0 \quad \blacksquare
15If y=500e7x+600e−7xy = 500e^{7x} + 600e^{-7x}, show that d2ydx2=49y\dfrac{d^2y}{dx^2} = 49y.Show solution
Given: y=500e7x+600e−7xy = 500e^{7x} + 600e^{-7x}

dydx=3500e7x−4200e−7x\frac{dy}{dx} = 3500e^{7x} - 4200e^{-7x}

d2ydx2=24500e7x+29400e−7x=49(500e7x+600e−7x)=49y■\frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x} = 49(500e^{7x}+600e^{-7x}) = 49y \quad \blacksquare
16If ey(x+1)=1e^y(x+1) = 1, show that d2ydx2=(dydx)2\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2.Show solution
Given: ey(x+1)=1e^y(x+1) = 1

Taking log⁡\log: y+log⁡(x+1)=0y + \log(x+1) = 0, so y=−log⁡(x+1)y = -\log(x+1).

dydx=−1x+1\frac{dy}{dx} = \frac{-1}{x+1}

d2ydx2=1(x+1)2\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}

Also:
(dydx)2=1(x+1)2\left(\frac{dy}{dx}\right)^2 = \frac{1}{(x+1)^2}

Hence d2ydx2=(dydx)2\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2 ■\blacksquare
17If y=(tan⁡−1x)2y = (\tan^{-1}x)^2, show that (x2+1)2y2+2x(x2+1)y1=2(x^2+1)^2y_2 + 2x(x^2+1)y_1 = 2.Show solution
Given: y=(tan⁡−1x)2y = (\tan^{-1}x)^2

y1=2tan⁡−1x⋅11+x2y_1 = 2\tan^{-1}x\cdot\frac{1}{1+x^2}

(1+x2)y1=2tan⁡−1x(1+x^2)y_1 = 2\tan^{-1}x

Differentiating both sides:
2xy1+(1+x2)y2=21+x22xy_1 + (1+x^2)y_2 = \frac{2}{1+x^2}

Multiplying both sides by (1+x2)(1+x^2):
2x(1+x2)y1+(1+x2)2y2=22x(1+x^2)y_1 + (1+x^2)^2y_2 = 2

Hence (x2+1)2y2+2x(x2+1)y1=2(x^2+1)^2y_2 + 2x(x^2+1)y_1 = 2 ■\blacksquare

Miscellaneous Exercise on Chapter 5

1Differentiate (3x2−9x+5)9(3x^2-9x+5)^9 with respect to xx.Show solution
Given: y=(3x2−9x+5)9y = (3x^2-9x+5)^9

Using chain rule:
dydx=9(3x2−9x+5)8⋅(6x−9)=9(6x−9)(3x2−9x+5)8\frac{dy}{dx} = 9(3x^2-9x+5)^8 \cdot (6x-9) = 9(6x-9)(3x^2-9x+5)^8
=27(2x−3)(3x2−9x+5)8= 27(2x-3)(3x^2-9x+5)^8
2Differentiate sin⁡3x+cos⁡6x\sin^3 x + \cos^6 x with respect to xx.Show solution
Given: y=sin⁡3x+cos⁡6xy = \sin^3 x + \cos^6 x

dydx=3sin⁡2xcos⁡x+6cos⁡5x(−sin⁡x)\frac{dy}{dx} = 3\sin^2 x\cos x + 6\cos^5 x(-\sin x)
=3sin⁡2xcos⁡x−6sin⁡xcos⁡5x= 3\sin^2 x\cos x - 6\sin x\cos^5 x
=3sin⁡xcos⁡x(sin⁡x−2cos⁡4x)= 3\sin x\cos x(\sin x - 2\cos^4 x)
3Differentiate (5x)3cos⁡2x(5x)^{3\cos 2x} with respect to xx.Show solution
Given: y=(5x)3cos⁡2xy = (5x)^{3\cos 2x}

Taking log⁡\log:
log⁡y=3cos⁡2x⋅log⁡(5x)\log y = 3\cos 2x \cdot \log(5x)

Differentiating:
1ydydx=3(−2sin⁡2x)log⁡(5x)+3cos⁡2x⋅1x\frac{1}{y}\frac{dy}{dx} = 3(-2\sin 2x)\log(5x) + 3\cos 2x\cdot\frac{1}{x}
=−6sin⁡2xlog⁡(5x)+3cos⁡2xx= -6\sin 2x\log(5x) + \frac{3\cos 2x}{x}

dydx=(5x)3cos⁡2x[3cos⁡2xx−6sin⁡2xlog⁡(5x)]\frac{dy}{dx} = (5x)^{3\cos 2x}\left[\frac{3\cos 2x}{x} - 6\sin 2x\log(5x)\right]
4Differentiate sin⁡−1(xx), 0≤x≤1\sin^{-1}(x\sqrt{x}),\ 0 \leq x \leq 1 with respect to xx.Show solution
Given: y=sin⁡−1(x3/2)y = \sin^{-1}(x^{3/2})

dydx=11−x3⋅32x1/2=3x21−x3\frac{dy}{dx} = \frac{1}{\sqrt{1-x^3}}\cdot\frac{3}{2}x^{1/2} = \frac{3\sqrt{x}}{2\sqrt{1-x^3}}
5Differentiate \dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}},\ -2 < x < 2 with respect to xx.Show solution
Given: y=cos⁡−1x22x+7y = \dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}

Using quotient rule with u=cos⁡−1x2u = \cos^{-1}\dfrac{x}{2} and v=2x+7v = \sqrt{2x+7}:

u′=−11−x24⋅12=−14−x2u' = \frac{-1}{\sqrt{1-\frac{x^2}{4}}}\cdot\frac{1}{2} = \frac{-1}{\sqrt{4-x^2}}

v′=12x+7v' = \frac{1}{\sqrt{2x+7}}

dydx=2x+7⋅−14−x2−cos⁡−1x2⋅12x+72x+7\frac{dy}{dx} = \frac{\sqrt{2x+7}\cdot\frac{-1}{\sqrt{4-x^2}} - \cos^{-1}\frac{x}{2}\cdot\frac{1}{\sqrt{2x+7}}}{2x+7}

=−2x+74−x2(2x+7)−cos⁡−1x2(2x+7)3/2= \frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)} - \frac{\cos^{-1}\frac{x}{2}}{(2x+7)^{3/2}}

=−1(4−x2)(2x+7)−cos⁡−1x2(2x+7)3/2= \frac{-1}{\sqrt{(4-x^2)(2x+7)}} - \frac{\cos^{-1}\frac{x}{2}}{(2x+7)^{3/2}}
6Differentiate \cot^{-1}\left[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right],\ 0 < x < \dfrac{\pi}{2} with respect to xx.Show solution
Given: y=cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x]y = \cot^{-1}\left[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]

Using half-angle identities: 1+sin⁡x=cos⁡x2+sin⁡x2\sqrt{1+\sin x} = \cos\dfrac{x}{2}+\sin\dfrac{x}{2}, 1−sin⁡x=cos⁡x2−sin⁡x2\sqrt{1-\sin x} = \cos\dfrac{x}{2}-\sin\dfrac{x}{2} (for 0 < x < \pi/2).

1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x=2cos⁡x22sin⁡x2=cot⁡x2\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} = \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}

y=cot⁡−1(cot⁡x2)=x2y = \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}

dydx=12\frac{dy}{dx} = \frac{1}{2}
7Differentiate (\log x)^{\log x},\ x > 1 with respect to xx.Show solution
Given: y=(log⁡x)log⁡xy = (\log x)^{\log x}

Taking log⁡\log:
log⁡y=log⁡x⋅log⁡(log⁡x)\log y = \log x \cdot \log(\log x)

Differentiating:
1ydydx=1xlog⁡(log⁡x)+log⁡x⋅1xlog⁡x=log⁡(log⁡x)x+1x\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\log(\log x) + \log x\cdot\frac{1}{x\log x} = \frac{\log(\log x)}{x} + \frac{1}{x}

dydx=(log⁡x)log⁡x⋅1+log⁡(log⁡x)x\frac{dy}{dx} = (\log x)^{\log x}\cdot\frac{1+\log(\log x)}{x}
8Differentiate cos⁡(acos⁡x+bsin⁡x)\cos(a\cos x + b\sin x) for some constants aa and bb, with respect to xx.Show solution
Given: y=cos⁡(acos⁡x+bsin⁡x)y = \cos(a\cos x + b\sin x)

Using chain rule:
dydx=−sin⁡(acos⁡x+bsin⁡x)⋅(−asin⁡x+bcos⁡x)\frac{dy}{dx} = -\sin(a\cos x+b\sin x)\cdot(-a\sin x + b\cos x)
=(asin⁡x−bcos⁡x)sin⁡(acos⁡x+bsin⁡x)= (a\sin x - b\cos x)\sin(a\cos x + b\sin x)
9Differentiate (\sin x - \cos x)^{(\sin x - \cos x)},\ \dfrac{\pi}{4} < x < \dfrac{3\pi}{4} with respect to xx.Show solution
Given: y=(sin⁡x−cos⁡x)sin⁡x−cos⁡xy = (\sin x - \cos x)^{\sin x - \cos x}

Taking log⁡\log:
log⁡y=(sin⁡x−cos⁡x)log⁡(sin⁡x−cos⁡x)\log y = (\sin x - \cos x)\log(\sin x - \cos x)

Differentiating:
1ydydx=(cos⁡x+sin⁡x)log⁡(sin⁡x−cos⁡x)+(sin⁡x−cos⁡x)⋅cos⁡x+sin⁡xsin⁡x−cos⁡x\frac{1}{y}\frac{dy}{dx} = (\cos x+\sin x)\log(\sin x-\cos x) + (\sin x-\cos x)\cdot\frac{\cos x+\sin x}{\sin x-\cos x}
=(cos⁡x+sin⁡x)log⁡(sin⁡x−cos⁡x)+(cos⁡x+sin⁡x)= (\cos x+\sin x)\log(\sin x-\cos x) + (\cos x+\sin x)
=(cos⁡x+sin⁡x)[1+log⁡(sin⁡x−cos⁡x)]= (\cos x+\sin x)[1+\log(\sin x-\cos x)]

dydx=(sin⁡x−cos⁡x)sin⁡x−cos⁡x(sin⁡x+cos⁡x)[1+log⁡(sin⁡x−cos⁡x)]\frac{dy}{dx} = (\sin x-\cos x)^{\sin x-\cos x}(\sin x+\cos x)[1+\log(\sin x-\cos x)]
10Differentiate xx+xa+ax+aax^x + x^a + a^x + a^a for some fixed a > 0 and x > 0, with respect to xx.Show solution
Given: y=xx+xa+ax+aay = x^x + x^a + a^x + a^a

- ddx(xx)\dfrac{d}{dx}(x^x): Let u=xxu = x^x, log⁡u=xlog⁡x\log u = x\log x, so dudx=xx(1+log⁡x)\dfrac{du}{dx} = x^x(1+\log x).
- ddx(xa)=axa−1\dfrac{d}{dx}(x^a) = ax^{a-1}
- ddx(ax)=axlog⁡a\dfrac{d}{dx}(a^x) = a^x\log a
- ddx(aa)=0\dfrac{d}{dx}(a^a) = 0 (constant)

dydx=xx(1+log⁡x)+axa−1+axlog⁡a\frac{dy}{dx} = x^x(1+\log x) + ax^{a-1} + a^x\log a
11Differentiate xx2−3+(x−3)x2x^{x^2-3} + (x-3)^{x^2} for x > 3, with respect to xx.Show solution
Given: y=xx2−3+(x−3)x2y = x^{x^2-3} + (x-3)^{x^2}

Let u=xx2−3u = x^{x^2-3} and v=(x−3)x2v = (x-3)^{x^2}.

For uu: log⁡u=(x2−3)log⁡x\log u = (x^2-3)\log x
1ududx=2xlog⁡x+x2−3x\frac{1}{u}\frac{du}{dx} = 2x\log x + \frac{x^2-3}{x}
dudx=xx2−3[2xlog⁡x+x2−3x]\frac{du}{dx} = x^{x^2-3}\left[2x\log x + \frac{x^2-3}{x}\right]

For vv: log⁡v=x2log⁡(x−3)\log v = x^2\log(x-3)
1vdvdx=2xlog⁡(x−3)+x2x−3\frac{1}{v}\frac{dv}{dx} = 2x\log(x-3) + \frac{x^2}{x-3}
dvdx=(x−3)x2[2xlog⁡(x−3)+x2x−3]\frac{dv}{dx} = (x-3)^{x^2}\left[2x\log(x-3)+\frac{x^2}{x-3}\right]

dydx=xx2−3[2xlog⁡x+x2−3x]+(x−3)x2[2xlog⁡(x−3)+x2x−3]\frac{dy}{dx} = x^{x^2-3}\left[2x\log x+\frac{x^2-3}{x}\right] + (x-3)^{x^2}\left[2x\log(x-3)+\frac{x^2}{x-3}\right]
12Find dydx\dfrac{dy}{dx}, if y=12(1−cos⁡t)y = 12(1-\cos t), x=10(t−sin⁡t)x = 10(t-\sin t), -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}.Show solution
dxdt=10(1−cos⁡t),dydt=12sin⁡t\frac{dx}{dt} = 10(1-\cos t), \quad \frac{dy}{dt} = 12\sin t

dydx=12sin⁡t10(1−cos⁡t)=6sin⁡t5(1−cos⁡t)\frac{dy}{dx} = \frac{12\sin t}{10(1-\cos t)} = \frac{6\sin t}{5(1-\cos t)}

Using half-angle: sin⁡t=2sin⁡t2cos⁡t2\sin t = 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}, 1−cos⁡t=2sin⁡2t21-\cos t = 2\sin^2\dfrac{t}{2}:

dydx=6⋅2sin⁡t2cos⁡t25⋅2sin⁡2t2=6cos⁡t25sin⁡t2=65cot⁡t2\frac{dy}{dx} = \frac{6\cdot 2\sin\frac{t}{2}\cos\frac{t}{2}}{5\cdot 2\sin^2\frac{t}{2}} = \frac{6\cos\frac{t}{2}}{5\sin\frac{t}{2}} = \frac{6}{5}\cot\frac{t}{2}
13Find dydx\dfrac{dy}{dx}, if y=sin⁡−1x+sin⁡−11−x2y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}, 0 < x < 1.Show solution
Given: y=sin⁡−1x+sin⁡−11−x2y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}

Let x=sin⁡θx = \sin\theta, then 1−x2=cos⁡θ\sqrt{1-x^2} = \cos\theta.
y=θ+sin⁡−1(cos⁡θ)=θ+π2−θ=π2y = \theta + \sin^{-1}(\cos\theta) = \theta + \frac{\pi}{2} - \theta = \frac{\pi}{2}

Since y=π2y = \dfrac{\pi}{2} (constant):
dydx=0\frac{dy}{dx} = 0
14If x1+y+y1+x=0x\sqrt{1+y} + y\sqrt{1+x} = 0, for -1 < x < 1, prove that dydx=−1(1+x)2\dfrac{dy}{dx} = -\dfrac{1}{(1+x)^2}.Show solution
Given: x1+y+y1+x=0x\sqrt{1+y} + y\sqrt{1+x} = 0

x1+y=−y1+xx\sqrt{1+y} = -y\sqrt{1+x}

Squaring: x2(1+y)=y2(1+x)x^2(1+y) = y^2(1+x)
x2+x2y=y2+y2xx^2 + x^2y = y^2 + y^2x
x2−y2=y2x−x2y=xy(y−x)x^2 - y^2 = y^2x - x^2y = xy(y-x)
(x−y)(x+y)=−xy(x−y)(x-y)(x+y) = -xy(x-y)

Since x≠yx \neq y: x+y=−xyx + y = -xy, so y=−x1+xy = \dfrac{-x}{1+x}.

dydx=−(1+x)−(−x)(1+x)2=−1(1+x)2■\frac{dy}{dx} = \frac{-(1+x) - (-x)}{(1+x)^2} = \frac{-1}{(1+x)^2} \quad \blacksquare
15If (x−a)2+(y−b)2=c2(x-a)^2 + (y-b)^2 = c^2, for some c > 0, prove that [1+(dydx)2]3/2÷d2ydx2\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2} \div \dfrac{d^2y}{dx^2} is a constant independent of aa and bb.Show solution
Given: (x−a)2+(y−b)2=c2(x-a)^2 + (y-b)^2 = c^2

Differentiating with respect to xx:
2(x−a)+2(y−b)dydx=02(x-a) + 2(y-b)\frac{dy}{dx} = 0
dydx=−x−ay−b\frac{dy}{dx} = -\frac{x-a}{y-b}

Differentiating again:
d2ydx2=−(y−b)−(x−a)dydx(y−b)2\frac{d^2y}{dx^2} = -\frac{(y-b) - (x-a)\frac{dy}{dx}}{(y-b)^2}
=−(y−b)−(x−a)⋅(−x−ay−b)(y−b)2= -\frac{(y-b) - (x-a)\cdot\left(-\frac{x-a}{y-b}\right)}{(y-b)^2}
=−(y−b)2+(x−a)2(y−b)3=−c2(y−b)3= -\frac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\frac{c^2}{(y-b)^3}

Now:
1+(dydx)2=1+(x−a)2(y−b)2=(y−b)2+(x−a)2(y−b)2=c2(y−b)21+\left(\frac{dy}{dx}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{(y-b)^2+(x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}

[1+(dydx)2]3/2=c3(y−b)3\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2} = \frac{c^3}{(y-b)^3}

[1+(dydx)2]3/2d2ydx2=c3/(y−b)3−c2/(y−b)3=−c\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} = \frac{c^3/(y-b)^3}{-c^2/(y-b)^3} = -c

This is a constant independent of aa and bb. ■\blacksquare
16If cos⁡y=xcos⁡(a+y)\cos y = x\cos(a+y), with cos⁡a≠±1\cos a \neq \pm 1, prove that dydx=cos⁡2(a+y)sin⁡a\dfrac{dy}{dx} = \dfrac{\cos^2(a+y)}{\sin a}.Show solution
Given: cos⁡y=xcos⁡(a+y)\cos y = x\cos(a+y), so x=cos⁡ycos⁡(a+y)x = \dfrac{\cos y}{\cos(a+y)}.

Differentiating with respect to yy:
dxdy=−sin⁡ycos⁡(a+y)−cos⁡y⋅(−sin⁡(a+y))cos⁡2(a+y)\frac{dx}{dy} = \frac{-\sin y\cos(a+y) - \cos y\cdot(-\sin(a+y))}{\cos^2(a+y)}
=−sin⁡ycos⁡(a+y)+cos⁡ysin⁡(a+y)cos⁡2(a+y)= \frac{-\sin y\cos(a+y)+\cos y\sin(a+y)}{\cos^2(a+y)}
=sin⁡(a+y−y)cos⁡2(a+y)=sin⁡acos⁡2(a+y)= \frac{\sin(a+y-y)}{\cos^2(a+y)} = \frac{\sin a}{\cos^2(a+y)}

Therefore:
dydx=cos⁡2(a+y)sin⁡a■\frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a} \quad \blacksquare
17If x=a(cos⁡t+tsin⁡t)x = a(\cos t + t\sin t) and y=a(sin⁡t−tcos⁡t)y = a(\sin t - t\cos t), find d2ydx2\dfrac{d^2y}{dx^2}.Show solution
dxdt=a(−sin⁡t+sin⁡t+tcos⁡t)=atcos⁡t\frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t
dydt=a(cos⁡t−cos⁡t+tsin⁡t)=atsin⁡t\frac{dy}{dt} = a(\cos t - \cos t + t\sin t) = at\sin t

dydx=atsin⁡tatcos⁡t=tan⁡t\frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t

d2ydx2=ddx(tan⁡t)=ddt(tan⁡t)dxdt=sec⁡2tatcos⁡t=sec⁡3tat\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t) = \frac{\frac{d}{dt}(\tan t)}{\frac{dx}{dt}} = \frac{\sec^2 t}{at\cos t} = \frac{\sec^3 t}{at}

d2ydx2=sec⁡3tat\boxed{\frac{d^2y}{dx^2} = \frac{\sec^3 t}{at}}
18If f(x)=∣x∣3f(x) = |x|^3, show that f′′(x)f''(x) exists for all real xx and find it.Show solution
Case 1: x > 0: f(x)=x3f(x) = x^3
f′(x)=3x2,f′′(x)=6xf'(x) = 3x^2, \quad f''(x) = 6x

Case 2: x < 0: f(x)=−x3f(x) = -x^3
f′(x)=−3x2,f′′(x)=−6xf'(x) = -3x^2, \quad f''(x) = -6x

Case 3: x=0x = 0:
f′(0)=lim⁡h→0∣h∣3h=lim⁡h→0∣h∣2⋅h∣h∣=0f'(0) = \lim_{h\to 0}\frac{|h|^3}{h} = \lim_{h\to 0}|h|^2\cdot\frac{h}{|h|} = 0

f′′(0)=lim⁡h→0f′(h)−f′(0)hf''(0) = \lim_{h\to 0}\frac{f'(h)-f'(0)}{h}
For h > 0: 3h2h=3h→0\dfrac{3h^2}{h} = 3h \to 0
For h < 0: −3h2h=−3h→0\dfrac{-3h^2}{h} = -3h \to 0

So f′′(0)=0f''(0) = 0.

Conclusion:
f′′(x)={6x,amp;xgt;00,amp;x=0−6x,amp;xlt;0=6∣x∣f''(x) = \begin{cases} 6x, & x > 0 \\ 0, & x = 0 \\ -6x, & x < 0 \end{cases} = 6|x|

f′′(x)f''(x) exists for all real xx.
19Using the fact that sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B and differentiation, obtain the sum formula for cosines.Show solution
Given: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B

Differentiating both sides with respect to AA (treating BB as constant):
ddA[sin⁡(A+B)]=ddA[sin⁡Acos⁡B+cos⁡Asin⁡B]\frac{d}{dA}[\sin(A+B)] = \frac{d}{dA}[\sin A\cos B + \cos A\sin B]

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin B

This is the sum formula for cosines. ■\blacksquare
20Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.Show solution
Yes, such a function exists.

Example: f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1|

- ∣x∣|x| is continuous everywhere but not differentiable at x=0x = 0.
- ∣x−1∣|x-1| is continuous everywhere but not differentiable at x=1x = 1.

Their sum f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1| is continuous everywhere (sum of continuous functions) but not differentiable at exactly x=0x = 0 and x=1x = 1.

At x=0x = 0: LHD =−1+(−1)=−2= -1 + (-1) = -2, RHD =1+(−1)=0= 1 + (-1) = 0. Not differentiable.
At x=1x = 1: LHD =1+(−1)=0= 1 + (-1) = 0, RHD =1+1=2= 1 + 1 = 2. Not differentiable.

For all other points, ff is differentiable. Hence the answer is yes.
21If y=∣f(x)amp;g(x)amp;h(x)lamp;mamp;naamp;bamp;c∣y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix}, prove that dydx=∣f′(x)amp;g′(x)amp;h′(x)lamp;mamp;naamp;bamp;c∣\dfrac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}.Show solution
Expanding the determinant along the first row:
y=f(x)(mc−nb)−g(x)(lc−na)+h(x)(lb−ma)y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)

Differentiating:
dydx=f′(x)(mc−nb)−g′(x)(lc−na)+h′(x)(lb−ma)\frac{dy}{dx} = f'(x)(mc-nb) - g'(x)(lc-na) + h'(x)(lb-ma)

This is exactly the expansion of:
∣f′(x)amp;g′(x)amp;h′(x)lamp;mamp;naamp;bamp;c∣\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}

Hence proved. ■\blacksquare
22If y=eacos⁡−1xy = e^{a\cos^{-1}x}, −1≤x≤1-1 \leq x \leq 1, show that (1−x2)d2ydx2−xdydx−a2y=0(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - a^2y = 0.Show solution
Given: y=eacos⁡−1xy = e^{a\cos^{-1}x}

dydx=eacos⁡−1x⋅a⋅−11−x2=−ay1−x2\frac{dy}{dx} = e^{a\cos^{-1}x}\cdot a\cdot\frac{-1}{\sqrt{1-x^2}} = \frac{-ay}{\sqrt{1-x^2}}

Squaring: (1−x2)(dydx)2=a2y2(1-x^2)\left(\dfrac{dy}{dx}\right)^2 = a^2y^2 ... (i)

Differentiating (i) with respect to xx:
−2x(dydx)2+(1−x2)⋅2dydxd2ydx2=2a2ydydx-2x\left(\frac{dy}{dx}\right)^2 + (1-x^2)\cdot 2\frac{dy}{dx}\frac{d^2y}{dx^2} = 2a^2y\frac{dy}{dx}

Dividing by 2dydx2\dfrac{dy}{dx} (assuming dydx≠0\dfrac{dy}{dx} \neq 0):
−xdydx+(1−x2)d2ydx2=a2y-x\frac{dy}{dx} + (1-x^2)\frac{d^2y}{dx^2} = a^2y

(1−x2)d2ydx2−xdydx−a2y=0■(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0 \quad \blacksquare

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