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Limits and Derivatives

CBSE · Class 11 · Mathematics

NCERT Solutions for Limits and Derivatives — CBSE Class 11 Mathematics.

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80 Questions Solved · 3 Sections

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EXERCISE 12.1

1lim⁡x→3x+3\lim_{x \to 3} x + 3Show solution
Substitute x=3x=3 since this is a polynomial expression:

lim⁡x→3(x+3)=3+3=6. \lim_{x\to 3}(x+3)=3+3=6.

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2lim⁡x→π(x−227)\lim_{x \to \pi} \left( x - \frac{22}{7} \right)Show solution
Since the expression is continuous, substitute x=πx=\pi:

lim⁡x→π(x−227)=π−227. \lim_{x\to \pi}\left(x-\frac{22}{7}\right)=\pi-\frac{22}{7}.

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3lim⁡r→1πr2\lim_{r \to 1} \pi r^2Show solution
Use direct substitution:

lim⁡r→1πr2=π(1)2=π. \lim_{r\to 1}\pi r^2=\pi(1)^2=\pi.

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4lim⁡x→44x+3x−2\lim_{x \to 4} \frac{4x + 3}{x - 2}Show solution
Substitute x=4x=4:

lim⁡x→44x+3x−2=4(4)+34−2=16+32=192. \lim_{x\to 4}\frac{4x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}.

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5lim⁡x→−1x10+x5+1x−1\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}Show solution
Substitute x=−1x=-1:

lim⁡x→−1x10+x5+1x−1=(−1)10+(−1)5+1−1−1=1−1+1−2=−12. \lim_{x\to -1}\frac{x^{10}+x^5+1}{x-1} =\frac{(-1)^{10}+(-1)^5+1}{-1-1} =\frac{1-1+1}{-2} = -\frac12.

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6lim⁡x→0(x+1)5−1x\lim_{x \to 0} \frac{(x + 1)^5 - 1}{x}Show solution
Use the standard limit for (x+1)5(x+1)^5 by expansion or direct simplification:

(x+1)5−1x=1+5x+⋯−1x. \frac{(x+1)^5-1}{x} =\frac{1+5x+\cdots-1}{x}.
The leading term is 5x5x, so the limit is

lim⁡x→0(x+1)5−1x=5. \lim_{x\to 0}\frac{(x+1)^5-1}{x}=5.

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7lim⁡x→23x2−x−10x2−4\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}Show solution
Factor the numerator and denominator:

3x2−x−10=(3x+5)(x−2),x2−4=(x−2)(x+2). 3x^2-x-10=(3x+5)(x-2), \qquad x^2-4=(x-2)(x+2).
So

lim⁡x→23x2−x−10x2−4=lim⁡x→2(3x+5)(x−2)(x−2)(x+2)=lim⁡x→23x+5x+2. \lim_{x\to 2}\frac{3x^2-x-10}{x^2-4} =\lim_{x\to 2}\frac{(3x+5)(x-2)}{(x-2)(x+2)} =\lim_{x\to 2}\frac{3x+5}{x+2}.
Now substitute x=2x=2:

3(2)+52+2=114. \frac{3(2)+5}{2+2}=\frac{11}{4}.

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8lim⁡x→3x4−812x2−5x−3\lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}Show solution
Factor both polynomials:

x4−81=(x2−9)(x2+9)=(x−3)(x+3)(x2+9), x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9),
2x2−5x−3=(2x+1)(x−3). 2x^2-5x-3=(2x+1)(x-3).
Cancel (x−3)(x-3):

lim⁡x→3x4−812x2−5x−3=lim⁡x→3(x+3)(x2+9)2x+1. \lim_{x\to 3}\frac{x^4-81}{2x^2-5x-3} =\lim_{x\to 3}\frac{(x+3)(x^2+9)}{2x+1}.
Substitute x=3x=3:

(6)(18)7=1087. \frac{(6)(18)}{7}=\frac{108}{7}.

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9lim⁡x→0ax+bcx+1\lim_{x \to 0} \frac{ax + b}{cx + 1}Show solution
At x=0x=0,

lim⁡x→0ax+bcx+1=a(0)+bc(0)+1=b. \lim_{x\to 0}\frac{ax+b}{cx+1}=\frac{a(0)+b}{c(0)+1}=b.

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10lim⁡z→1z13−1z16−1\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}Show solution
Let y=z1/6y=z^{1/6}. Then z1/3=y2z^{1/3}=y^2 and as z→1z\to 1, y→1y\to 1.

So
lim⁡z→1z1/3−1z1/6−1=lim⁡y→1y2−1y−1=lim⁡y→1(y+1)=2. \lim_{z\to 1}\frac{z^{1/3}-1}{z^{1/6}-1} =\lim_{y\to 1}\frac{y^2-1}{y-1} =\lim_{y\to 1}(y+1)=2.

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11lim⁡x→1ax2+bx+ccx2+bx+a,a+b+c≠0\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0Show solution
At x=1x=1,

ax2+bx+ccx2+bx+a=a+b+cc+b+a=1, \frac{ax^2+bx+c}{cx^2+bx+a} =\frac{a+b+c}{c+b+a}=1,
provided a+b+c≠0a+b+c\ne 0 so the expression is defined. Hence the limit is 11.

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12lim⁡x→−21x+12x+2\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}Show solution
Simplify first:

1x+12x+2=2+x2xx+2=2+x2x(x+2). \frac{\frac1x+\frac12}{x+2}= \frac{\frac{2+x}{2x}}{x+2}=\frac{2+x}{2x(x+2)}.
Now substitute x=−2x=-2 by cancelling the common factor (x+2)(x+2) from the original form:

1x+12=2+x2x. \frac{1}{x}+\frac12=\frac{2+x}{2x}.
So the expression becomes

2+x2xx+2=12x. \frac{\frac{2+x}{2x}}{x+2}=\frac{1}{2x}.
Thus
lim⁡x→−21x+12x+2=12(−2)=−14. \lim_{x\to -2}\frac{\frac1x+\frac12}{x+2}=\frac{1}{2(-2)}=-\frac14.

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13lim⁡x→0sin⁡axbx\lim_{x \to 0} \frac{\sin ax}{bx}Show solution
Use the standard limit lim⁡u→0sin⁡uu=1\lim_{u\to 0}\frac{\sin u}{u}=1:

lim⁡x→0sin⁡axbx=ablim⁡x→0sin⁡axax=ab⋅1=ab. \lim_{x\to 0}\frac{\sin ax}{bx} =\frac{a}{b}\lim_{x\to 0}\frac{\sin ax}{ax} =\frac{a}{b}\cdot 1 =\frac{a}{b}.

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14lim⁡x→0sin⁡axsin⁡bx,a,b≠0\lim_{x \to 0} \frac{\sin ax}{\sin bx}, a, b \neq 0Show solution
Write

sin⁡axsin⁡bx=sin⁡axax⋅bxsin⁡bx⋅ab. \frac{\sin ax}{\sin bx}=\frac{\sin ax}{ax}\cdot \frac{bx}{\sin bx}\cdot \frac{a}{b}.
As x→0x\to 0, each of the first two factors tends to 11. Therefore,

lim⁡x→0sin⁡axsin⁡bx=ab. \lim_{x\to 0}\frac{\sin ax}{\sin bx}=\frac{a}{b}.

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15lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)}Show solution
Since sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x and π(π−x)→π⋅0=0\pi(\pi-x)\to \pi\cdot 0=0, but we use the standard form:

lim⁡x→πsin⁡(π−x)π(π−x)=1πlim⁡x→πsin⁡(π−x)π−x. \lim_{x\to \pi}\frac{\sin(\pi-x)}{\pi(\pi-x)} =\frac{1}{\pi}\lim_{x\to \pi}\frac{\sin(\pi-x)}{\pi-x}.
Let u=π−xu=\pi-x. Then u→0u\to 0 and
1πlim⁡u→0sin⁡uu=1π⋅1=1π. \frac{1}{\pi}\lim_{u\to 0}\frac{\sin u}{u}=\frac{1}{\pi}\cdot 1=\frac{1}{\pi}.
So the limit is 1π\frac{1}{\pi}.

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16lim⁡x→0cos⁡xπ−x\lim_{x\to 0}\frac{\cos x}{\pi - x}Show solution
Direct substitution gives

lim⁡x→0cos⁡xπ−x=cos⁡0π−0=1π. \lim_{x\to 0}\frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}.

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17lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x\to 0}\frac{\cos 2x - 1}{\cos x - 1}Show solution
Use identities:

cos⁡2x−1=−2sin⁡2x,cos⁡x−1=−2sin⁡2x2. \cos 2x-1 = -2\sin^2 x, \qquad \cos x-1 = -2\sin^2\frac{x}{2}.
Then
cos⁡2x−1cos⁡x−1=−2sin⁡2x−2sin⁡2(x/2)=(sin⁡xsin⁡(x/2))2. \frac{\cos 2x-1}{\cos x-1}= \frac{-2\sin^2 x}{-2\sin^2(x/2)} =\left(\frac{\sin x}{\sin(x/2)}\right)^2.
Now
sin⁡x=2sin⁡x2cos⁡x2, \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2},
so
(sin⁡xsin⁡(x/2))2=(2cos⁡(x/2))2. \left(\frac{\sin x}{\sin(x/2)}\right)^2 = (2\cos(x/2))^2.
As x→0x\to 0, this tends to 44.

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18lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x\to 0}\frac{ax + x\cos x}{b\sin x}Show solution
Factor xx from the numerator:

ax+xcos⁡xbsin⁡x=x(a+cos⁡x)bsin⁡x. \frac{ax+x\cos x}{b\sin x}=\frac{x(a+\cos x)}{b\sin x}.
Now use lim⁡x→0xsin⁡x=1\lim_{x\to 0}\frac{x}{\sin x}=1 and cos⁡0=1\cos 0=1:

lim⁡x→0x(a+cos⁡x)bsin⁡x=a+1b. \lim_{x\to 0}\frac{x(a+\cos x)}{b\sin x}=\frac{a+1}{b}.

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19lim⁡x→0xsec⁡x\lim_{x\to 0}x\sec xShow solution
Since sec⁡x\sec x is continuous at 00 and sec⁡0=1\sec 0=1,

lim⁡x→0xsec⁡x=(lim⁡x→0x)(lim⁡x→0sec⁡x)=0⋅1=0. \lim_{x\to 0} x\sec x = \left(\lim_{x\to 0}x\right)\left(\lim_{x\to 0}\sec x\right)=0\cdot 1=0.

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20lim⁡x→0sin⁡ax+bxax+sin⁡bxa,b,a+b≠0,\lim_{x\to 0}\frac{\sin ax + bx}{ax + \sin bx} a,b,a + b\neq 0,Show solution
Use sin⁡kx∼kx\sin kx \sim kx as x→0x\to 0:

lim⁡x→0sin⁡ax+bxax+sin⁡bx=lim⁡x→0ax+bxax+bx=1, \lim_{x\to 0}\frac{\sin ax + bx}{ax + \sin bx} =\lim_{x\to 0}\frac{ax+bx}{ax+bx}=1,
provided a+b≠0a+b\neq 0 so the cancellation is valid in the limit.

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21lim⁡x→0(cosec⁡x−cot⁡x)\lim_{x\to 0}(\operatorname{cosec}x - \cot x)Show solution
Use the standard identity

csc⁡x−cot⁡x=1−cos⁡xsin⁡x. \csc x-\cot x=\frac{1-\cos x}{\sin x}.
Then

lim⁡x→0(csc⁡x−cot⁡x)=lim⁡x→01−cos⁡xsin⁡x. \lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}.
Now
1−cos⁡x→0andsin⁡x∼x, 1-\cos x\to 0 \quad\text{and}\quad \sin x\sim x,
so the limit is 00. More directly, using 1−cos⁡x=2sin⁡2(x/2)1-\cos x=2\sin^2(x/2),

1−cos⁡xsin⁡x=2sin⁡2(x/2)sin⁡x→0. \frac{1-\cos x}{\sin x}=\frac{2\sin^2(x/2)}{\sin x}\to 0.

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22lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\frac{\tan 2x}{x - \frac{\pi}{2}}Show solution
Let u=x−π2u=x-\frac{\pi}{2}, so u→0u\to 0. Then

tan⁡2x=tan⁡(π+2u)=tan⁡(2u), \tan 2x = \tan\left(\pi+2u\right)=\tan(2u),
since tangent has period π\pi. Therefore

lim⁡x→π2tan⁡2xx−π2=lim⁡u→0tan⁡(2u)u=2lim⁡u→0tan⁡(2u)2u=2⋅1=2. \lim_{x\to \frac{\pi}{2}}\frac{\tan 2x}{x-\frac{\pi}{2}} =\lim_{u\to 0}\frac{\tan(2u)}{u} =2\lim_{u\to 0}\frac{\tan(2u)}{2u}=2\cdot 1=2.
But because tan⁡(π+2u)=tan⁡(2u)\tan(\pi+2u)=\tan(2u), the computation gives 22, not −4-4.

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23Find lim⁡x→0f(x)\lim_{x\to 0}f(x) and lim⁡x→1f(x)\lim_{x\to 1}f(x), where f(x)={2x+3,x≤03(x+1),x>0f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}Show solution
For
f(x)={2x+3,x≤03(x+1),x>0 f(x)=\begin{cases}2x+3,&x\le 0\\3(x+1),&x>0\end{cases}
At x→0x\to 0:
- left side gives 2(0)+3=32(0)+3=3
- right side gives 3(0+1)=33(0+1)=3

So
lim⁡x→0f(x)=3. \lim_{x\to 0}f(x)=3.
At x→1x\to 1, since 1>01>0, use f(x)=3(x+1)f(x)=3(x+1):

lim⁡x→1f(x)=3(1+1)=6. \lim_{x\to 1}f(x)=3(1+1)=6.

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24Find lim⁡x→1f(x)\lim_{x\to 1}f(x), where f(x)={x2−1,x≤1−x2−1,x>1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}Show solution
At x=1x=1,
- from the left: x2−1→1−1=0x^2-1\to 1-1=0
- from the right: −x2−1→−1−1=−2-x^2-1\to -1-1=-2

The left and right limits are different, so the limit does not exist.

But the question asks for lim⁡x→1f(x)\lim_{x\to 1}f(x), and since the one-sided limits are unequal, the limit does not exist.

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25Evaluate lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={∣x∣x,x≠00,x=0f(x) = \begin{cases} \frac{|x|}{x}, & x\neq 0 \\ 0, & x = 0 \end{cases}Show solution
For x≠0x\ne 0,
f(x)=∣x∣x={1,x>0,−1,x<0. f(x)=\frac{|x|}{x}= \begin{cases} 1,&x>0,\\ -1,&x<0. \end{cases}
So
lim⁡x→0+f(x)=1,lim⁡x→0−f(x)=−1. \lim_{x\to 0^+}f(x)=1,\qquad \lim_{x\to 0^-}f(x)=-1.
Since the left and right limits are not equal, the limit at 00 does not exist.

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26Find lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={x∣x∣,x≠00,x=0f(x) = \begin{cases} \frac{x}{|x|}, & x\neq 0 \\ 0, & x = 0 \end{cases}Show solution
For x≠0x\neq 0, we have

- if x>0x>0, then ∣x∣=x|x|=x, so x∣x∣=1\dfrac{x}{|x|}=1
- if x<0x<0, then ∣x∣=−x|x|=-x, so x∣x∣=−1\dfrac{x}{|x|}=-1

Thus the left hand limit at 00 is −1-1 and the right hand limit is 11. Since they are not equal, the limit at x=0x=0 does not exist.

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27Find lim⁡x→5f(x)\lim_{x\to 5}f(x), where f(x)=∣x∣−5f(x) = |x| - 5Show solution
Since ∣x∣|x| is continuous, as x→5x\to 5 we get

lim⁡x→5(∣x∣−5)=∣5∣−5=5−5=0.\lim_{x\to 5} (|x|-5)=|5|-5=5-5=0.

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28Suppose f(x)={a+bx,x<14,x=1b−ax,x>1f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}

and if lim⁡x→1f(x)=f(1)\lim_{x\to 1}f(x) = f(1) what are possible values of aa and bb?Show solution
For the limit to equal f(1)=4f(1)=4, the left hand limit and right hand limit at x=1x=1 must both be 4.

For x<1x<1:
lim⁡x→1−f(x)=a+b(1)=a+b \lim_{x\to 1^-}f(x)=a+b(1)=a+b
So,
a+b=4 a+b=4

For x>1x>1:
lim⁡x→1+f(x)=b−a(1)=b−a \lim_{x\to 1^+}f(x)=b-a(1)=b-a
So,
b−a=4 b-a=4

Solving:
(a+b)+(b−a)=4+4⇒2b=8⇒b=4 (a+b)+(b-a)=4+4 \Rightarrow 2b=8 \Rightarrow b=4
Then
a+b=4⇒a+4=4⇒a=0 a+b=4 \Rightarrow a+4=4 \Rightarrow a=0

So the possible values are **a=0a=0 and b=4b=4**.

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29Let a1,a2,…,ana_1, a_2, \ldots, a_n be fixed real numbers and define a function

f(x)=(x−a1)(x−a2)…(x−an).f(x) = (x - a_1)(x - a_2)\ldots(x - a_n).

What is lim⁡x→aif(x)\lim_{x \to a_i} f(x)? For some a≠a1,a2,…,ana \neq a_1, a_2, \ldots, a_n, compute lim⁡x→af(x)\lim_{x \to a} f(x).Show solution
If x→aix\to a_i for any one of the fixed numbers aia_i, then one factor becomes 00:
lim⁡x→ai(x−a1)(x−a2)⋯(x−an)=0. \lim_{x\to a_i}(x-a_1)(x-a_2)\cdots(x-a_n)=0.

If a≠a1,a2,…,ana\neq a_1,a_2,\ldots,a_n, then none of the factors is zero at x=ax=a, so the product is a polynomial and its limit equals its value at aa:
lim⁡x→af(x)=f(a)=(a−a1)(a−a2)⋯(a−an). \lim_{x\to a}f(x)=f(a)=(a-a_1)(a-a_2)\cdots(a-a_n).

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30If f(x)={∣x∣+1,x<00,x=0∣x∣−1,x>0f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}.Show solution
For x<0x<0, ∣x∣=−x|x|=-x, so
f(x)=∣x∣+1=−x+1. f(x)=|x|+1=-x+1.
Hence as x→0−x\to 0^-,
f(x)→1. f(x)\to 1.
For x>0x>0, ∣x∣=x|x|=x, so
f(x)=∣x∣−1=x−1. f(x)=|x|-1=x-1.
Hence as x→0+x\to 0^+,
f(x)→−1. f(x)\to -1.
Since the left and right hand limits are different, the limit at x=0x=0 does not exist.

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31If the function f(x)f(x) satisfies lim⁡x→1f(x)−2x2−1=π\lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi, evaluate lim⁡x→1f(x)\lim_{x \to 1} f(x).Show solution
Given
lim⁡x→1f(x)−2x2−1=π. \lim_{x\to 1}\frac{f(x)-2}{x^2-1}=\pi.
As x→1x\to 1, we have x2−1→0x^2-1\to 0. For the quotient to have a finite limit, the numerator must also approach 00:
lim⁡x→1(f(x)−2)=0. \lim_{x\to 1}(f(x)-2)=0.
Therefore
lim⁡x→1f(x)=2. \lim_{x\to 1} f(x)=2.

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32If f(x)={mx2+n,x<0nx+m,0≤x≤1nx3+m,x>1f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}. For what integers mm and nn does both lim⁡x→0f(x)\lim_{x \to 0} f(x)Show solution
We need both limits to exist.

At x→0x\to 0:
- from x<0x<0, f(x)=mx2+n→nf(x)=mx^2+n \to n
- from 0≤x≤10\le x\le 1, f(x)=nx+m→mf(x)=nx+m \to m

So for lim⁡x→0f(x)\lim_{x\to 0}f(x) to exist,
n=m. n=m.

At x→1x\to 1:
- from 0≤x≤10\le x\le 1, f(x)=nx+m→n+mf(x)=nx+m \to n+m
- from x>1x>1, f(x)=nx3+m→n+mf(x)=nx^3+m \to n+m

These are automatically equal, so the limit at x=1x=1 exists for any integers once m=nm=n.

Hence the required condition is
m=n \boxed{m=n}
for integers m,nm,n.

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EXERCISE 12.2

1Find the derivative of x2−2x^2 - 2 at x=10x = 10.Show solution
Let f(x)=x2−2f(x)=x^2-2. Its derivative is
f′(x)=2x. f'(x)=2x.
At x=10x=10:
f′(10)=2⋅10=20. f'(10)=2\cdot 10=20.

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2Find the derivative of xx at x=1x = 1.Show solution
For f(x)=xf(x)=x, the derivative is
f′(x)=1. f'(x)=1.
So at x=1x=1 it is still 11.

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3Find the derivative of 99x99x at x=100x = 100.Show solution
For f(x)=99xf(x)=99x, the derivative is the constant
f′(x)=99. f'(x)=99.
So at x=100x=100, the derivative is 9999.

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4(i)Find the derivative of the following functions from first principle.Show solution
Using the first principle results from the chapter:

1. For f(x)=x3−27f(x)=x^3-27,
ddx(x3−27)=3x2. \frac{d}{dx}(x^3-27)=3x^2.

2. For f(x)=(x−1)(x−2)=x2−3x+2f(x)=(x-1)(x-2)=x^2-3x+2,
ddx(x2−3x+2)=2x−3. \frac{d}{dx}(x^2-3x+2)=2x-3.

3. For f(x)=1x2=x−2f(x)=\frac{1}{x^2}=x^{-2},
ddx(x−2)=−2x−3=−2x3. \frac{d}{dx}(x^{-2})=-2x^{-3}=-\frac{2}{x^3}.

4. For f(x)=x+1x−1f(x)=\frac{x+1}{x-1}, using the quotient rule,
ddx(x+1x−1)=(1)(x−1)−(x+1)(1)(x−1)2=x−1−x−1(x−1)2=−2(x−1)2. \frac{d}{dx}\left(\frac{x+1}{x-1}\right)=\frac{(1)(x-1)-(x+1)(1)}{(x-1)^2} =\frac{x-1-x-1}{(x-1)^2} =-\frac{2}{(x-1)^2}.

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4(ii)Find the derivative of the following functions from first principle.Show solution
For
f(x)=x100100+x9999+⋯+x22+x+1, f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1,
we differentiate term by term:
f′(x)=x99+x98+⋯+x+1. f'(x)=x^{99}+x^{98}+\cdots+x+1.
Now,
f′(1)=1+1+⋯+1=100 f'(1)=1+1+\cdots+1=100
and
f′(0)=0+0+⋯+0+1=1. f'(0)=0+0+\cdots+0+1=1.
Hence
f′(1)=100f′(0). f'(1)=100f'(0).

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4(iii)Find the derivative of the following functions from first principle.Show solution
Using standard differentiation rules:

(i)
ddx[(x−a)(x−b)]=(x−a)′(x−b)+(x−a)(x−b)′ \frac{d}{dx}[(x-a)(x-b)] = (x-a)'(x-b)+(x-a)(x-b)'
=1⋅(x−b)+(x−a)⋅1=2x−a−b. =1\cdot(x-b)+(x-a)\cdot 1=2x-a-b.

(ii)
ddx[(ax+b)(cx+d)2] \frac{d}{dx}[(ax+b)(cx+d)^2]
Use product rule:
=ax+b⇒(ax+b)′=a,(cx+d)2⇒2(cx+d)⋅c. =ax+b \Rightarrow (ax+b)'=a, \quad (cx+d)^2 \Rightarrow 2(cx+d)\cdot c.
So
=a(cx+d)2+(ax+b)⋅2c(cx+d) = a(cx+d)^2 + (ax+b)\cdot 2c(cx+d)
=(cx+d)(a(cx+d)+2c(ax+b)). =(cx+d)\big(a(cx+d)+2c(ax+b)\big).

(iii)
ddx(x−ax−b)=(1)(x−b)−(x−a)(1)(x−b)2=x−b−x+a(x−b)2=a−b(x−b)2. \frac{d}{dx}\left(\frac{x-a}{x-b}\right) =\frac{(1)(x-b)-(x-a)(1)}{(x-b)^2} =\frac{x-b-x+a}{(x-b)^2} =\frac{a-b}{(x-b)^2}.
So the derivative is
a−b(x−b)2. \frac{a-b}{(x-b)^2}.

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4(iv)Find the derivative of the following functions from first principle.Show solution
Let
f(x)=xn−anx−a. f(x)=\frac{x^n-a^n}{x-a}.
Using the quotient rule with numerator u=xn−anu=x^n-a^n and denominator v=x−av=x-a:
u′=nxn−1,v′=1. u'=nx^{n-1},\quad v'=1.
So
f′(x)=u′v−uv′v2=nxn−1(x−a)−(xn−an)(x−a)2. f'(x)=\frac{u'v-uv'}{v^2} =\frac{nx^{n-1}(x-a)-(x^n-a^n)}{(x-a)^2}.
Expanding the numerator:
=nxn−anxn−1−xn+an(x−a)2=(n−1)xn−naxn−1+an(x−a)2. =\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2} =\frac{(n-1)x^n-nax^{n-1}+a^n}{(x-a)^2}.

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5For the function

f(x)=x100100+x9999+⋯+x22+x+1.f (x) = \frac {x ^ {1 0 0}}{1 0 0} + \frac {x ^ {9 9}}{9 9} + \dots + \frac {x ^ {2}}{2} + x + 1.Show solution
For
f(x)=x100100+x9999+⋯+x22+x+1, f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1,
differentiate term by term:
f′(x)=x99+x98+⋯+x+1. f'(x)=x^{99}+x^{98}+\cdots+x+1.
This is the required derivative.

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6Find the derivative of xn+axn−1+a2xn−2+⋯+an−1x+anx^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n for some fixed real number aa.
7(i)For some constants aa and bb, find the derivative of
7(ii)For some constants aa and bb, find the derivative of
7(iii)For some constants aa and bb, find the derivative of
8Find the derivative of xn−anx−a\frac{x^n - a^n}{x-a} for some constant aa.
9(i)Find the derivative of
9(vi)Find the derivative of
10Find the derivative of cos⁡x\cos x from first principle.
11(ii)Find the derivative of the following functions:
11(v)Find the derivative of the following functions:
11(vii)Find the derivative of the following functions:

EXERCISE 12.1

2(x+a)(x + a)
3(px+q)(rx+s)(px + q)\left(\frac{r}{x} + s\right)
4(ax+b)(cx+d)2(ax + b)(cx + d)^2
5ax+bcx+d\frac{ax + b}{cx + d}
61+1x1−1x\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}
71ax2+bx+c\frac{1}{ax^2 + bx + c}
8ax+bpx2+qx+r\frac{ax + b}{px^2 + qx + r}
9px2+qx+rax+b\frac{px^2 + qx + r}{ax + b}
10ax4−bx2+cos⁡x\frac{a}{x^4} - \frac{b}{x^2} + \cos x
114x−24\sqrt{x} - 2
12(ax+b)n(ax + b)^n
13(ax+b)n(cx+d)m(ax + b)^n (cx + d)^m
14sin⁡(x+a)\sin(x + a)
15csc⁡xcot⁡x\csc x \cot x
16cos⁡x1+sin⁡x\frac{\cos x}{1 + \sin x}
17sin⁡x+cos⁡xsin⁡x−cos⁡x\frac{\sin x + \cos x}{\sin x - \cos x}
18sec⁡x−1sec⁡x+1\frac{\sec x - 1}{\sec x + 1}
19sin⁡nx\sin^n x
20a+bsin⁡xc+dcos⁡x\frac{a + b \sin x}{c + d \cos x}
21sin⁡(x+a)cos⁡x\frac{\sin(x + a)}{\cos x}
22x4(5sin⁡x−3cos⁡x)x^4 (5 \sin x - 3 \cos x)
23(x2+1)cos⁡x(x^2 + 1) \cos x
24(ax2+sin⁡x)(p+qcos⁡x)(ax^2 + \sin x)(p + q \cos x)
25(x+cos⁡x)(x−tan⁡x)(x + \cos x)(x - \tan x)
264x+5sin⁡x3x+7cos⁡x\frac{4x + 5 \sin x}{3x + 7 \cos x}
27x2cos⁡(π4)sin⁡x\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}
28x1+tan⁡x\frac{x}{1 + \tan x}
29(x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x)
30xsin⁡nx\frac{x}{\sin^n x}

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