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Limits and Derivatives

Jharkhand Board · Class 11 · Mathematics

NCERT Solutions for Limits and Derivatives — Jharkhand Board Class 11 Mathematics.

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73 Questions Solved · 3 Sections

Exercise 12.1

1lim⁡x→3(x+3)\lim_{x\to 3}(x + 3)Show solution
Given: lim⁡x→3(x+3)\lim_{x\to 3}(x + 3)

Concept: For a polynomial function, the limit is found by direct substitution.

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

Answer: 66
2lim⁡x→π(x−227)\lim_{x\to \pi}\left(x - \dfrac{22}{7}\right)Show solution
Given: lim⁡x→π(x−227)\lim_{x\to \pi}\left(x - \dfrac{22}{7}\right)

Concept: Direct substitution for a polynomial/linear function.

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

Answer: π−227\pi - \dfrac{22}{7}
3lim⁡r→1πr2\lim_{r\to 1}\pi r^2Show solution
Given: lim⁡r→1πr2\lim_{r\to 1}\pi r^2

Concept: Direct substitution.

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

Answer: π\pi
4lim⁡x→44x+3x−2\lim_{x\to 4}\dfrac{4x + 3}{x - 2}Show solution
Given: lim⁡x→44x+3x−2\lim_{x\to 4}\dfrac{4x + 3}{x - 2}

Concept: Direct substitution (denominator ≠0\neq 0 at x=4x = 4).

Working:
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}

Answer: 192\dfrac{19}{2}
5lim⁡x→−1x10+x5+1x−1\lim_{x\to -1}\dfrac{x^{10} + x^5 + 1}{x - 1}Show solution
Given: lim⁡x→−1x10+x5+1x−1\lim_{x\to -1}\dfrac{x^{10} + x^5 + 1}{x - 1}

Concept: Direct substitution (denominator ≠0\neq 0 at x=−1x = -1).

Working:
lim⁡x→−1x10+x5+1x−1=(−1)10+(−1)5+1−1−1=1−1+1−2=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} = \frac{1}{-2} = -\frac{1}{2}

Answer: −12-\dfrac{1}{2}
6lim⁡x→0(x+1)5−1x\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}Show solution
Given: lim⁡x→0(x+1)5−1x\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}

Concept: Use the standard limit lim⁡x→axn−anx−a=nan−1\lim_{x\to a}\dfrac{x^n - a^n}{x - a} = na^{n-1}.

Let y=x+1y = x + 1, so as x→0x \to 0, y→1y \to 1.

Working:
lim⁡x→0(x+1)5−1x=lim⁡y→1y5−15y−1=5⋅15−1=5×1=5\lim_{x\to 0}\frac{(x+1)^5 - 1}{x} = \lim_{y\to 1}\frac{y^5 - 1^5}{y - 1} = 5 \cdot 1^{5-1} = 5 \times 1 = 5

Answer: 55
7lim⁡x→23x2−x−10x2−4\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}Show solution
Given: lim⁡x→23x2−x−10x2−4\lim_{x\to 2}\dfrac{3x^2 - x - 10}{x^2 - 4}

At x=2x = 2: numerator =12−2−10=0= 12 - 2 - 10 = 0, denominator =0= 0. So we factorise.

Factorising the numerator:
3x2−x−10=(x−2)(3x+5)3x^2 - x - 10 = (x - 2)(3x + 5)

Factorising the denominator:
x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)

Working:
lim⁡x→2(x−2)(3x+5)(x−2)(x+2)=lim⁡x→23x+5x+2=3(2)+52+2=114\lim_{x\to 2}\frac{(x-2)(3x+5)}{(x-2)(x+2)} = \lim_{x\to 2}\frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{11}{4}

Answer: 114\dfrac{11}{4}
8lim⁡x→3x4−812x2−5x−3\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}Show solution
Given: lim⁡x→3x4−812x2−5x−3\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}

At x=3x = 3: numerator =81−81=0= 81 - 81 = 0, denominator =18−15−3=0= 18 - 15 - 3 = 0. Factorise.

Factorising numerator:
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)

Factorising denominator:
2x2−5x−3=(x−3)(2x+1)2x^2 - 5x - 3 = (x - 3)(2x + 1)

Working:
lim⁡x→3(x−3)(x+3)(x2+9)(x−3)(2x+1)=lim⁡x→3(x+3)(x2+9)2x+1=(6)(18)7=1087\lim_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} = \lim_{x\to 3}\frac{(x+3)(x^2+9)}{2x+1} = \frac{(6)(18)}{7} = \frac{108}{7}

Answer: 1087\dfrac{108}{7}
9lim⁡x→0ax+bcx+1\lim_{x\to 0}\dfrac{ax + b}{cx + 1}Show solution
Given: lim⁡x→0ax+bcx+1\lim_{x\to 0}\dfrac{ax + b}{cx + 1}

Concept: Direct substitution (denominator =1≠0= 1 \neq 0 at x=0x = 0).

Working:
lim⁡x→0ax+bcx+1=a(0)+bc(0)+1=b1=b\lim_{x\to 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{b}{1} = b

Answer: bb
10lim⁡z→1z3−1z6−1\lim_{z\to 1}\dfrac{z^3 - 1}{z^6 - 1}Show solution
Given: lim⁡z→1z3−1z6−1\lim_{z\to 1}\dfrac{z^3 - 1}{z^6 - 1}

At z=1z = 1: both numerator and denominator are 00. Factorise.

Working:
lim⁡z→1z3−1z6−1=lim⁡z→1z3−1(z3)2−12=lim⁡z→1z3−1(z3−1)(z3+1)\lim_{z\to 1}\frac{z^3 - 1}{z^6 - 1} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3)^2 - 1^2} = \lim_{z\to 1}\frac{z^3 - 1}{(z^3 - 1)(z^3 + 1)}
=lim⁡z→11z3+1=11+1=12= \lim_{z\to 1}\frac{1}{z^3 + 1} = \frac{1}{1 + 1} = \frac{1}{2}

Answer: 12\dfrac{1}{2}
11lim⁡x→1ax2+bx+ccx2+bx+a,a+b+c≠0\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a},\quad a + b + c \neq 0Show solution
Given: lim⁡x→1ax2+bx+ccx2+bx+a\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a}, where a+b+c≠0a + b + c \neq 0.

Concept: At x=1x = 1, numerator =a+b+c≠0= a + b + c \neq 0 and denominator =c+b+a=a+b+c≠0= c + b + a = a + b + c \neq 0. So direct substitution applies.

Working:
lim⁡x→1ax2+bx+ccx2+bx+a=a(1)+b(1)+cc(1)+b(1)+a=a+b+ca+b+c=1\lim_{x\to 1}\frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1) + b(1) + c}{c(1) + b(1) + a} = \frac{a + b + c}{a + b + c} = 1

Answer: 11
12lim⁡x→−21x+12x+2\lim_{x\to -2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}Show solution
Given: lim⁡x→−21x+12x+2\lim_{x\to -2}\dfrac{\dfrac{1}{x} + \dfrac{1}{2}}{x + 2}

At x=−2x = -2: numerator =−12+12=0= -\frac{1}{2} + \frac{1}{2} = 0, denominator =0= 0. Simplify.

Working:
1x+12x+2=2+x2xx+2=x+22x(x+2)=12x\frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{\frac{2 + x}{2x}}{x + 2} = \frac{x + 2}{2x(x + 2)} = \frac{1}{2x}

lim⁡x→−212x=12(−2)=−14\lim_{x\to -2}\frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}

Answer: −14-\dfrac{1}{4}
13lim⁡x→0sin⁡axbx\lim_{x\to 0}\dfrac{\sin ax}{bx}Show solution
Given: lim⁡x→0sin⁡axbx\lim_{x\to 0}\dfrac{\sin ax}{bx}

Concept: Use the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1.

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

Answer: ab\dfrac{a}{b}
14lim⁡x→0sin⁡axsin⁡bx,a,b≠0\lim_{x\to 0}\dfrac{\sin ax}{\sin bx},\quad a, b \neq 0Show solution
Given: lim⁡x→0sin⁡axsin⁡bx\lim_{x\to 0}\dfrac{\sin ax}{\sin bx}, a,b≠0a, b \neq 0.

Concept: Use lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1.

Working:
lim⁡x→0sin⁡axsin⁡bx=lim⁡x→0sin⁡axax⋅axsin⁡bxbx⋅bx=lim⁡x→0sin⁡axaxlim⁡x→0sin⁡bxbx⋅ab=11⋅ab=ab\lim_{x\to 0}\frac{\sin ax}{\sin bx} = \lim_{x\to 0}\frac{\dfrac{\sin ax}{ax} \cdot ax}{\dfrac{\sin bx}{bx} \cdot bx} = \frac{\lim_{x\to 0}\dfrac{\sin ax}{ax}}{\lim_{x\to 0}\dfrac{\sin bx}{bx}} \cdot \frac{a}{b} = \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}

Answer: ab\dfrac{a}{b}
15lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}Show solution
Given: lim⁡x→πsin⁡(π−x)π(π−x)\lim_{x\to \pi}\dfrac{\sin(\pi - x)}{\pi(\pi - x)}

Concept: Let y=π−xy = \pi - x. As x→πx \to \pi, y→0y \to 0.

Working:
lim⁡x→πsin⁡(π−x)π(π−x)=lim⁡y→0sin⁡yπy=1πlim⁡y→0sin⁡yy=1π⋅1=1π\lim_{x\to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)} = \lim_{y\to 0}\frac{\sin y}{\pi y} = \frac{1}{\pi}\lim_{y\to 0}\frac{\sin y}{y} = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}

Answer: 1π\dfrac{1}{\pi}
16lim⁡x→0cos⁡xπ−x\lim_{x\to 0}\dfrac{\cos x}{\pi - x}Show solution
Given: lim⁡x→0cos⁡xπ−x\lim_{x\to 0}\dfrac{\cos x}{\pi - x}

Concept: Direct substitution (denominator =π≠0= \pi \neq 0 at x=0x = 0).

Working:
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}

Answer: 1π\dfrac{1}{\pi}
17lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x\to 0}\dfrac{\cos 2x - 1}{\cos x - 1}Show solution
Given: lim⁡x→0cos⁡2x−1cos⁡x−1\lim_{x\to 0}\dfrac{\cos 2x - 1}{\cos x - 1}

Concept: Use 1−cos⁡θ=2sin⁡2θ21 - \cos\theta = 2\sin^2\dfrac{\theta}{2}.

Working:
cos⁡2x−1=−2sin⁡2x,cos⁡x−1=−2sin⁡2x2\cos 2x - 1 = -2\sin^2 x, \quad \cos x - 1 = -2\sin^2\frac{x}{2}

lim⁡x→0−2sin⁡2x−2sin⁡2x2=lim⁡x→0sin⁡2xsin⁡2x2\lim_{x\to 0}\frac{-2\sin^2 x}{-2\sin^2\frac{x}{2}} = \lim_{x\to 0}\frac{\sin^2 x}{\sin^2\frac{x}{2}}

Using sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}:
=lim⁡x→04sin⁡2x2cos⁡2x2sin⁡2x2=lim⁡x→04cos⁡2x2=4cos⁡20=4×1=4= \lim_{x\to 0}\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}} = \lim_{x\to 0}4\cos^2\frac{x}{2} = 4\cos^2 0 = 4 \times 1 = 4

Answer: 44
18lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x\to 0}\dfrac{ax + x\cos x}{b\sin x}Show solution
Given: lim⁡x→0ax+xcos⁡xbsin⁡x\lim_{x\to 0}\dfrac{ax + x\cos x}{b\sin x}

Working:
lim⁡x→0x(a+cos⁡x)bsin⁡x=lim⁡x→0xsin⁡x⋅a+cos⁡xb\lim_{x\to 0}\frac{x(a + \cos x)}{b\sin x} = \lim_{x\to 0}\frac{x}{\sin x} \cdot \frac{a + \cos x}{b}
=lim⁡x→01sin⁡xx⋅a+cos⁡xb=11⋅a+cos⁡0b=a+1b= \lim_{x\to 0}\frac{1}{\dfrac{\sin x}{x}} \cdot \frac{a + \cos x}{b} = \frac{1}{1} \cdot \frac{a + \cos 0}{b} = \frac{a + 1}{b}

Answer: a+1b\dfrac{a + 1}{b}
19lim⁡x→0xsec⁡x\lim_{x\to 0}x\sec xShow solution
Given: lim⁡x→0xsec⁡x\lim_{x\to 0}x\sec x

Concept: Direct substitution.

Working:
lim⁡x→0xsec⁡x=lim⁡x→0xcos⁡x=0cos⁡0=01=0\lim_{x\to 0}x\sec x = \lim_{x\to 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0

Answer: 00
20lim⁡x→0sin⁡ax+bxax+sin⁡bx,a,b,a+b≠0\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx},\quad a, b, a+b \neq 0Show solution
Given: lim⁡x→0sin⁡ax+bxax+sin⁡bx\lim_{x\to 0}\dfrac{\sin ax + bx}{ax + \sin bx}

Concept: Divide numerator and denominator by xx.

Working:
lim⁡x→0sin⁡axx+ba+sin⁡bxx=lim⁡x→0a⋅sin⁡axax+ba+b⋅sin⁡bxbx\lim_{x\to 0}\frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}} = \lim_{x\to 0}\frac{a\cdot\dfrac{\sin ax}{ax} + b}{a + b\cdot\dfrac{\sin bx}{bx}}
=a⋅1+ba+b⋅1=a+ba+b=1= \frac{a \cdot 1 + b}{a + b \cdot 1} = \frac{a + b}{a + b} = 1

Answer: 11
21lim⁡x→0(csc⁡x−cot⁡x)\lim_{x\to 0}(\csc x - \cot x)Show solution
Given: lim⁡x→0(csc⁡x−cot⁡x)\lim_{x\to 0}(\csc x - \cot x)

Working:
lim⁡x→0(1sin⁡x−cos⁡xsin⁡x)=lim⁡x→01−cos⁡xsin⁡x\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) = \lim_{x\to 0}\frac{1 - \cos x}{\sin x}

Multiply numerator and denominator by xx:
=lim⁡x→01−cos⁡xx⋅xsin⁡x=0⋅1=0= \lim_{x\to 0}\frac{1 - \cos x}{x} \cdot \frac{x}{\sin x} = 0 \cdot 1 = 0

(Using standard limits: lim⁡x→01−cos⁡xx=0\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0 and lim⁡x→0xsin⁡x=1\lim_{x\to 0}\dfrac{x}{\sin x} = 1.)

Answer: 00
22lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \dfrac{\pi}{2}}Show solution
Given: lim⁡x→π2tan⁡2xx−π2\lim_{x\to \frac{\pi}{2}}\dfrac{\tan 2x}{x - \dfrac{\pi}{2}}

Concept: Let y=x−π2y = x - \dfrac{\pi}{2}, so as x→π2x \to \dfrac{\pi}{2}, y→0y \to 0, and x=y+π2x = y + \dfrac{\pi}{2}.

Working:
tan⁡2x=tan⁡2(y+π2)=tan⁡(2y+π)=tan⁡2y\tan 2x = \tan 2\left(y + \frac{\pi}{2}\right) = \tan(2y + \pi) = \tan 2y

lim⁡y→0tan⁡2yy=lim⁡y→0sin⁡2yycos⁡2y=lim⁡y→0sin⁡2y2y⋅2cos⁡2y=1⋅21=2\lim_{y\to 0}\frac{\tan 2y}{y} = \lim_{y\to 0}\frac{\sin 2y}{y \cos 2y} = \lim_{y\to 0}\frac{\sin 2y}{2y} \cdot \frac{2}{\cos 2y} = 1 \cdot \frac{2}{1} = 2

Answer: 22
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,amp;x≤03(x+1),amp;xgt;0f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}Show solution
Given: f(x)={2x+3,amp;x≤03(x+1),amp;xgt;0f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}

Finding lim⁡x→0f(x)\lim_{x\to 0}f(x):

Left-hand limit (LHL):
lim⁡x→0−f(x)=lim⁡x→0−(2x+3)=2(0)+3=3\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(2x + 3) = 2(0) + 3 = 3

Right-hand limit (RHL):
lim⁡x→0+f(x)=lim⁡x→0+3(x+1)=3(0+1)=3\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}3(x + 1) = 3(0 + 1) = 3

Since LHL == RHL =3= 3:
lim⁡x→0f(x)=3\lim_{x\to 0}f(x) = 3

Finding lim⁡x→1f(x)\lim_{x\to 1}f(x):

For xx near 11 (both sides), x > 0, so f(x)=3(x+1)f(x) = 3(x+1).

LHL:
lim⁡x→1−f(x)=3(1+1)=6\lim_{x\to 1^-}f(x) = 3(1 + 1) = 6

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

Since LHL == RHL =6= 6:
lim⁡x→1f(x)=6\lim_{x\to 1}f(x) = 6
24Find lim⁡x→1f(x)\lim_{x\to 1}f(x), where f(x)={x2−1,amp;x≤1−x2−1,amp;xgt;1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}Show solution
Given: f(x)={x2−1,amp;x≤1−x2−1,amp;xgt;1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}

LHL:
lim⁡x→1−f(x)=lim⁡x→1−(x2−1)=1−1=0\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(x^2 - 1) = 1 - 1 = 0

RHL:
lim⁡x→1+f(x)=lim⁡x→1+(−x2−1)=−1−1=−2\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(-x^2 - 1) = -1 - 1 = -2

Since LHL ≠\neq RHL, lim⁡x→1f(x)\lim_{x\to 1}f(x) does not exist.
25Evaluate lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={∣x∣x,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{cases}Show solution
Given: f(x)={∣x∣x,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{cases}

For x > 0: ∣x∣=x|x| = x, so f(x)=xx=1f(x) = \dfrac{x}{x} = 1.

For x < 0: ∣x∣=−x|x| = -x, so f(x)=−xx=−1f(x) = \dfrac{-x}{x} = -1.

LHL:
lim⁡x→0−f(x)=−1\lim_{x\to 0^-}f(x) = -1

RHL:
lim⁡x→0+f(x)=1\lim_{x\to 0^+}f(x) = 1

Since LHL ≠\neq RHL, lim⁡x→0f(x)\lim_{x\to 0}f(x) does not exist.
26Find lim⁡x→0f(x)\lim_{x\to 0}f(x), where f(x)={x∣x∣,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{x}{|x|}, & x\neq 0\\ 0, & x = 0 \end{cases}Show solution
Given: f(x)={x∣x∣,amp;x≠00,amp;x=0f(x) = \begin{cases} \dfrac{x}{|x|}, & x\neq 0\\ 0, & x = 0 \end{cases}

For x > 0: f(x)=xx=1f(x) = \dfrac{x}{x} = 1.

For x < 0: f(x)=x−x=−1f(x) = \dfrac{x}{-x} = -1.

LHL:
lim⁡x→0−f(x)=−1\lim_{x\to 0^-}f(x) = -1

RHL:
lim⁡x→0+f(x)=1\lim_{x\to 0^+}f(x) = 1

Since LHL ≠\neq RHL, lim⁡x→0f(x)\lim_{x\to 0}f(x) does not exist.
27Find lim⁡x→5f(x)\lim_{x\to 5}f(x), where f(x)=∣x∣−5f(x) = |x| - 5Show solution
Given: f(x)=∣x∣−5f(x) = |x| - 5

For xx near 55 (both sides), x > 0, so ∣x∣=x|x| = x.

LHL:
lim⁡x→5−f(x)=lim⁡x→5−(x−5)=5−5=0\lim_{x\to 5^-}f(x) = \lim_{x\to 5^-}(x - 5) = 5 - 5 = 0

RHL:
lim⁡x→5+f(x)=lim⁡x→5+(x−5)=5−5=0\lim_{x\to 5^+}f(x) = \lim_{x\to 5^+}(x - 5) = 5 - 5 = 0

Since LHL == RHL =0= 0:
lim⁡x→5f(x)=0\lim_{x\to 5}f(x) = 0
28Suppose f(x)={a+bx,amp;xlt;14,amp;x=1b−ax,amp;xgt;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
Given: f(1)=4f(1) = 4 and lim⁡x→1f(x)=f(1)=4\lim_{x\to 1}f(x) = f(1) = 4.

For the limit to exist, LHL must equal RHL.

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

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

For the limit to exist:
a+b=b−a  ⟹  2a=0  ⟹  a=0a + b = b - a \implies 2a = 0 \implies a = 0

For the limit to equal f(1)=4f(1) = 4:
a+b=4  ⟹  0+b=4  ⟹  b=4a + b = 4 \implies 0 + b = 4 \implies b = 4

Answer: a=0a = 0 and b=4b = 4.
29Let a1,a2,…,ana_1, a_2, \ldots, a_n be fixed real numbers and define f(x)=(x−a1)(x−a2)⋯(x−an)f(x) = (x - a_1)(x - a_2)\cdots(x - a_n). What is lim⁡x→a1f(x)\lim_{x\to a_1}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
Given: f(x)=(x−a1)(x−a2)⋯(x−an)f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)

Finding lim⁡x→a1f(x)\lim_{x\to a_1}f(x):

Since f(x)f(x) is a polynomial, the limit equals the value at x=a1x = a_1:
lim⁡x→a1f(x)=(a1−a1)(a1−a2)⋯(a1−an)=0\lim_{x\to a_1}f(x) = (a_1 - a_1)(a_1 - a_2)\cdots(a_1 - a_n) = 0

Finding lim⁡x→af(x)\lim_{x\to a}f(x) for a≠a1,a2,…,ana \neq a_1, a_2, \ldots, a_n:

Again by direct substitution:
lim⁡x→af(x)=(a−a1)(a−a2)⋯(a−an)\lim_{x\to a}f(x) = (a - a_1)(a - a_2)\cdots(a - a_n)

This is a non-zero finite value since aa is different from all aia_i.
30If f(x)={∣x∣+1,amp;xlt;00,amp;x=0∣x∣−1,amp;xgt;0f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}. For what value(s) of aa does lim⁡x→af(x)\lim_{x\to a}f(x) exist?Show solution
Given: f(x)={∣x∣+1,amp;xlt;00,amp;x=0∣x∣−1,amp;xgt;0f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}

Case 1: a=0a = 0

LHL: lim⁡x→0−f(x)=lim⁡x→0−(∣x∣+1)=0+1=1\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(|x| + 1) = 0 + 1 = 1

RHL: lim⁡x→0+f(x)=lim⁡x→0+(∣x∣−1)=0−1=−1\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(|x| - 1) = 0 - 1 = -1

LHL ≠\neq RHL, so the limit does not exist at a=0a = 0.

Case 2: a < 0

For xx near a < 0, f(x)=∣x∣+1=−x+1f(x) = |x| + 1 = -x + 1 (since x < 0).

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

LHL == RHL, so the limit exists for all a < 0.

Case 3: a > 0

For xx near a > 0, f(x)=∣x∣−1=x−1f(x) = |x| - 1 = x - 1.

lim⁡x→af(x)=a−1\lim_{x\to a}f(x) = a - 1

LHL == RHL, so the limit exists for all a > 0.

Conclusion: lim⁡x→af(x)\lim_{x\to a}f(x) exists for all a≠0a \neq 0, i.e., for all a∈(−∞,0)∪(0,∞)a \in (-\infty, 0) \cup (0, \infty).
31If the function f(x)f(x) satisfies lim⁡x→1f(x)−2x2−1=π\lim_{x\to 1}\dfrac{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}\dfrac{f(x) - 2}{x^2 - 1} = \pi

Working:

As x→1x \to 1, the denominator x2−1→0x^2 - 1 \to 0. For the limit to be finite (equal to π\pi), the numerator must also →0\to 0.

Therefore:
lim⁡x→1[f(x)−2]=lim⁡x→1f(x)−2x2−1⋅lim⁡x→1(x2−1)=π×0=0\lim_{x\to 1}[f(x) - 2] = \lim_{x\to 1}\frac{f(x)-2}{x^2-1} \cdot \lim_{x\to 1}(x^2-1) = \pi \times 0 = 0

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

lim⁡x→1f(x)=2\lim_{x\to 1}f(x) = 2

Answer: lim⁡x→1f(x)=2\lim_{x\to 1}f(x) = 2
32If f(x)={mx2+n,amp;xlt;0nx+m,amp;0≤x≤1nx3+m,amp;xgt;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) and lim⁡x→1f(x)\lim_{x\to 1}f(x) exist?Show solution
Given: f(x)={mx2+n,amp;xlt;0nx+m,amp;0≤x≤1nx3+m,amp;xgt;1f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}

Condition for lim⁡x→0f(x)\lim_{x\to 0}f(x) to exist:

LHL: lim⁡x→0−f(x)=lim⁡x→0−(mx2+n)=n\lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(mx^2 + n) = n

RHL: lim⁡x→0+f(x)=lim⁡x→0+(nx+m)=m\lim_{x\to 0^+}f(x) = \lim_{x\to 0^+}(nx + m) = m

For limit to exist: LHL == RHL ⇒\Rightarrow n=mn = m.

Condition for lim⁡x→1f(x)\lim_{x\to 1}f(x) to exist:

LHL: lim⁡x→1−f(x)=lim⁡x→1−(nx+m)=n+m\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}(nx + m) = n + m

RHL: lim⁡x→1+f(x)=lim⁡x→1+(nx3+m)=n+m\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}(nx^3 + m) = n + m

LHL == RHL =n+m= n + m for all values of mm and nn. So lim⁡x→1f(x)\lim_{x\to 1}f(x) exists for all integers mm and nn.

Conclusion: Both limits exist when m=nm = n (where mm and nn are any equal integers).

Exercise 12.2

1Find the derivative of x2−2x^2 - 2 at x=10x = 10.Show solution
Given: f(x)=x2−2f(x) = x^2 - 2

Formula: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}

Working:
f′(x)=lim⁡h→0(x+h)2−2−(x2−2)h=lim⁡h→0x2+2xh+h2−x2hf'(x) = \lim_{h\to 0}\frac{(x+h)^2 - 2 - (x^2 - 2)}{h} = \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h}
=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2x= \lim_{h\to 0}\frac{2xh + h^2}{h} = \lim_{h\to 0}(2x + h) = 2x

At x=10x = 10:
f′(10)=2(10)=20f'(10) = 2(10) = 20

Answer: 2020
2Find the derivative of xx at x=1x = 1.Show solution
Given: f(x)=xf(x) = x

Working:
f′(x)=lim⁡h→0(x+h)−xh=lim⁡h→0hh=1f'(x) = \lim_{h\to 0}\frac{(x+h) - x}{h} = \lim_{h\to 0}\frac{h}{h} = 1

At x=1x = 1: f′(1)=1f'(1) = 1

Answer: 11
3Find the derivative of 99x99x at x=100x = 100.Show solution
Given: f(x)=99xf(x) = 99x

Working:
f′(x)=lim⁡h→099(x+h)−99xh=lim⁡h→099hh=99f'(x) = \lim_{h\to 0}\frac{99(x+h) - 99x}{h} = \lim_{h\to 0}\frac{99h}{h} = 99

At x=100x = 100: f′(100)=99f'(100) = 99

Answer: 9999
4Find the derivative of the following functions from first principle.
(i) x3−27x^3 - 27
(ii) (x−1)(x−2)(x-1)(x-2)
(iii) 1x2\dfrac{1}{x^2}
(iv) x+1x−1\dfrac{x+1}{x-1}Show solution
(i) f(x)=x3−27f(x) = x^3 - 27

f′(x)=lim⁡h→0(x+h)3−27−(x3−27)hf'(x) = \lim_{h\to 0}\frac{(x+h)^3 - 27 - (x^3 - 27)}{h}
=lim⁡h→0x3+3x2h+3xh2+h3−x3h= \lim_{h\to 0}\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}
=lim⁡h→0(3x2+3xh+h2)=3x2= \lim_{h\to 0}(3x^2 + 3xh + h^2) = 3x^2

Answer: f′(x)=3x2f'(x) = 3x^2

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(ii) f(x)=(x−1)(x−2)=x2−3x+2f(x) = (x-1)(x-2) = x^2 - 3x + 2

f′(x)=lim⁡h→0(x+h)2−3(x+h)+2−(x2−3x+2)hf'(x) = \lim_{h\to 0}\frac{(x+h)^2 - 3(x+h) + 2 - (x^2 - 3x + 2)}{h}
=lim⁡h→02xh+h2−3hh=lim⁡h→0(2x+h−3)=2x−3= \lim_{h\to 0}\frac{2xh + h^2 - 3h}{h} = \lim_{h\to 0}(2x + h - 3) = 2x - 3

Answer: f′(x)=2x−3f'(x) = 2x - 3

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(iii) f(x)=1x2f(x) = \dfrac{1}{x^2}

f′(x)=lim⁡h→01(x+h)2−1x2h=lim⁡h→0x2−(x+h)2h⋅x2(x+h)2f'(x) = \lim_{h\to 0}\frac{\dfrac{1}{(x+h)^2} - \dfrac{1}{x^2}}{h} = \lim_{h\to 0}\frac{x^2 - (x+h)^2}{h \cdot x^2(x+h)^2}
=lim⁡h→0−2xh−h2h⋅x2(x+h)2=lim⁡h→0−(2x+h)x2(x+h)2=−2xx4=−2x3= \lim_{h\to 0}\frac{-2xh - h^2}{h \cdot x^2(x+h)^2} = \lim_{h\to 0}\frac{-(2x + h)}{x^2(x+h)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}

Answer: f′(x)=−2x3f'(x) = -\dfrac{2}{x^3}

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(iv) f(x)=x+1x−1f(x) = \dfrac{x+1}{x-1}

f′(x)=lim⁡h→0x+h+1x+h−1−x+1x−1hf'(x) = \lim_{h\to 0}\frac{\dfrac{x+h+1}{x+h-1} - \dfrac{x+1}{x-1}}{h}
=lim⁡h→0(x+h+1)(x−1)−(x+1)(x+h−1)h(x+h−1)(x−1)= \lim_{h\to 0}\frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)}

Expanding numerator:
(x+h+1)(x−1)=x2−x+hx−h+x−1=x2+hx−h−1(x+h+1)(x-1) = x^2 - x + hx - h + x - 1 = x^2 + hx - h - 1
(x+1)(x+h−1)=x2+xh−x+x+h−1=x2+xh+h−1(x+1)(x+h-1) = x^2 + xh - x + x + h - 1 = x^2 + xh + h - 1

Numerator =(x2+hx−h−1)−(x2+xh+h−1)=−2h= (x^2 + hx - h - 1) - (x^2 + xh + h - 1) = -2h

f′(x)=lim⁡h→0−2hh(x+h−1)(x−1)=lim⁡h→0−2(x+h−1)(x−1)=−2(x−1)2f'(x) = \lim_{h\to 0}\frac{-2h}{h(x+h-1)(x-1)} = \lim_{h\to 0}\frac{-2}{(x+h-1)(x-1)} = \frac{-2}{(x-1)^2}

Answer: f′(x)=−2(x−1)2f'(x) = \dfrac{-2}{(x-1)^2}
5For the function f(x)=x100100+x9999+⋯+x22+x+1f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1. Prove that f′(1)=100f′(0)f'(1) = 100f'(0).Show solution
Given: f(x)=x100100+x9999+⋯+x22+x+1f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1

Finding f′(x)f'(x):

Using ddx(xn)=nxn−1\dfrac{d}{dx}(x^n) = nx^{n-1}:
f′(x)=100x99100+99x9899+⋯+2x2+1f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \cdots + \frac{2x}{2} + 1
f′(x)=x99+x98+⋯+x+1f'(x) = x^{99} + x^{98} + \cdots + x + 1

Finding f′(1)f'(1):
f′(1)=199+198+⋯+1+1=1+1+⋯+1⏟100 terms=100f'(1) = 1^{99} + 1^{98} + \cdots + 1 + 1 = \underbrace{1 + 1 + \cdots + 1}_{100 \text{ terms}} = 100

Finding f′(0)f'(0):
f′(0)=0+0+⋯+0+1=1f'(0) = 0 + 0 + \cdots + 0 + 1 = 1

Verification:
100⋅f′(0)=100×1=100=f′(1)100 \cdot f'(0) = 100 \times 1 = 100 = f'(1)

Hence, f′(1)=100f′(0)f'(1) = 100f'(0). ■\hspace{2cm}\blacksquare
6Find the derivative of xn+axn−1+a2xn−2+…+an−1x+anx^n + ax^{n-1} + a^2x^{n-2} + \ldots + a^{n-1}x + a^n for some fixed real number aa.Show solution
Given: f(x)=xn+axn−1+a2xn−2+⋯+an−1x+anf(x) = x^n + ax^{n-1} + a^2x^{n-2} + \cdots + a^{n-1}x + a^n

Using ddx(xk)=kxk−1\dfrac{d}{dx}(x^k) = kx^{k-1} and ddx(constant)=0\dfrac{d}{dx}(\text{constant}) = 0:

f′(x)=nxn−1+a(n−1)xn−2+a2(n−2)xn−3+⋯+an−1f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \cdots + a^{n-1}

f′(x)=nxn−1+(n−1)axn−2+(n−2)a2xn−3+⋯+an−1\boxed{f'(x) = nx^{n-1} + (n-1)ax^{n-2} + (n-2)a^2x^{n-3} + \cdots + a^{n-1}}
7For some constants aa and bb, find the derivative of:
(i) (x−a)(x−b)(x-a)(x-b)
(ii) (ax2+b)2(ax^2+b)^2
(iii) x−ax−b\dfrac{x-a}{x-b}Show solution
(i) f(x)=(x−a)(x−b)=x2−(a+b)x+abf(x) = (x-a)(x-b) = x^2 - (a+b)x + ab

f′(x)=2x−(a+b)f'(x) = 2x - (a+b)

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(ii) f(x)=(ax2+b)2=a2x4+2abx2+b2f(x) = (ax^2 + b)^2 = a^2x^4 + 2abx^2 + b^2

f′(x)=4a2x3+4abx=4ax(ax2+b)f'(x) = 4a^2x^3 + 4abx = 4ax(ax^2 + b)

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(iii) f(x)=x−ax−bf(x) = \dfrac{x-a}{x-b}

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

u=x−a⇒u′=1u = x - a \Rightarrow u' = 1; v=x−b⇒v′=1v = x - b \Rightarrow v' = 1

f′(x)=(1)(x−b)−(x−a)(1)(x−b)2=x−b−x+a(x−b)2=a−b(x−b)2f'(x) = \frac{(1)(x-b) - (x-a)(1)}{(x-b)^2} = \frac{x - b - x + a}{(x-b)^2} = \frac{a - b}{(x-b)^2}
8Find the derivative of xn−anx−a\dfrac{x^n - a^n}{x - a} for some constant aa.Show solution
Given: f(x)=xn−anx−af(x) = \dfrac{x^n - a^n}{x - a}

Using the quotient rule:

u=xn−an⇒u′=nxn−1u = x^n - a^n \Rightarrow u' = nx^{n-1}; v=x−a⇒v′=1v = x - a \Rightarrow v' = 1

f′(x)=nxn−1(x−a)−(xn−an)(1)(x−a)2f'(x) = \frac{nx^{n-1}(x-a) - (x^n - a^n)(1)}{(x-a)^2}

=nxn−naxn−1−xn+an(x−a)2= \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2}

=(n−1)xn−naxn−1+an(x−a)2= \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}
9Find the derivative of:
(i) 2x−342x - \dfrac{3}{4}
(ii) (5x3+3x−1)(x−1)(5x^3 + 3x - 1)(x-1)
(iii) x−3(5+3x)x^{-3}(5 + 3x)
(iv) x5(3−6x−9)x^5(3 - 6x^{-9})
(v) x−4(3−4x−5)x^{-4}(3 - 4x^{-5})
(vi) 2x+1−x23x−1\dfrac{2}{x+1} - \dfrac{x^2}{3x-1}Show solution
(i) f(x)=2x−34f(x) = 2x - \dfrac{3}{4}

f′(x)=2f'(x) = 2

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(ii) f(x)=(5x3+3x−1)(x−1)f(x) = (5x^3 + 3x - 1)(x - 1)

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

u=5x3+3x−1, u′=15x2+3u = 5x^3 + 3x - 1,\ u' = 15x^2 + 3
v=x−1, v′=1v = x - 1,\ v' = 1

f′(x)=(15x2+3)(x−1)+(5x3+3x−1)(1)f'(x) = (15x^2 + 3)(x-1) + (5x^3 + 3x - 1)(1)
=15x3−15x2+3x−3+5x3+3x−1= 15x^3 - 15x^2 + 3x - 3 + 5x^3 + 3x - 1
=20x3−15x2+6x−4= 20x^3 - 15x^2 + 6x - 4

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(iii) f(x)=x−3(5+3x)=5x−3+3x−2f(x) = x^{-3}(5 + 3x) = 5x^{-3} + 3x^{-2}

f′(x)=−15x−4−6x−3=−15x4−6x3f'(x) = -15x^{-4} - 6x^{-3} = -\frac{15}{x^4} - \frac{6}{x^3}

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(iv) f(x)=x5(3−6x−9)=3x5−6x−4f(x) = x^5(3 - 6x^{-9}) = 3x^5 - 6x^{-4}

f′(x)=15x4+24x−5=15x4+24x5f'(x) = 15x^4 + 24x^{-5} = 15x^4 + \frac{24}{x^5}

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(v) f(x)=x−4(3−4x−5)=3x−4−4x−9f(x) = x^{-4}(3 - 4x^{-5}) = 3x^{-4} - 4x^{-9}

f′(x)=−12x−5+36x−10=−12x5+36x10f'(x) = -12x^{-5} + 36x^{-10} = -\frac{12}{x^5} + \frac{36}{x^{10}}

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(vi) f(x)=2x+1−x23x−1f(x) = \dfrac{2}{x+1} - \dfrac{x^2}{3x-1}

Differentiate each term using quotient rule.

For 2x+1\dfrac{2}{x+1}:
ddx(2x+1)=0⋅(x+1)−2⋅1(x+1)2=−2(x+1)2\frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{0 \cdot (x+1) - 2 \cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2}

For x23x−1\dfrac{x^2}{3x-1}:
ddx(x23x−1)=2x(3x−1)−x2⋅3(3x−1)2=6x2−2x−3x2(3x−1)2=3x2−2x(3x−1)2\frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{2x(3x-1) - x^2 \cdot 3}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2}

f′(x)=−2(x+1)2−3x2−2x(3x−1)2f'(x) = \frac{-2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2}
10Find the derivative of cos⁡x\cos x from first principle.Show solution
Given: f(x)=cos⁡xf(x) = \cos x

Using first principle:
f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \lim_{h\to 0}\frac{\cos(x+h) - \cos x}{h}

Using cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

cos⁡(x+h)−cos⁡x=−2sin⁡(x+h2)sin⁡(h2)\cos(x+h) - \cos x = -2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)

f′(x)=lim⁡h→0−2sin⁡(x+h2)sin⁡(h2)hf'(x) = \lim_{h\to 0}\frac{-2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}

=lim⁡h→0[−sin⁡(x+h2)⋅sin⁡(h/2)h/2]= \lim_{h\to 0}\left[-\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2}\right]

=−sin⁡(x+0)⋅1=−sin⁡x= -\sin(x + 0) \cdot 1 = -\sin x

Answer: ddx(cos⁡x)=−sin⁡x\dfrac{d}{dx}(\cos x) = -\sin x
11Find the derivative of the following functions:
(i) sin⁡xcos⁡x\sin x\cos x
(ii) sec⁡x\sec x
(iii) 5sec⁡x+4cos⁡x5\sec x + 4\cos x
(iv) csc⁡x\csc x
(v) 3cot⁡x+5csc⁡x3\cot x + 5\csc x
(vi) 5sin⁡x−6cos⁡x+75\sin x - 6\cos x + 7
(vii) 2tan⁡x−7sec⁡x2\tan x - 7\sec xShow solution
(i) f(x)=sin⁡xcos⁡xf(x) = \sin x \cos x

Using product rule:
f′(x)=cos⁡x⋅cos⁡x+sin⁡x⋅(−sin⁡x)=cos⁡2x−sin⁡2x=cos⁡2xf'(x) = \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x = \cos 2x

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(ii) f(x)=sec⁡x=1cos⁡xf(x) = \sec x = \dfrac{1}{\cos x}

Using quotient rule:
f′(x)=0⋅cos⁡x−1⋅(−sin⁡x)cos⁡2x=sin⁡xcos⁡2x=sec⁡xtan⁡xf'(x) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x \tan x

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(iii) f(x)=5sec⁡x+4cos⁡xf(x) = 5\sec x + 4\cos x

f′(x)=5sec⁡xtan⁡x+4(−sin⁡x)=5sec⁡xtan⁡x−4sin⁡xf'(x) = 5\sec x\tan x + 4(-\sin x) = 5\sec x\tan x - 4\sin x

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(iv) f(x)=csc⁡x=1sin⁡xf(x) = \csc x = \dfrac{1}{\sin x}

Using quotient rule:
f′(x)=0⋅sin⁡x−1⋅cos⁡xsin⁡2x=−cos⁡xsin⁡2x=−csc⁡xcot⁡xf'(x) = \frac{0 \cdot \sin x - 1 \cdot \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = -\csc x \cot x

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(v) f(x)=3cot⁡x+5csc⁡xf(x) = 3\cot x + 5\csc x

Using ddx(cot⁡x)=−csc⁡2x\dfrac{d}{dx}(\cot x) = -\csc^2 x and ddx(csc⁡x)=−csc⁡xcot⁡x\dfrac{d}{dx}(\csc x) = -\csc x\cot x:
f′(x)=3(−csc⁡2x)+5(−csc⁡xcot⁡x)=−3csc⁡2x−5csc⁡xcot⁡xf'(x) = 3(-\csc^2 x) + 5(-\csc x\cot x) = -3\csc^2 x - 5\csc x\cot x

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(vi) f(x)=5sin⁡x−6cos⁡x+7f(x) = 5\sin x - 6\cos x + 7

f′(x)=5cos⁡x−6(−sin⁡x)+0=5cos⁡x+6sin⁡xf'(x) = 5\cos x - 6(-\sin x) + 0 = 5\cos x + 6\sin x

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(vii) f(x)=2tan⁡x−7sec⁡xf(x) = 2\tan x - 7\sec x

Using ddx(tan⁡x)=sec⁡2x\dfrac{d}{dx}(\tan x) = \sec^2 x and ddx(sec⁡x)=sec⁡xtan⁡x\dfrac{d}{dx}(\sec x) = \sec x\tan x:
f′(x)=2sec⁡2x−7sec⁡xtan⁡xf'(x) = 2\sec^2 x - 7\sec x\tan x

Miscellaneous Exercise on Chapter 12

1Find the derivative of the following functions from first principle:
(i) −x-x
(ii) (−x)−1(-x)^{-1}
(iii) sin⁡(x+1)\sin(x+1)
(iv) cos⁡(x−π8)\cos\left(x - \dfrac{\pi}{8}\right)Show solution
(i) f(x)=−xf(x) = -x

f′(x)=lim⁡h→0−(x+h)−(−x)h=lim⁡h→0−hh=−1f'(x) = \lim_{h\to 0}\frac{-(x+h)-(-x)}{h} = \lim_{h\to 0}\frac{-h}{h} = -1

Answer: f′(x)=−1f'(x) = -1

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(ii) f(x)=(−x)−1=−1xf(x) = (-x)^{-1} = -\dfrac{1}{x}

f′(x)=lim⁡h→0−1x+h−(−1x)h=lim⁡h→01x−1x+hhf'(x) = \lim_{h\to 0}\frac{-\dfrac{1}{x+h} - \left(-\dfrac{1}{x}\right)}{h} = \lim_{h\to 0}\frac{\dfrac{1}{x} - \dfrac{1}{x+h}}{h}
=lim⁡h→0hx(x+h)h=lim⁡h→01x(x+h)=1x2= \lim_{h\to 0}\frac{\dfrac{h}{x(x+h)}}{h} = \lim_{h\to 0}\frac{1}{x(x+h)} = \frac{1}{x^2}

Answer: f′(x)=1x2f'(x) = \dfrac{1}{x^2}

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(iii) f(x)=sin⁡(x+1)f(x) = \sin(x+1)

f′(x)=lim⁡h→0sin⁡(x+h+1)−sin⁡(x+1)hf'(x) = \lim_{h\to 0}\frac{\sin(x+h+1) - \sin(x+1)}{h}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→02cos⁡(x+1+h2)sin⁡(h2)h= \lim_{h\to 0}\frac{2\cos\left(x+1+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}
=lim⁡h→0cos⁡(x+1+h2)⋅sin⁡(h/2)h/2=cos⁡(x+1)⋅1=cos⁡(x+1)= \lim_{h\to 0}\cos\left(x+1+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2} = \cos(x+1) \cdot 1 = \cos(x+1)

Answer: f′(x)=cos⁡(x+1)f'(x) = \cos(x+1)

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(iv) f(x)=cos⁡(x−π8)f(x) = \cos\left(x - \dfrac{\pi}{8}\right)

f′(x)=lim⁡h→0cos⁡(x+h−π8)−cos⁡(x−π8)hf'(x) = \lim_{h\to 0}\frac{\cos\left(x+h-\dfrac{\pi}{8}\right) - \cos\left(x-\dfrac{\pi}{8}\right)}{h}

Using cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→0−2sin⁡(x−π8+h2)sin⁡(h2)h= \lim_{h\to 0}\frac{-2\sin\left(x - \dfrac{\pi}{8} + \dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}
=−sin⁡(x−π8)⋅1=−sin⁡(x−π8)= -\sin\left(x - \frac{\pi}{8}\right) \cdot 1 = -\sin\left(x - \frac{\pi}{8}\right)

Answer: f′(x)=−sin⁡(x−π8)f'(x) = -\sin\left(x - \dfrac{\pi}{8}\right)
2Find the derivative of (x+a)(x + a).Show solution
Given: f(x)=x+af(x) = x + a

f′(x)=ddx(x)+ddx(a)=1+0=1f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1

Answer: 11
3Find the derivative of (px+q)(rx+s)(px + q)\left(\dfrac{r}{x} + s\right).Show solution
Given: f(x)=(px+q)(rx+s)=pr+psx+qrx+qsf(x) = (px + q)\left(\dfrac{r}{x} + s\right) = pr + psx + \dfrac{qr}{x} + qs

f′(x)=ps−qrx2f'(x) = ps - \frac{qr}{x^2}

Alternatively using product rule:
u=px+q, u′=pu = px + q,\ u' = p; v=rx+s, v′=−rx2v = \dfrac{r}{x} + s,\ v' = -\dfrac{r}{x^2}

f′(x)=p(rx+s)+(px+q)(−rx2)=prx+ps−prx−qrx2=ps−qrx2f'(x) = p\left(\frac{r}{x} + s\right) + (px + q)\left(-\frac{r}{x^2}\right) = \frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2} = ps - \frac{qr}{x^2}

Answer: f′(x)=ps−qrx2f'(x) = ps - \dfrac{qr}{x^2}
4Find the derivative of (ax+b)(cx+d)2(ax + b)(cx + d)^2.Show solution
Given: f(x)=(ax+b)(cx+d)2f(x) = (ax + b)(cx + d)^2

Using product rule: u=ax+b, u′=au = ax + b,\ u' = a; v=(cx+d)2, v′=2c(cx+d)v = (cx+d)^2,\ v' = 2c(cx+d)

f′(x)=a(cx+d)2+(ax+b)⋅2c(cx+d)f'(x) = a(cx+d)^2 + (ax+b) \cdot 2c(cx+d)
=(cx+d)[a(cx+d)+2c(ax+b)]= (cx+d)\left[a(cx+d) + 2c(ax+b)\right]
=(cx+d)(acx+ad+2acx+2bc)= (cx+d)(acx + ad + 2acx + 2bc)
=(cx+d)(3acx+ad+2bc)= (cx+d)(3acx + ad + 2bc)

Answer: f′(x)=(cx+d)(3acx+ad+2bc)f'(x) = (cx+d)(3acx + ad + 2bc)
5Find the derivative of ax+bcx+d\dfrac{ax + b}{cx + d}.Show solution
Given: f(x)=ax+bcx+df(x) = \dfrac{ax+b}{cx+d}

Using quotient rule: u=ax+b, u′=au = ax+b,\ u' = a; v=cx+d, v′=cv = cx+d,\ v' = c

f′(x)=a(cx+d)−(ax+b)c(cx+d)2=acx+ad−acx−bc(cx+d)2=ad−bc(cx+d)2f'(x) = \frac{a(cx+d) - (ax+b)c}{(cx+d)^2} = \frac{acx + ad - acx - bc}{(cx+d)^2} = \frac{ad - bc}{(cx+d)^2}

Answer: f′(x)=ad−bc(cx+d)2f'(x) = \dfrac{ad - bc}{(cx+d)^2}
6Find the derivative of 1+1x1−1x\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}.Show solution
Given: f(x)=1+1x1−1x=x+1xx−1x=x+1x−1f(x) = \dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}} = \dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}} = \dfrac{x+1}{x-1}

Using quotient rule: u=x+1, u′=1u = x+1,\ u' = 1; v=x−1, v′=1v = x-1,\ v' = 1

f′(x)=1⋅(x−1)−(x+1)⋅1(x−1)2=x−1−x−1(x−1)2=−2(x−1)2f'(x) = \frac{1 \cdot (x-1) - (x+1) \cdot 1}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2}

Answer: f′(x)=−2(x−1)2f'(x) = \dfrac{-2}{(x-1)^2}
7Find the derivative of 1ax2+bx+c\dfrac{1}{ax^2 + bx + c}.Show solution
Given: f(x)=1ax2+bx+cf(x) = \dfrac{1}{ax^2 + bx + c}

Using quotient rule (or chain rule): u=1, u′=0u = 1,\ u' = 0; v=ax2+bx+c, v′=2ax+bv = ax^2+bx+c,\ v' = 2ax+b

f′(x)=0⋅(ax2+bx+c)−1⋅(2ax+b)(ax2+bx+c)2=−(2ax+b)(ax2+bx+c)2f'(x) = \frac{0 \cdot (ax^2+bx+c) - 1 \cdot (2ax+b)}{(ax^2+bx+c)^2} = \frac{-(2ax+b)}{(ax^2+bx+c)^2}

Answer: f′(x)=−(2ax+b)(ax2+bx+c)2f'(x) = \dfrac{-(2ax+b)}{(ax^2+bx+c)^2}
8Find the derivative of ax+bpx2+qx+r\dfrac{ax + b}{px^2 + qx + r}.Show solution
Given: f(x)=ax+bpx2+qx+rf(x) = \dfrac{ax+b}{px^2+qx+r}

u=ax+b, u′=au = ax+b,\ u' = a; v=px2+qx+r, v′=2px+qv = px^2+qx+r,\ v' = 2px+q

f′(x)=a(px2+qx+r)−(ax+b)(2px+q)(px2+qx+r)2f'(x) = \frac{a(px^2+qx+r) - (ax+b)(2px+q)}{(px^2+qx+r)^2}

Expanding numerator:
=apx2+aqx+ar−(2apx2+aqx+2bpx+bq)= apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)
=apx2+aqx+ar−2apx2−aqx−2bpx−bq= apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq
=−apx2+ar−2bpx−bq= -apx^2 + ar - 2bpx - bq
=−(apx2+2bpx+bq−ar)= -(apx^2 + 2bpx + bq - ar)

f′(x)=−(apx2+2bpx+bq−ar)(px2+qx+r)2f'(x) = \frac{-(apx^2 + 2bpx + bq - ar)}{(px^2+qx+r)^2}
9Find the derivative of px2+qx+rax+b\dfrac{px^2 + qx + r}{ax + b}.Show solution
Given: f(x)=px2+qx+rax+bf(x) = \dfrac{px^2+qx+r}{ax+b}

u=px2+qx+r, u′=2px+qu = px^2+qx+r,\ u' = 2px+q; v=ax+b, v′=av = ax+b,\ v' = a

f′(x)=(2px+q)(ax+b)−(px2+qx+r)⋅a(ax+b)2f'(x) = \frac{(2px+q)(ax+b) - (px^2+qx+r) \cdot a}{(ax+b)^2}

Expanding numerator:
=2apx2+2bpx+aqx+bq−apx2−aqx−ar= 2apx^2 + 2bpx + aqx + bq - apx^2 - aqx - ar
=apx2+2bpx+bq−ar= apx^2 + 2bpx + bq - ar

f′(x)=apx2+2bpx+bq−ar(ax+b)2f'(x) = \frac{apx^2 + 2bpx + bq - ar}{(ax+b)^2}
10Find the derivative of ax4−bx2+cos⁡x\dfrac{a}{x^4} - \dfrac{b}{x^2} + \cos x.Show solution
Given: f(x)=ax−4−bx−2+cos⁡xf(x) = ax^{-4} - bx^{-2} + \cos x

f′(x)=−4ax−5+2bx−3−sin⁡x=−4ax5+2bx3−sin⁡xf'(x) = -4ax^{-5} + 2bx^{-3} - \sin x = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x
11Find the derivative of 4x−24\sqrt{x} - 2.Show solution
Given: f(x)=4x1/2−2f(x) = 4x^{1/2} - 2

f′(x)=4⋅12x−1/2−0=2xf'(x) = 4 \cdot \frac{1}{2}x^{-1/2} - 0 = \frac{2}{\sqrt{x}}

Answer: f′(x)=2xf'(x) = \dfrac{2}{\sqrt{x}}
12Find the derivative of (ax+b)n(ax + b)^n.Show solution
Given: f(x)=(ax+b)nf(x) = (ax+b)^n

Using chain rule (or first principles with binomial theorem):

f′(x)=n(ax+b)n−1⋅a=na(ax+b)n−1f'(x) = n(ax+b)^{n-1} \cdot a = na(ax+b)^{n-1}

Answer: f′(x)=na(ax+b)n−1f'(x) = na(ax+b)^{n-1}
13Find the derivative of (ax+b)n(cx+d)m(ax + b)^n(cx + d)^m.Show solution
Given: f(x)=(ax+b)n(cx+d)mf(x) = (ax+b)^n(cx+d)^m

Using product rule: u=(ax+b)n, u′=na(ax+b)n−1u = (ax+b)^n,\ u' = na(ax+b)^{n-1}; v=(cx+d)m, v′=mc(cx+d)m−1v = (cx+d)^m,\ v' = mc(cx+d)^{m-1}

f′(x)=na(ax+b)n−1(cx+d)m+(ax+b)n⋅mc(cx+d)m−1f'(x) = na(ax+b)^{n-1}(cx+d)^m + (ax+b)^n \cdot mc(cx+d)^{m-1}
=(ax+b)n−1(cx+d)m−1[na(cx+d)+mc(ax+b)]= (ax+b)^{n-1}(cx+d)^{m-1}\left[na(cx+d) + mc(ax+b)\right]
14Find the derivative of sin⁡(x+a)\sin(x + a).Show solution
Given: f(x)=sin⁡(x+a)f(x) = \sin(x+a)

Using first principle:
f′(x)=lim⁡h→0sin⁡(x+h+a)−sin⁡(x+a)hf'(x) = \lim_{h\to 0}\frac{\sin(x+h+a) - \sin(x+a)}{h}

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}:

=lim⁡h→02cos⁡(x+a+h2)sin⁡(h2)h=cos⁡(x+a)⋅1=cos⁡(x+a)= \lim_{h\to 0}\frac{2\cos\left(x+a+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h} = \cos(x+a) \cdot 1 = \cos(x+a)

Answer: f′(x)=cos⁡(x+a)f'(x) = \cos(x+a)
15Find the derivative of csc⁡xcot⁡x\csc x \cot x.Show solution
Given: f(x)=csc⁡xcot⁡xf(x) = \csc x \cot x

Using product rule: u=csc⁡x, u′=−csc⁡xcot⁡xu = \csc x,\ u' = -\csc x\cot x; v=cot⁡x, v′=−csc⁡2xv = \cot x,\ v' = -\csc^2 x

f′(x)=(−csc⁡xcot⁡x)(cot⁡x)+csc⁡x(−csc⁡2x)f'(x) = (-\csc x\cot x)(\cot x) + \csc x(-\csc^2 x)
=−csc⁡xcot⁡2x−csc⁡3x= -\csc x\cot^2 x - \csc^3 x
=−csc⁡x(cot⁡2x+csc⁡2x)= -\csc x(\cot^2 x + \csc^2 x)
16Find the derivative of cos⁡x1+sin⁡x\dfrac{\cos x}{1 + \sin x}.Show solution
Given: f(x)=cos⁡x1+sin⁡xf(x) = \dfrac{\cos x}{1 + \sin x}

u=cos⁡x, u′=−sin⁡xu = \cos x,\ u' = -\sin x; v=1+sin⁡x, v′=cos⁡xv = 1 + \sin x,\ v' = \cos x

f′(x)=−sin⁡x(1+sin⁡x)−cos⁡x⋅cos⁡x(1+sin⁡x)2f'(x) = \frac{-\sin x(1+\sin x) - \cos x \cdot \cos x}{(1+\sin x)^2}
=−sin⁡x−sin⁡2x−cos⁡2x(1+sin⁡x)2= \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}
=−sin⁡x−(sin⁡2x+cos⁡2x)(1+sin⁡x)2= \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1+\sin x)^2}
=−sin⁡x−1(1+sin⁡x)2=−(1+sin⁡x)(1+sin⁡x)2=−11+sin⁡x= \frac{-\sin x - 1}{(1+\sin x)^2} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x}
17Find the derivative of sin⁡x+cos⁡xsin⁡x−cos⁡x\dfrac{\sin x + \cos x}{\sin x - \cos x}.Show solution
Given: f(x)=sin⁡x+cos⁡xsin⁡x−cos⁡xf(x) = \dfrac{\sin x + \cos x}{\sin x - \cos x}

u=sin⁡x+cos⁡x, u′=cos⁡x−sin⁡xu = \sin x + \cos x,\ u' = \cos x - \sin x
v=sin⁡x−cos⁡x, v′=cos⁡x+sin⁡xv = \sin x - \cos x,\ v' = \cos x + \sin x

f′(x)=(cos⁡x−sin⁡x)(sin⁡x−cos⁡x)−(sin⁡x+cos⁡x)(cos⁡x+sin⁡x)(sin⁡x−cos⁡x)2f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}

Numerator:
=−(cos⁡x−sin⁡x)2−(sin⁡x+cos⁡x)2= -(\cos x - \sin x)^2 - (\sin x + \cos x)^2
=−(cos⁡2x−2sin⁡xcos⁡x+sin⁡2x)−(sin⁡2x+2sin⁡xcos⁡x+cos⁡2x)= -(\cos^2 x - 2\sin x\cos x + \sin^2 x) - (\sin^2 x + 2\sin x\cos x + \cos^2 x)
=−(1−2sin⁡xcos⁡x)−(1+2sin⁡xcos⁡x)= -(1 - 2\sin x\cos x) - (1 + 2\sin x\cos x)
=−1+2sin⁡xcos⁡x−1−2sin⁡xcos⁡x=−2= -1 + 2\sin x\cos x - 1 - 2\sin x\cos x = -2

f′(x)=−2(sin⁡x−cos⁡x)2f'(x) = \frac{-2}{(\sin x - \cos x)^2}
18Find the derivative of sec⁡x−1sec⁡x+1\dfrac{\sec x - 1}{\sec x + 1}.Show solution
Given: f(x)=sec⁡x−1sec⁡x+1f(x) = \dfrac{\sec x - 1}{\sec x + 1}

Rewrite: f(x)=1cos⁡x−11cos⁡x+1=1−cos⁡x1+cos⁡xf(x) = \dfrac{\dfrac{1}{\cos x} - 1}{\dfrac{1}{\cos x} + 1} = \dfrac{1 - \cos x}{1 + \cos x}

u=1−cos⁡x, u′=sin⁡xu = 1 - \cos x,\ u' = \sin x; v=1+cos⁡x, v′=−sin⁡xv = 1 + \cos x,\ v' = -\sin x

f′(x)=sin⁡x(1+cos⁡x)−(1−cos⁡x)(−sin⁡x)(1+cos⁡x)2f'(x) = \frac{\sin x(1+\cos x) - (1-\cos x)(-\sin x)}{(1+\cos x)^2}
=sin⁡x+sin⁡xcos⁡x+sin⁡x−sin⁡xcos⁡x(1+cos⁡x)2= \frac{\sin x + \sin x\cos x + \sin x - \sin x\cos x}{(1+\cos x)^2}
=2sin⁡x(1+cos⁡x)2= \frac{2\sin x}{(1+\cos x)^2}
19Find the derivative of sin⁡nx\sin^n x.Show solution
Given: f(x)=sin⁡nxf(x) = \sin^n x

Using chain rule:
f′(x)=nsin⁡n−1x⋅cos⁡xf'(x) = n\sin^{n-1}x \cdot \cos x

Answer: f′(x)=nsin⁡n−1xcos⁡xf'(x) = n\sin^{n-1}x\cos x
20Find the derivative of a+bsin⁡xc+dcos⁡x\dfrac{a + b\sin x}{c + d\cos x}.Show solution
Given: f(x)=a+bsin⁡xc+dcos⁡xf(x) = \dfrac{a + b\sin x}{c + d\cos x}

u=a+bsin⁡x, u′=bcos⁡xu = a + b\sin x,\ u' = b\cos x; v=c+dcos⁡x, v′=−dsin⁡xv = c + d\cos x,\ v' = -d\sin x

f′(x)=bcos⁡x(c+dcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2f'(x) = \frac{b\cos x(c + d\cos x) - (a + b\sin x)(-d\sin x)}{(c + d\cos x)^2}
=bccos⁡x+bdcos⁡2x+adsin⁡x+bdsin⁡2x(c+dcos⁡x)2= \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c + d\cos x)^2}
=bccos⁡x+bd(cos⁡2x+sin⁡2x)+adsin⁡x(c+dcos⁡x)2= \frac{bc\cos x + bd(\cos^2 x + \sin^2 x) + ad\sin x}{(c + d\cos x)^2}
=bccos⁡x+bd+adsin⁡x(c+dcos⁡x)2= \frac{bc\cos x + bd + ad\sin x}{(c + d\cos x)^2}
21Find the derivative of sin⁡(x+a)cos⁡x\dfrac{\sin(x + a)}{\cos x}.Show solution
Given: f(x)=sin⁡(x+a)cos⁡xf(x) = \dfrac{\sin(x+a)}{\cos x}

u=sin⁡(x+a), u′=cos⁡(x+a)u = \sin(x+a),\ u' = \cos(x+a); v=cos⁡x, v′=−sin⁡xv = \cos x,\ v' = -\sin x

f′(x)=cos⁡(x+a)cos⁡x−sin⁡(x+a)(−sin⁡x)cos⁡2xf'(x) = \frac{\cos(x+a)\cos x - \sin(x+a)(-\sin x)}{\cos^2 x}
=cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡xcos⁡2x= \frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}
=cos⁡(x+a−x)cos⁡2x=cos⁡acos⁡2x=cos⁡asec⁡2x= \frac{\cos(x+a-x)}{\cos^2 x} = \frac{\cos a}{\cos^2 x} = \cos a \sec^2 x
22Find the derivative of x4(5sin⁡x−3cos⁡x)x^4(5\sin x - 3\cos x).Show solution
Given: f(x)=x4(5sin⁡x−3cos⁡x)f(x) = x^4(5\sin x - 3\cos x)

Using product rule: u=x4, u′=4x3u = x^4,\ u' = 4x^3; v=5sin⁡x−3cos⁡x, v′=5cos⁡x+3sin⁡xv = 5\sin x - 3\cos x,\ v' = 5\cos x + 3\sin x

f′(x)=4x3(5sin⁡x−3cos⁡x)+x4(5cos⁡x+3sin⁡x)f'(x) = 4x^3(5\sin x - 3\cos x) + x^4(5\cos x + 3\sin x)
=x3[4(5sin⁡x−3cos⁡x)+x(5cos⁡x+3sin⁡x)]= x^3[4(5\sin x - 3\cos x) + x(5\cos x + 3\sin x)]
=x3[20sin⁡x−12cos⁡x+5xcos⁡x+3xsin⁡x]= x^3[20\sin x - 12\cos x + 5x\cos x + 3x\sin x]
23Find the derivative of (x2+1)cos⁡x(x^2 + 1)\cos x.Show solution
Given: f(x)=(x2+1)cos⁡xf(x) = (x^2 + 1)\cos x

Using product rule: u=x2+1, u′=2xu = x^2+1,\ u' = 2x; v=cos⁡x, v′=−sin⁡xv = \cos x,\ v' = -\sin x

f′(x)=2xcos⁡x+(x2+1)(−sin⁡x)=2xcos⁡x−(x2+1)sin⁡xf'(x) = 2x\cos x + (x^2+1)(-\sin x) = 2x\cos x - (x^2+1)\sin x
24Find the derivative of (ax2+sin⁡x)(p+qcos⁡x)(ax^2 + \sin x)(p + q\cos x).Show solution
Given: f(x)=(ax2+sin⁡x)(p+qcos⁡x)f(x) = (ax^2 + \sin x)(p + q\cos x)

u=ax2+sin⁡x, u′=2ax+cos⁡xu = ax^2 + \sin x,\ u' = 2ax + \cos x; v=p+qcos⁡x, v′=−qsin⁡xv = p + q\cos x,\ v' = -q\sin x

f′(x)=(2ax+cos⁡x)(p+qcos⁡x)+(ax2+sin⁡x)(−qsin⁡x)f'(x) = (2ax + \cos x)(p + q\cos x) + (ax^2 + \sin x)(-q\sin x)
=(2ax+cos⁡x)(p+qcos⁡x)−qsin⁡x(ax2+sin⁡x)= (2ax + \cos x)(p + q\cos x) - q\sin x(ax^2 + \sin x)
25Find the derivative of (x+cos⁡x)(x−tan⁡x)(x + \cos x)(x - \tan x).Show solution
Given: f(x)=(x+cos⁡x)(x−tan⁡x)f(x) = (x + \cos x)(x - \tan x)

u=x+cos⁡x, u′=1−sin⁡xu = x + \cos x,\ u' = 1 - \sin x; v=x−tan⁡x, v′=1−sec⁡2x=−tan⁡2xv = x - \tan x,\ v' = 1 - \sec^2 x = -\tan^2 x

f′(x)=(1−sin⁡x)(x−tan⁡x)+(x+cos⁡x)(1−sec⁡2x)f'(x) = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)
=(1−sin⁡x)(x−tan⁡x)−(x+cos⁡x)tan⁡2x= (1 - \sin x)(x - \tan x) - (x + \cos x)\tan^2 x
26Find the derivative of 4x+5sin⁡x3x+7cos⁡x\dfrac{4x + 5\sin x}{3x + 7\cos x}.Show solution
Given: f(x)=4x+5sin⁡x3x+7cos⁡xf(x) = \dfrac{4x + 5\sin x}{3x + 7\cos x}

u=4x+5sin⁡x, u′=4+5cos⁡xu = 4x + 5\sin x,\ u' = 4 + 5\cos x; v=3x+7cos⁡x, v′=3−7sin⁡xv = 3x + 7\cos x,\ v' = 3 - 7\sin x

f′(x)=(4+5cos⁡x)(3x+7cos⁡x)−(4x+5sin⁡x)(3−7sin⁡x)(3x+7cos⁡x)2f'(x) = \frac{(4 + 5\cos x)(3x + 7\cos x) - (4x + 5\sin x)(3 - 7\sin x)}{(3x + 7\cos x)^2}

Expanding numerator:
=12x+28cos⁡x+15xcos⁡x+35cos⁡2x−(12x−28xsin⁡x+15sin⁡x−35sin⁡2x)= 12x + 28\cos x + 15x\cos x + 35\cos^2 x - (12x - 28x\sin x + 15\sin x - 35\sin^2 x)
=12x+28cos⁡x+15xcos⁡x+35cos⁡2x−12x+28xsin⁡x−15sin⁡x+35sin⁡2x= 12x + 28\cos x + 15x\cos x + 35\cos^2 x - 12x + 28x\sin x - 15\sin x + 35\sin^2 x
=28cos⁡x+15xcos⁡x+35(cos⁡2x+sin⁡2x)+28xsin⁡x−15sin⁡x= 28\cos x + 15x\cos x + 35(\cos^2 x + \sin^2 x) + 28x\sin x - 15\sin x
=28cos⁡x+15xcos⁡x+35+28xsin⁡x−15sin⁡x= 28\cos x + 15x\cos x + 35 + 28x\sin x - 15\sin x

f′(x)=35+28cos⁡x−15sin⁡x+x(15cos⁡x+28sin⁡x)(3x+7cos⁡x)2f'(x) = \frac{35 + 28\cos x - 15\sin x + x(15\cos x + 28\sin x)}{(3x + 7\cos x)^2}
27Find the derivative of x2cos⁡(π4)sin⁡x\dfrac{x^2\cos\left(\dfrac{\pi}{4}\right)}{\sin x}.Show solution
Given: f(x)=x2cos⁡(π4)sin⁡xf(x) = \dfrac{x^2\cos\left(\dfrac{\pi}{4}\right)}{\sin x}

Note: cos⁡(π4)=12\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}} is a constant. So f(x)=12⋅x2sin⁡xf(x) = \dfrac{1}{\sqrt{2}} \cdot \dfrac{x^2}{\sin x}.

u=x2, u′=2xu = x^2,\ u' = 2x; v=sin⁡x, v′=cos⁡xv = \sin x,\ v' = \cos x

f′(x)=12⋅2xsin⁡x−x2cos⁡xsin⁡2x=x2cos⁡(π4)(2sin⁡x−xcos⁡x)xsin⁡2xf'(x) = \frac{1}{\sqrt{2}} \cdot \frac{2x\sin x - x^2\cos x}{\sin^2 x} = \frac{x^2\cos\left(\dfrac{\pi}{4}\right)(2\sin x - x\cos x)}{x\sin^2 x}

f′(x)=cos⁡(π4)(2xsin⁡x−x2cos⁡x)sin⁡2x\boxed{f'(x) = \frac{\cos\left(\dfrac{\pi}{4}\right)(2x\sin x - x^2\cos x)}{\sin^2 x}}
28Find the derivative of x1+tan⁡x\dfrac{x}{1 + \tan x}.Show solution
Given: f(x)=x1+tan⁡xf(x) = \dfrac{x}{1 + \tan x}

u=x, u′=1u = x,\ u' = 1; v=1+tan⁡x, v′=sec⁡2xv = 1 + \tan x,\ v' = \sec^2 x

f′(x)=1⋅(1+tan⁡x)−x⋅sec⁡2x(1+tan⁡x)2=1+tan⁡x−xsec⁡2x(1+tan⁡x)2f'(x) = \frac{1 \cdot (1 + \tan x) - x \cdot \sec^2 x}{(1 + \tan x)^2} = \frac{1 + \tan x - x\sec^2 x}{(1 + \tan x)^2}
29Find the derivative of (x+sec⁡x)(x−tan⁡x)(x + \sec x)(x - \tan x).Show solution
Given: f(x)=(x+sec⁡x)(x−tan⁡x)f(x) = (x + \sec x)(x - \tan x)

u=x+sec⁡x, u′=1+sec⁡xtan⁡xu = x + \sec x,\ u' = 1 + \sec x\tan x; v=x−tan⁡x, v′=1−sec⁡2x=−tan⁡2xv = x - \tan x,\ v' = 1 - \sec^2 x = -\tan^2 x

f′(x)=(1+sec⁡xtan⁡x)(x−tan⁡x)+(x+sec⁡x)(1−sec⁡2x)f'(x) = (1 + \sec x\tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)
=(1+sec⁡xtan⁡x)(x−tan⁡x)−(x+sec⁡x)tan⁡2x= (1 + \sec x\tan x)(x - \tan x) - (x + \sec x)\tan^2 x
30Find the derivative of xsin⁡nx\dfrac{x}{\sin^n x}.Show solution
Given: f(x)=xsin⁡nxf(x) = \dfrac{x}{\sin^n x}

u=x, u′=1u = x,\ u' = 1; v=sin⁡nx, v′=nsin⁡n−1xcos⁡xv = \sin^n x,\ v' = n\sin^{n-1}x\cos x

f′(x)=1⋅sin⁡nx−x⋅nsin⁡n−1xcos⁡xsin⁡2nxf'(x) = \frac{1 \cdot \sin^n x - x \cdot n\sin^{n-1}x\cos x}{\sin^{2n} x}
=sin⁡n−1x(sin⁡x−nxcos⁡x)sin⁡2nx= \frac{\sin^{n-1}x(\sin x - nx\cos x)}{\sin^{2n} x}
=sin⁡x−nxcos⁡xsin⁡n+1x= \frac{\sin x - nx\cos x}{\sin^{n+1} x}

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