# NCERT Solutions for Class 11 Maths Chapter 12 - Limits and Derivatives

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Exercise 13.1

Evaluate the following limits in Exercises 1 to 22

$$\textbf{1.}\space \lim\limits_{x \to 3}\space x + 3.$$

$$\textbf{Sol.}\space\lim\limits_{x \to 3}\space\text{x+3 = 3+3=6.}$$

$$\textbf{2.}\space\lim\limits_{x\to\pi}\bigg(x-\frac{22}{7}\bigg).\\\textbf{Sol.}\space\lim\limits_{x \to \pi}\bigg(x-\frac{22}{7}\bigg)=\pi-\frac{22}{7}.$$

$$\textbf{3.}\space\lim\limits_{r \to1}\space \pi r^2.\\\textbf{Sol.}\space\lim\limits_{x\to1}\space\pi r^2=\pi×(1)^2=\pi.$$

$$\textbf{4.}\space\lim\limits_{x\to4}\space\frac{4x+3}{x-2}.\\\textbf{Sol.}\space\lim\limits_{x \to 4}\space\frac{4x+3}{x-2}\\=\frac{4×4+3}{4-2}=\frac{19}{2}.$$

$$\textbf{5.}\space\lim\limits_{x\to-1}\space\frac{x^{10}+x^5+1}{x-1}.\\\textbf{Sol.}\space\lim\limits_{x \to -1}\frac{x^{10}+x^5+1}{x+1}\\=\frac{(-1)^{10}+(-1)^5+1}{-1-1}\\=\frac{1-1+1}{-1-1}=\frac{1}{-2}=\frac{-1}{2}.$$

$$\textbf{6.}\space\lim\limits_{x\to 0}\space\frac{(x+1)^5-1}{x}.\\\textbf{Sol.}\space\lim\limits_{x\to 0}\space\frac{(x+1)^5-1}{x}$$

Let x + 1 = y, x = y – 1

When x → 0, then y → 1

$$\therefore\space\lim\limits_{y \to 1}\frac{y^5-1}{y-1}=\lim\limits_{h\to 1}\frac{y^5-1^5}{y-1}\\=\lim\limits_{h\to 1}\space 5(y)^{5-1}$$

= 5 × 14 = 5.

$$\textbf{7.}\space\lim\limits_{x\to 2}\space\frac{3x^2-x-10}{x^2-4}.$$

$$\textbf{Sol.}\space\lim\limits_{x\to 2}\frac{3x^2-x-10}{x^2-4}\\=\lim\limits_{x\to 2}\frac{3x^2-6x+5x-10}{x^2-2^2}\\\lbrack\because\space a^2-b^2=(a-b)(a+b)\rbrack\\=\lim\limits_{ x\to 2}\space\frac{3x(x-2)+5(x-2)}{(x-2)(x+2)}\\=\lim\limits_{x\to2}\frac{(x-2)(3x+5)}{(x-2)(x+2)}\\=\frac{3×2+5}{2+2}=\frac{11}{2}.$$

$$\textbf{8.}\space\lim\limits_{x\to 3}\frac{x^4-81}{2x^2-5x-3}.\\\textbf{Sol.}\space\lim\limits_{x\to 3}\frac{x^4-81}{2x^2-5x-3}\\=\frac{(x^2)^2-9^2}{2x^2-(6-1)x-3}\\\bigg(\frac{0}{0}\text{form}\bigg)\\=\lim\limits_{x\to 3}\frac{(x^2-9)(x^2+9)}{2x^2-6x+x-3}\\(\because\space a^2-b^2=(a+b)(a-b))\\=\lim\limits_{x\to3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$$

$$=\lim\limits_{x \to 3}\space\frac{(x+3)(x^2+9)}{2x+1}\\=\frac{(3+3)(3^2+9)}{2×3+1}\\=\frac{6×18}{7}=\frac{108}{7}.$$

$$\textbf{9.}\space\lim\limits_{x \to 0}\space\frac{ax+b}{cx+1}.\\\textbf{Sol.}\space\lim\limits_{x\to 0}\frac{ax+b}{cx+1}\\=\frac{a×0+b}{c×0+1}=\frac{b}{1}=b.$$

$$\textbf{10.}\space\lim\limits_{x\to 1}\space\frac{z^{1/3}-1}{z^{1/6}-1}.\\\textbf{Sol.}\space\lim\limits_{z\to1}\space\frac{z^{1/3}-1}{z^{1/6}-1}\\=\lim\limits_{z\to1}\frac{z^{1/3}-1}{z-1}\div\frac{z^{1/6}-1}{z-1}\\=\lim\limits_{z\to 1}\bigg[\frac{z^{1/3}-1}{z-1}\bigg]\div\lim\limits_{z\to 1}\bigg[\frac{z^{1/6}-1}{z-1}\bigg]\\=\frac{1}{3}(1)^{1/3-1}\div\frac{1}{6}(1)^{1/6-1}\\=\frac{1}{3}\div\frac{1}{6}=\frac{6}{3}=2.$$

$$\textbf{11.}\space\lim\limits_{x\to1}\frac{ax^2+bx+c}{cx^2+bx+a},\text{a+b+c}\neq0.\\\textbf{Sol.}\space\lim\limits_{x\to1}\frac{ax^2+bx+c}{cx^2+bx+a}\\=\frac{a×(1)^2+b×1+c}{c×(1)^2+b×1+a}\\=\frac{a+b+c}{c+b+a}=1.$$

$$\textbf{12.}\space\lim\limits_{x\to -2}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}.\\\textbf{Sol.}\space\lim\limits_{x\to-2}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\\=\lim\limits_{x\to-2}\frac{\frac{2+x}{2x}}{(x+2)}\\\Rarr\space\lim\limits_{x\to -2}\frac{(2+x)}{2x(x+2)}=\lim\limits_{x\to-2}\frac{1}{2x}\\=\frac{1}{2(-2)}=-\frac{1}{4}.$$

$$\textbf{13.}\space\lim\limits_{x\to 0}\frac{\text{sin ax}}{bx}.\\\textbf{Sol.}\space \lim\limits_{x\to0}\frac{\text{sin}\space ax}{bx}\\=\lim\limits_{x\to0}\frac{(a)\text{sin\space ax}}{b(ax)}\\\text{(Dividing and multiplying by a)}\\=\frac{a}{b}\lim\limits_{x\to0}\frac{\text{sin ax}}{ax}\\=\frac{a}{b}×1=\frac{a}{b}\\\bigg(\because\space\lim\limits_{x\to0}\frac{\text{sin ax}}{\text{ax}}=1\bigg)$$

$$\textbf{14.}\space\lim\limits_{x\to0}\frac{\text{sin ax}}{\text{sin bx}}.\\\textbf{Sol.}\space\lim_{x\to 0}\frac{\text{sin ax}}{\text{sin bx}}$$

Multiplying and dividing by (ax) (bx), we get

$$=\lim\limits_{x\to 0}\bigg(\frac{sin \space ax}{ax}×\frac{bx}{sin \space bx}×\frac{ax}{bx}\bigg)\\=\lim\limits_{x\to0}\frac{sin \space ax}{ax}×\lim\limits_{x\to0}\frac{bx}{sin\space bx}×\lim\limits_{x\to0}\frac{ax}{bx}\\\bigg(\because\space\lim\limits_{x\to0}\frac{sin \space ax}{ax}=\lim\limits_{x\to0}\frac{bx}{sin\space bx}=1\bigg)\\=1×1×\frac{a}{b}=\frac{a}{b}$$

$$\textbf{15.\space}\lim\limits_{x\to\pi}\frac{\text{sin}(\pi-x)}{\pi(\pi-x)}.\\\textbf{Sol.}\space\lim\limits_{x\to\pi}\frac{\text{sin}(\pi-x)}{\pi(\pi-x)}$$

$$\text{Let} \pi\text{ – x = y, when x} → \pi, \text{then y = 0}\\\Rarr\space\lim_{y\to0}\frac{\text{sin y}}{\pi y}=\frac{1}{\pi}\space\lim\limits_{y\to0}\frac{\text{sin y}}{y}=\frac{1}{\pi}×1=\frac{1}{\pi}\\\bigg(\because\space \lim\limits _{x\to0}\space\frac{sin \space x}{x}\bigg)$$

$$\textbf{16.}\space\lim_{x\to 0}\frac{\text{cosx}}{\pi-x}.\\\textbf{Sol.}\space\lim\limits_{ x \to 0}\frac{cos x}{\pi-x}$$

$$\text{Let}\space\pi-x=y\space\text{when}\space x\to0.\space\\\text{Then y=}\pi\space\text{and\space}x=\pi-y\\=\lim\limits_{y\to\pi}\frac{\text{cos}(\pi-y)}{y}\\\lbrack\because\space(\pi-\theta)=-\text{cos}\space\theta\rbrack\\=\lim\limits_{y\to\pi}\space\frac{-\text{cos y}}{y}\\=\frac{-cos y}{y}=\frac{1}{\pi}\space(\because\space\text{cos}\pi=-1)$$

$$\textbf{17.}\space\frac{\text{cos 2x-1}}{\text{cos x-1}}.\\\textbf{Sol.}\space\lim\limits_{x\to 0}\frac{cos 2x-1}{cos x-1}\\=\lim\limits_{x\to0}\frac{1-2\space\text{sin}^2 x-1}{1-2\text{sin}^2\frac{x}{2}-1}\\(\because\space\text{cos 2x}=2 cos^2 x-1)\\=\lim\limits_{x\to0}\frac{-2\space\text{sin}^2x}{-2\text{sin}^2\frac{x}{2}}$$

$$=\lim\limits_{x\to0}\space\frac{\bigg(2 \text{sin}\frac{x}{2}\text{cos}\frac{x}{2}\bigg)^2}{\text{sin}^2\frac{x}{2}}\\\bigg(\because\space \text{sin x = 2 sin}\frac{x}{2}\text{cos}\frac{x}{2}\bigg)\\=\lim\limits_{x\to0}\frac{4 sin^2\frac{x}{2}\text{cos}^2\frac{x}{2}}{\text{sin}^2\frac{x}{2}}\\=\lim\limits_{x\to 0}\space 4\text{cos}^2\frac{x}{2}$$

= 4 × cos2 0

= 4 × 12

= 4

$$\textbf{18.}\space\lim\limits_{x\to0}\frac{ax+x cos x}{b\space sin\space x}.\\\textbf{Sol.}\lim\limits_{x\to0}\frac{ax+x cos x}{b sin x}\\=\lim_{x\to 0}\frac{x(a+ cos x)}{b sin x}\\\bigg(\because\space \lim_{x\to0}\space\frac{sin x}{x}=1\bigg)\\=\frac{a+ cos\space0}{b×1}=\frac{a+1}{b}$$

$$\textbf{19.}\space\lim\limits_{x\to0}\space\text{x sec x.}\\\textbf{Sol.}\space\lim_{x\to0}\space x sec x= 0×sec\space 0\\ =0×1=0.$$

$$\textbf{20.}\space\lim_{x\to0}\space\frac{\text{sin ax+bx}}{\text{ax+ sin bx}}\space,\text{a,b and \space a+b}\neq0.$$

$$\textbf{Sol.}\space\lim\limits_{x\to0}\space\frac{\text{sin ax + bx}}{\text{ax + sin bx}}\\\lim\limits_{x\to0}\frac{x\bigg(\frac{sin\space ax}{x}+b\bigg)}{x\bigg(a+\frac{sin\space bx}{x}\bigg)}\\=\lim\limits_{x\to0}\frac{\frac{a \space sin \space ax}{ax}+b}{a+\frac{b \space sinx }{bx}}\\=\frac{a×1+b}{a+b×1}\space\bigg(\because\space \lim_{x\to0}\frac{sin x}{x}=1\bigg)\\=\frac{a+b}{a+b}\\=1.$$

$$\textbf{21}\space\lim\limits_{x\to0}(\text{cosec x - cot x}).\\\textbf{Sol.}\space\lim\limits_{x\to0}(\text{cosec x- cot x})\\=\lim\limits_{x\to0}\bigg(\frac{1}{\text{sin x}}-\frac{\text{cos x}}{\text{sin x}}\bigg)\\=\lim\limits_{x\to0}\space\frac{\text{1- cos x}}{\text{sin x}}\\=\lim\limits_{x\to0}\frac{2\space\text{sin}^2\frac{x}{2}}{\text{2 sin}\frac{x}{2}\text{cos}\frac{x}{2}}\\=\lim\limits_{x\to0}\space\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}\\=\lim\limits_{x\to0}\space\text{tan}\frac{x}{2}=tan\space0=0.$$

$$\textbf{22.}\space \lim\limits_{x\to \frac{\pi}{2}}\frac{\text{tan 2x}}{x-\frac{\pi}{2}}.\\\textbf{Sol.}\space\text{Given},\space \lim\limits_{x\to\frac{x}{2}}\frac{\text{tan 2x}}{x-\frac{\pi}{2}}\\\text{Let}\space x-\frac{\pi}{2}=y\\\text{when}\space x\to \frac{\pi}{2},\text{Then y}\to 0$$

Therefore, given limit

$$=\lim\limits_{h\to0}\frac{\text{tan}\space2\bigg(\frac{\pi}{2}+y\bigg)}{y}\\=\lim\limits_{h\to0}\space\frac{\text{tan}(\pi+2y)}{y}\\=\lim\limits_{h\to0}\space\frac{\text{tan 2y}}{y}\\(\because\space \text{tan}(\pi+\theta)=\text{tan}\space\theta)\\=\lim\limits_{h\to 0}\frac{2\space\text{tan 2y}}{2y}\\=2×1=2\space\bigg(\because\space\lim\limits_{x\to0}\frac{tan\space x}{x}=1\bigg)$$

$$\textbf{23.}\space \text{Find}\space\lim\limits_{x\to 0}\text{f(x) and}\lim\limits_{x\to1}\text{f(x), where}\\\text{f(x)}=\begin{dcases}2x+3,\space x\leq0\\3(x+1),x\gt0\end{dcases}.\\\textbf{Sol.}\space\text{We have}\space f(x)=\begin{dcases}2x+3, x\leq0 \\3(x+1),x\gt0\end{dcases}\\\text{At x = 0,}\\\text{RHL}=\lim\limits_{x\to0^+}\space f(x)=\lim\limits_{h\to0}\space(0+h)\\=\lim\limits_{h\to0}\space 3(0+h+1)$$

= 3(0 + 0 + 1) = 3 × 1 = 3

$$\text{LHL}=\lim\limits_{h\to0^-}\space f(x)=\lim\limits_{h\to0}f(0-h)\\=\lim\limits_{h\to0}\space 2(0-h)+3$$

= 2(0 – 0) + 3 = 3

⇒ RHL = LHL = f(0)

Here, at x = 0 limit exists

$$\text{At x=1},\space\text{RHL}=\lim\limits_{x\to 1^+}\space f(x)=\lim\limits_{h\to0}\space f(1+h)\\=\lim\limits_{h\to0}\space 3(1+h+1)\\\text{(Putting x = 1 + h)}\\\text{=3(1+0+1)= 6}\\\text{LHL}=\lim\limits_{x\to1^-}\space f(x)=\lim\limits_{h\to0}f (1-h)\\\text{(Putting x = 1 – h)}\\=\lim\limits_{h\to0}\space 3(1-h+1)$$

and f(1) = 3(1 + 1) = 6

Therefore, RHL = LHL and hence limit exists.

$$\textbf{24.}\space\text{Find}\space\lim\limits_{x\to1}\space\text{f(x),}\text{where}\\f(x)=\begin{cases}x^2-1,x\leq1\\-x^2-1,x\gt1\end{cases}.\\\textbf{Sol.}\space\text{Given,}\space f(x)=\begin{cases}x^2-1, x\leq1\\-x^2-1,x\gt1\end{cases}\\\text{At x=1,}\\\text{LHL}=\lim\limits_{x\to1^-}\space\text{f(x)}=\lim\limits_{h\to0}\space f(1-h)\\=\lim\limits_{h\to0}(1-h^2)-1\space\text{(Put x=1-h)}$$

= (1 – 0)2 – 1 = 1 – 1 = 0

$$\text{RHL}=\lim\limits_{x\to1^+}\space f(x)=\lim\limits_{h\to0}\space f(1+h)\\=\lim\limits_{h\to0}=(1+h^2)-1\space\\\text{(Put x=1+h)}$$

= – (1 + 0)2 – 1 = – 1 – 1 = – 2

⇒ LHL ≠ RHL

Hence, at x = 1, limit doesn’t exist.

$$\textbf{25.}\space\text{Evaluate}\space\lim\limits_{x\to0}\space\text{f(x)},\text{where}\\\text{f(x)}=\begin{cases}\frac{|x|}{x},\space x\neq0\\0,\space x=0\end{cases}.\\\textbf{Sol.}\space\text{Given,}\space f(x)=\begin{cases}\frac{|x|}{x},x\neq0\\x_0,\space x=0\end{cases}\\\text{At\space x=0,}\\\text{LHL}=\lim\limits_{x\to0^-}\space f(x)=\lim\limits_{h\to0}\space f(0-h)\\=\lim\limits_{h\to0}\frac{|0-h|}{(0-h)}=\lim\limits_{h\to0}\space\frac{-(0-h)}{(0-h)}=-1\\\text{RHL}=\lim\limits_{x\to0^+}\space\text{f(x)}=\lim\limits_{h\to0}\space f(0+h)\\=\lim\limits_{h\to0}\space\frac{|0+h|}{(0+h)}=\lim\limits_{h\to0}\space\frac{(0+h)}{(0+h)}=1$$

⇒ LHL ≠ RHL

Hence, at x = 0, limit does not exist.

$$\textbf{26.}\space\lim\limits_{x\to0}\space\text{f(x),}\space\text{where f(x)}=\begin{cases}\frac{x}{|x|},\space x\neq0\\0,\space x=0\end{cases}.\\\textbf{Sol.}\space\text{Given,}\space f(x)=\begin{cases} \frac{x}{|x|},x\neq0\\0,\space x=0\end{cases}\\\text{At x = 0,}\\\text{LHL}=\lim\limits_{x\to0^-}\space f(x)=\lim\limits_{h\to0}\space f(0-h)\\=\lim\limits_{h\to0}\frac{(0-h)}{|0-h|}\\=\lim\limits_{h\to0}\space\frac{(0-h)}{-(0-h)}=-1\\\text{RHL}=\lim\limits_{x\to0^+}\space f(x)=\lim\limits_{h\to0}f(0+h)$$

$$=\lim\limits_{h\to0}\space\frac{0+h}{|0+h|}\\=\lim\limits_{h\to0}\space\frac{0+h}{(0+h)}=1$$

⇒ LHL ≠ RHL

Hence, at x = 0 limit does not exist.

$$\textbf{27.}\space\text{Find}\space\lim\limits_{x\to5}\space f(x),\space\text{where f(x)=|x|-5.}$$

Sol. At x = 5

$$\text{LHL}=\lim\limits_{x\to5^-}\space f(x)\\=\lim\limits_{h\to0}\space f(5-h)\\=\lim\limits_{h\to0}|5-h|-5\\=\lim\limits_{h\to0}\space(5-h)-5\\=\lim\limits_{h\to0}\space(-h)\\=0\\\text{RHL}=\lim\limits_{x\to5^+}\space\text{f(x)}\\=\lim\limits_{h\to0}\space f(5+h)\\=\lim\limits_{h\to0}\space(5+h)-5\\=\lim\limits_{h\to0}\space h=0$$

Hence LHL = RHL = 0

$$\therefore\space\lim\limits_{x\to5}\space\text{f(x)=0.}$$

$$\textbf{28.}\space\text{Suppose}\space f(x)=\begin{cases}a+bx,\space x\lt1\\ 4,\space x=1\\ b-ax, x\gt1\end{cases}\space\\\text{if}\space\lim\limits_{x\to1}\space f(x)=f(1)\space\text{what are possible }\\\text{values of a and b?}$$

$$\textbf{Sol.}\space\text{At x=1,}\\\space\text{LHL=}\lim\limits_{x\to1^-}\space\text{f(x)}\\=\lim_{x\to1^-}\space(a+bx)$$

= a + b

According to question

$$\text{LHL}=\lim\limits_{x\to1^-}\space\text{f(x)=f(1)}$$

a + b = 4 …(i)

$$\text{RHL}=\lim\limits_{x\to1^+}\space\text{f(x)}\\=\lim\limits_{x\to1^+}\space(b-ax)$$

= b – a

According to question

RHL = f(1)

b – a = 4

– a + b = 4 …(ii)

From equation (i) and (ii), we get

a = 0, b = 4.

29. Let a1, a2, a3, …, an be fixed real numbers and define a function

f(x) = (x – a1) (x – a2) … (x – an)

$$\text{What is}\space\lim\limits_{x\to a_1}\space\text{f(x)}?\\\text{For some a = a}_1, \text{a}_2, …, \text{a}_\text{n}\space\text{compute}\space\lim\limits_{x\to a}\space\text{f(x)}.$$

Sol. Given, f(x) = (x – a1) (x – a2) … (x – an)

$$\lim\limits_{x\to a_1}\space\text{f(x)}=\lim_{x\to a_1}\space(x-a_1)(x-a_2)...(x-a_n)$$

= (a1 – a1) (a1 – a2) … (a1 – an)

= 0 × (a1 – a2) … (a1 – an) = 0

$$\text{Again,}\space\lim\limits_{x\to a}\space f(x)\\=\lim\limits_{x\to a}(x-a_1)(x-a_2)...(x-a_n)$$

= (a – a1) (a – a2) … (a – an)

= (a – a1) (a – a2) … (a – an).

$$\textbf{30.}\space\text{If f(x)}=\begin{cases}|x|+1, x\lt0\\0,\space x=0\\ |x|-1, x\gt0\end{cases},\\\text{for what values(s) of a does}\\\lim_{x\to a}\space\text{f(x) exist?}$$

$$\textbf{Sol.}\space\text{f(x)}=\begin{cases}|x|+1, x\lt0\\0,\space x=0\\ |x|-1, x\gt0\end{cases}$$

At x = 0,

$$\text{LHL}=\lim\limits_{h\to0^-}\space\text{f(x)}=\lim\limits_{h\to0}\space f|0-h|\\=\lim\limits_{h\to0}\space|0-h|+1\\=\lim\limits_{h\to0}-(0-h)+1=\lim\limits_{h\to0}\space h+1$$

= 0 + 1 = 1

$$\text{RHL}=\lim\limits_{x\to0^+}\space f(x)=\lim\limits_{h\to0}\space f(0+h)\\=\lim\limits_{h\to0}|0+h|-1\\=\lim\limits_{h\to0}\space h-1=0-1=-1$$

⇒ LHL ≠ RHL

⇒ At x = 0, limit does not exist.

Hence exists for all a ≠ 0.

31. If the function f(x) satisfies

$$\lim\limits_{x\to1}\space\frac{f(x)-2}{x^2-1}=\pi,\text{evaluate}\space\lim\limits_{x\to1}\space f(x).$$

$$\textbf{Sol.}\space\text{Given,}\space\lim_{x\to1}\frac{f(x)-2}{x^2-1}=\pi\\\Rarr\space\frac{\lim\limits_{x\to1}[f(x)-2]}{\lim\limits_{x\to1}(x^2-1)}=\pi\\\Rarr\space\lim\limits_{x\to1}[f(x)-2]=\pi\space\lim\limits_{x\to1}(x^2-1)\\\Rarr\space\lim\limits_{x\to1}\space f(x)-2=\pi(1^2-1)\\\Rarr\space\lim\limits_{x\to1}\space\text{f(x)-2}=\pi×0\\\Rarr\space\lim\limits_{x\to1}\space f(x)-2=0\\\Rarr\space\lim\limits_{x\to1}\space\text{f(x)=2.}$$

$$\textbf{32.}\space\text{If f(x) =}\begin{cases}mx^2+n,\space x\lt0\\ nx+m,0\leq x\leq 1\\nx^3+m,\space x\gt1\end{cases}.\\\text{For what integers m and n does both}\\\lim\limits_{x\to0}\space f(x)\space\text{and}\lim\limits_{x\to1}\space\text{f(x) exist?}\\\textbf{Sol.}\space\because\space\text{At}\space x=0,\space\text{limit exists.}$$

Then,  LHL = RHL

$$\Rarr\space\lim\limits_{x\to0^-}\space\text{f(x)}=\lim\limits_{x\to0^+}\space\text{f(x)}\\\Rarr\space\lim\limits_{h\to0}\space f(0-h)=\lim\limits_{h\to0}\space f(0+h)\\\Rarr\space\lim_{h\to0}\space m(0-h)^2+n=\lim\limits_{h\to0}\space n(0+h)+m$$

⇒ m(0 – 0)2 + n = n(0 + 0) + m

⇒ n = m …(i)

Again, at x = 1 limit exsts.

⇒ LHL = RHL

$$\Rarr\space\lim_{x\to1^+}\space f(x)=\lim_{x\to1^-}\space f(x)$$

$$\Rarr\space\lim\limits_{h\to0}\space f(1+h)=\lim\limits_{h\to0}\space f(1-h)\\\Rarr\space\lim\limits_{h\to0}\space n(1+h)^3+m\\=\lim\limits_{h\to0}\space n(1-h)+m$$

⇒ n(1 + 0)2 + m = n(1 – 0) + m

⇒ n + m = m + n …(II)

$$\text{Hence, from equations (i) and (ii) for}\space\lim\limits_{x\to0}\space\text{f(x)}\\\text{to be existed, we need n = m and}\lim\limits_{x\to1}\space\text{f(x)}\\\text{exists for any integral value of m and n.}$$

Exercise 13.2

1. Find the derivative of x2 – 2 at x = 10.

Sol. Let f(x) = x2 – 2, we have

$$\text{f'(x)}=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{[(x+h)^2-2]-(x^2-2)}{h}\\=\lim\limits_{h\to0}\space\frac{x^2+h^2+2xh-2-x^2+2}{h}\\=\lim\limits_{h\to0}\space\frac{h^2+2xh}{h}\\=\lim\limits_{h\to0}\space\frac{h(h+2x)}{h}=\lim\limits_{h\to0}(h+2x)$$

= 2x

At x = 10, f′(10) = 2 × 10 = 20.

2. Find the derivative of 99x at x = 100.

Sol. Let f(x) = 99x

$$\text{We have,}\\\space f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\\\Rarr\space f'(x)=\lim_{h\to0}\frac{99(x+h)-99x}{h}\\=\lim\limits_{h\to0}\space\frac{99x+99h-99x}{h}\\=\lim\limits_{h\to0}\space\frac{99h}{h}=99$$

At x = 100, f ′(100) = 99.

3. Find the derivative of x at x = 1.

Sol. Let f(x) = x

$$\text{We have,}\\\space f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\Rarr\space f'(x)=\lim\limits_{x\to0}\frac{x+h-x}{h}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\frac{h}{h}=1$$

At  x = 1, f ′(1) = 1.

4. Find the derivative of the following functions from first principle:

(i) x3 – 27

(ii) (x – 1) (x – 2)

$$\text{(iii)}\space\frac{1}{x^3}\\\text{(iv)}\space\frac{x+1}{x-1}$$

Sol. (i) Let f(x) = x3 – 27

We have

$$\text{f'(x)}=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\Rarr\space\text{f'(x)}=\lim\limits_{h\to0}\space\frac{[(x+h)^3-27]-(x^3-27)}{h}\\=\lim\limits_{h\to0}\space\frac{(x+h)^3-27-x^3+27}{h}\\=\lim\limits_{h\to0}\space\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}$$

(a + b)3 = a3 + b3 + 3a2b + 3ab2

$$=\lim\limits_{h\to0}\space\frac{h^3+3x^2h+3xh^2}{h}\\=\lim\limits_{h\to0}\space\frac{h(h^2+3x^2-3xh)}{h}$$

= 0 + 3x2 – 0

= 3x2.

(ii) Let f(x) = (x – 1) (x – 2) = x2 – 2x – x + 2

⇒ f(x) = x2 – 3x + 2

We have,

$$f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}$$

[(x+h)2-3(x+h)+2]

$$\Rarr\space f'(x)=\lim\limits_{h\to0}\frac{-(x^2-3x+2)}{h}$$

( f(x) = x2 – 3x + 2)

$$=\lim\limits_{h\to0}\\\space\frac{x^2+xh^2+2xh-3x-3h+2-x^2+3x-2}{h}$$

$$=\lim\limits_{h\to0}\space\frac{h^2+2xh-3h}{h}\\=\lim\limits_{h\to0}\space\frac{h(h+2x-3)}{h}\\=\lim\limits_{h\to0}\space h+2x-3=0+2x-3=2x-3\\\therefore\space\text{f'(x)=2x-3.}$$

$$\text{(iii)\space}\text{f(x)}=\frac{1}{x^2}\\f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\=\lim\limits_{h\to0}\space\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\\=\lim\limits_{h\to0}\space\frac{x^2-(x+h)^2}{hx^2(x+h)^2}\\=\lim\limits_{h\to0}\space\frac{x^2-(x+h)^2}{hx^2(x+h)^2}\\\lim\limits_{x\to0}\space\frac{x^2-x^2-h^2-2xh}{hx^2(x+h)^2}\\=\lim\limits_{h\to0}\space\frac{-h(h+2x)}{hx^2(x+h)^2}$$

$$=\frac{-2x}{x^2×x^2}=\frac{-2}{x^3}\\=\lim\limits_{h\to0}|5+h|-5\\=\lim\limits_{h\to0}(5+h)-5\\\lim\limits_{h\to0}h$$

= 0

Then LHL = RHL

$$\text{Hence}\space\lim\limits_{x\to5}\space f(x)=0.$$

$$\text{(iv)\space Let}\space f(x)=\frac{x+1}{x-1}\\\text{we have,}\\\text{f'(x)}=\lim_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}\\\bigg(\because\space \text{f(x)}=\frac{x+1}{x-1}\bigg)\\=\lim\limits_{h\to0}\\\space\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x+h-1)(x-1)h}\\(x^2-x+hx-h+x-1)$$

$$=\lim\limits_{h\to0}\space\frac{-(x^2+hx-x+x+h-1)}{(x+h-1)(x-1)h}\\=\lim\limits_{h\to0}\space\frac{x^2+hx-x-1-x^2-hx-h+1}{(x+h-1)(x-1)h}\\=\lim\limits_{h\to0}\space\frac{-2h}{(x+h-1)(x-1)h}\\=\frac{-2}{(x+0-1)(x-1)}=\frac{2}{(x-1)^2}$$

5. For the function

$$\text{f(x)}=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1.$$

Prove that f ′(1) = 100f ′(0).

$$\textbf{Sol.}\space\text{Given},\\\space\text{f(x)}=\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1$$

$$\Rarr\space\text{f'(x)}=\frac{100x^{99}}{100}+\frac{99x^{98}}{99}+...+\frac{2x}{2}+1+0$$

(∵ f(x) = xn ⇒ f ′(x) = nxn – 1)

⇒ f′(x) = x99 + x98 + … + x + 1 …(i)

Putting x = 1, we get

$$\text{f'(1)}=\underbrace{(1)^{99}+1^{98}+...+1+1}_{100\text{ times}}\\\underbrace{1+1+1+...+1+1}_{100\text{ times}}$$

⇒ f ′(1) = 100 …(ii)

Again putting x = 0, we get

f ′(0) = 0 + 0 + … + 0 + 1

⇒ f ′(0) = 1 …(iii)

From equations (ii) and (ii),

f ′(1) = 100f ′(0).   Hence Proved.

6. Find the derivative of

xn + axn – 1 + a2xn – 2 + … + an – 1x + an

for some fixed real number a.

Sol. Let f(x) = xn + axn – 1 + a2xn – 2 + … + an – 1x + an

Differentiating w.r.t. x, we get

f ′(x) = nxn – 1 + a(n – 1)xn – 1 – 1 + a2(n – 2)xn – 2 – 1 + … + an – 1 (1) + 0

[∵ f(x) = axn ⇒ f(x) = anxn – 1)]

⇒ f ′(x) = nxn – 1 + a(n – 1)xn – 2 + a2(n – 2)xn – 3 + … + an – 1.

7. For some constants a and b, find the derivative of:

(i) (x – a) (x – b)

(ii) (ax2 + b)2

$$\text{(iii)\space}\frac{x-a}{x-b}$$

Sol. (i) Let f(x) = (x – a) (x – b)

$$\frac{d}{dx}f(x)=\frac{d}{dx}(x-a)(x-b)\\\text{f'(x)=}(x-a)\frac{d}{dx}(x-b)+(x-b)\frac{d}{dx}(x-a)$$

f ′(x) = (x – a) [1 – 0] + (x – b) (1 – 0)

f ′(x) = x – a + x – b

= 2x – (a + b).

(ii) Let f(x) = (ax2 + b)2

⇒ f(x) = a2x4 + b2 + 2abx2

Differentiating w.r.t. x

f ′(x) = 4a2x3 + 0 + 2ab(2x)

$$\bigg[\because\space\frac{d}{dx}x^n=nx^{n-1}\bigg]$$

= 4a2x3 + 4abx = 4ax(ax2 + b).

$$\text{(iii)\space \text{Let}\space f(x)}=\frac{x-a}{x-b}$$

Differentiating w.r.t. x,

$$\text{f'(x)}=\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{(x-b)^2}\\\bigg[\because\space\frac{d}{dx}\frac{u}{v}=\frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}\bigg]\\=\frac{(x-b)(1-0)-(x-a)(1-0)}{(x-b)^2}\\=\frac{(x-b)-(x-a)}{(x-b)^2}\\=\frac{x-b-x+a}{(x-b)^2}=\frac{a-b}{(x-b)^2}$$

$$\textbf{8.}\space\text{Find the derivative of}\space\frac{x^n-a^n}{x-a}\\\space\text{for some constant a.}$$

$$\textbf{Sol.}\space\text{Let}\space y=\frac{x^n-a^n}{x-a}$$

Differentiating y w.r.t. x,

$$\frac{dy}{dx}=\frac{(x-a)\frac{d}{dx}(x^n-a^n)-(x^n-a^n)\frac{d}{dx}(x-a)}{(x-a)^2}\\=\frac{(x-a)[nx^{n-1}-0]-[x^n-a^n](1-0)}{(x-a)^2}$$

$$=\frac{(x-a)nx^{n-1}-x^n+a^n}{(x-a)^2}\\=\frac{x×nx^{n-1}-anx^{n-1}-x^n+a^n}{(x-a)^2}\\=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$$

9. Find the derivative of:

$$\text{(i)\space} 2x-\frac{3}{4}$$

(ii) (5x3 + 3x – 1) (x – 1)

(iii) x– 3 (5 + 3x)

(iv) x5 (3 – 6x– 9)

(v) x– 4 (3 – 4x– 5)

$$\text{(vi)}\space \frac{2}{x+1}-\frac{x^2}{3x-1}\\\textbf{Sol.\space}\text{(i) Let}\space y=2x-\frac{3}{4}\\\frac{dy}{dx}=2\frac{d}{dx}x-\frac{d}{dx}\frac{3}{4}\\\frac{dy}{dx}=2×1-0=2.\\\bigg(\because\space\frac{d}{dx} x=1,\frac{d}{dx}\text{constant=0}\bigg)$$

(ii) Let y = (5x3 + 3x – 1) (x – 1)

y = (5x4 + 3x2 – x – 5x3 – 3x + 1)

y = 5x4 – 5x3 + 3x2 – 4x + 1

$$\frac{dy}{dx}=\frac{d}{dx}(5x^4-5x^3+3x^2-4x+1)\\\frac{dy}{dx}=20x^3-15x^2+6x-4.$$

(iii) Let y = x– 3 (5 + 3x)

y = 5x– 3 + 3x– 2

$$\frac{dy}{dx}=\frac{d}{dx}(5x^{-3}+3x^{-2})$$

= 5 × – 3x– 4 + 3 × – 2x– 3

$$=\frac{-15}{x^4}-\frac{6}{x^3}\\=\frac{-3}{x^4}(5+2x).$$

(iv) Let y = x5 (3 – 6x– 9)

$$\frac{dy}{dx}=x^5\frac{d}{dx}(3-6x^{-9})+(3-6x^{-9})\frac{x}{dx}x^5\\\bigg(\because\space\frac{du}{dx}v=u\frac{d}{dx}v+v\frac{d}{dx}u\bigg)$$

$$\frac{dy}{dx}=x^5(0-6×-9x^{-10})+(3-6x^{-9}).5x^4\\\frac{dy}{dx}=54x^{-5}+15x^4-30x^{-5}\\\frac{dy}{dx}=24x^{-5}+15x^4\\=15x^4+\frac{24}{x^5}.$$

(v) Let y = x– 4 (3 – 4x– 5)

⇒ y = 3x– 4 – 4x– 9

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=3(-4)x^{-4-1}-4(-9)x^{-9-1}\\= – 12x^{\normalsize– 5} + 36x^{\normalsize– 10}\\=\frac{-12}{x^5}+\frac{36}{x^{10}}.$$

$$\text{(vi)\space Let}\space y=\frac{2}{x+1}-\frac{x^2}{3x-1}\\\text{y} = 2(x + 1)^{\normalsize – 1} – x^2(3x – 1)^{– 1}\\\frac{dy}{dx}=\frac{d}{dx}2(x+1)^{\normalsize-1}-\frac{d}{dx}x^2(3x-1)^{\normalsize-1}\\\frac{dy}{dx}=-2(x+1)^{\normalsize-2}\\-\bigg[x^2\frac{d}{dx}(3x-1)^{\normalsize-1}+(3x-1)^{\normalsize-1}\frac{d}{dx}x^2\bigg]\\=\frac{-2}{(x+1)^2}\\-\bigg[x^2×-1[3x-1]^2×\frac{d}{dx}3x+(3x-1)^{-1}×2x\bigg]\\=\frac{-2}{(x+1)^2}-\bigg[\frac{-x^2}{(3x-1)^2}×3×1+\frac{2x}{(3x-1)}\bigg]$$

$$=\frac{-2}{(x+1)^2}-\bigg[\frac{-3x^2}{(3x-1)^2}+\frac{2x}{3x-1}\bigg]\\=\frac{-2}{(x+1)^2}-\bigg[\frac{-3x^2+2x(3x-1)}{(3x-1)^2}\bigg]\\=\frac{-2}{(x+1)^2}-\bigg[\frac{-3x^2+6x^2-2x}{(3x-1)^2}\bigg]\\=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}\\=\frac{-2}{(x+1)^2}-\frac{x(3x-2)}{(3x-1)}.$$

10. Find the derivative of cos x from first principle.

Sol. Let f(x) = cos x

$$\text{we have,}\space\text{f'(x)}=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\text{(By first principle)}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{\text{cos}(x+h)-\text{cos x}}{h}\\=\lim\limits_{h\to0}\space\frac{-2 \text{sin}\frac{x+h+x}{2}\text{sin}\frac{x+h-x}{2}}{h}\\\bigg(\because\space\text{cos C - cos D}=-2\text{sin}\frac{(\text{C+D})}{2}\text{sin}\frac{\text{(C-D)}}{2}\bigg)$$

$$=\lim\limits_{h\to0}\space\frac{-2 \text{sin}\frac{2x+h}{2}\text{sin}\frac{h}{2}}{2×\frac{h}{2}}\\=\lim\limits_{h\to0}\begin{Bmatrix}\text{- sin}\bigg(\frac{2x+h}{2}\bigg)\end{Bmatrix}\begin{Bmatrix}\frac{\text{sin}\frac{h}{2}}{\frac{h}{2}}\end{Bmatrix}\\=-\text{sin}\frac{2x}{2}×1\\\bigg(\because\space \lim\limits_{h\to0}\frac{\text{sin}\frac{h}{2}}{\frac{h}{2}}=1\bigg)$$

⇒ f ′(x) = – sin x.

11. Find the derivative of the following functions:

(i) sin x cos x

(ii) sec x

(iii) 5 sec x + 4 cos x

(iv) cosec x

(v) 3 cot x + 5 cosec x

(vi) 5 sin x – 6 cos x + 7

(vii) 2 tan x – 7 sec x

Sol. (i) Let y = sin x cos x

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\text{sin x}\frac{d}{dx}\text{cos x+cos x}\frac{d}{dx}\text{sin x}$$

(By product formula)

= sin x (– sin x) + cos x cos x

= – sin2 x + cos2 x

= cos2 x – sin2 x

= cos 2x.

(ii) Let y = sec x

$$\Rarr\space y=\frac{1}{\text{cos x}}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\frac{\text{cosx}\frac{d}{dx}(1)-1×\frac{d}{dx}\text{cos x}}{\text{cos}^2x}\\\bigg(\because\space\frac{d}{dx}\frac{u}{v}=\frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}\bigg)\\=\frac{\text{cos x×0 - 1×(-sin x)}}{\text{cos}^2x}\\=\frac{0+ \text{sin x}}{\text{cos}^2x}\\=\frac{\text{sin x}}{\text{cos x}}×\frac{1}{\text{cos x}}\\\frac{dy}{dx}=\text{tan x sec x = sec x tan x}$$

(iii) Let y = 5 sec x + 4 cos x

$$\frac{dy}{dx}=5\frac{d}{dx}\text{sec x}+4\frac{d}{dx}\text{cos x}\\\frac{dy}{dx}=\text{5 sec x tanx - 4 sin x}.$$

$$\text{(iv) Let}\space\text{y = cosec x}=\frac{1}{sin x}$$

Differentiating y w.r.t.x, we get

$$\frac{dy}{dx}=\frac{\text{sin x}\frac{d}{dx}(1)-(1)\frac{d}{dx}(sin x)}{\text{sin}^2x}\\=\frac{\text{sin x×0 - 1×\text{cos x}}}{\text{sin}^2x}\\\bigg[\because\space\frac{d}{dx}\frac{v}{u}=\frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}\bigg]\\=\frac{-cos x}{sin x}×\frac{1}{sin x}\\\Rarr\space\frac{dy}{dx}=-\text{cot x cosec x}$$

= – cosec x cot x.

(v) Let y = 3 cot x + 5 cosec x

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=-3\space\text{cosec}^2x-5\text{cosec x cot x.}$$

(vi) Let y = 5 sin x – 6 cos x + 7

$$\frac{dy}{dx}=5\frac{d}{dx}\text{sin x}-6\frac{d}{dx}\text{cos x}+\frac{d}{dx}7\\=\frac{dy}{dx}=\text{5 cos x – 6(– sin x) + 0}$$

= 5 cos x + 6 sin x.

(vii) Let y = 2 tan x – 7 sec x

$$\frac{d}{dx}=2\frac{d}{dx}\text{tan x}-7\frac{d}{dx}\text{sec x}\\\frac{dy}{dx}=2\text{sec}^2x-\text{7 sec x tan x}.$$

Miscellaneous Exercise

1. Find the derivative of the following functions from the first principle:

(i) – x

(ii) (– x)– 1

(iii) sin (x + 1)

$$\text{(iv)}\space\text{cos}\bigg(\text{x}-\frac{\pi}{\text{8}}\bigg)$$

Sol. (i) Let f(x) = – x

We have,

$$\text{f'(x)=}\lim\limits_{h\to0}\frac{f(x-h)-f(x)}{h}\\=\lim\limits_{h\to0}\space\frac{-(x+h)-(-x)}{h}\\\text{(By first principle)}\\=\lim\limits_{h\to0}\space\frac{-(x+h-x)}{h}\space[\because\space\text{f(x)=-x}]\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{-h}{h}=-1$$

(ii) Let f(x) = (– x)– 1

$$\Rarr\space f(x)=-\frac{1}{x}$$

We have,

$$f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\text{(By first principle)}\\\Rarr\space\text{f'(x)}=\lim\limits_{h\to0}\frac{-\frac{1}{x+h}+\frac{1}{x}}{h}\\\bigg(\because\space\text{f(x)=}\frac{1}{x}\bigg)\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{-x+x+h}{x(x+h)h}\\=\lim\limits_{h\to0}\space\frac{h}{x(x+h)h}=\frac{1}{x(x+0)}=\frac{1}{x^2}$$

(iii) f(x) = sin (x + 1)

We have,

$$\text{f'(x)}=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\text{(By first principle)}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{\text{sin}(x+h+1)-\text{sin}(x+1)}{h}\\\lbrack\because\text{f(x)= sin(x+1)}\rbrack\\\text{2 cos}\space\frac{x+h+1+x+1}{2}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{\text{sin}\frac{x+h+1-x-1}{2}}{h}$$

$$\begin{bmatrix}\because\space\text{sin C- sin D}=\\ \text{2 cos}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\end{bmatrix}\\=\lim\limits_{h\to0}\space\frac{2\space\text{cos}\frac{2x+2+h}{2}\text{sin}\frac{h}{2}}{2×\frac{h}{2}}\\=\text{cos}\space\frac{2x+2}{1}×1\\\bigg(\because\space\lim\limits_{h\to0}\frac{\text{sin}\frac{h}{2}}{\frac{h}{2}}=1\bigg)$$

⇒ f ′(x) = cos (x + 1).

$$\text{(iv)\space Let}\space\text{f(x)= cos}\bigg(x-\frac{\pi}{8}\bigg)$$

We have,

$$f'(x)=\lim\limits_{h\to 0}\space\frac{f(x+h)-f(x)}{h}$$

(By first principle)

$$\Rarr\space\text{f'(x)}=\lim\limits_{h\to0}\space\frac{\text{cos}\bigg(x+h-\frac{\pi}{8}\bigg)-\text{cos}\bigg(x-\frac{\pi}{8}\bigg)}{h}$$

$$\bigg(\because\space\text{f(x)= cos}\bigg(x-\frac{\pi}{8}\bigg)\bigg)\\-2\space\text{sin}\space\frac{x+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}\\=\lim\limits_{h\to0}\space\frac{\text{sin}\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2}}{h}$$

$$\begin{bmatrix}\because\space\text{cos C- cos D}=\\-2\space\text{sin}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\end{bmatrix}\\=\lim\limits_{h\to0}\space\frac{-2 \text{sin}\frac{2x-2\bigg(\frac{\pi}{8}\bigg)+h}{2}\text{sin}\frac{h}{2}}{2×\frac{h}{2}}$$

$$=-\space\text{sin}\space\frac{2x-2\bigg(\frac{\pi}{8}\bigg)+0}{2}×1\\\bigg(\therefore\space\lim\limits_{x\to0}\frac{\text{sin}\frac{h}{2}}{\frac{h}{2}}=1\bigg)\\=-\text{sin}\space\frac{2\bigg(x-\frac{\pi}{8}\bigg)}{2}\\\Rarr\space f'(x)=-\text{sin}\bigg(x-\frac{\pi}{8}\bigg)$$

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constant and m and n are integers.

2. x + a.

Sol. Let y = x + a

$$\frac{dy}{dx}=\frac{d}{dx}(x+a)\\\frac{dy}{dx}=1+0=1.$$

$$\textbf{3.}\space(px+q)\bigg(\frac{r}{x}+s\bigg).\\\textbf{Sol.}\space\text{Let}\space y=\text{(px+q)}\bigg(\frac{r}{x}+s\bigg)$$

Differentiating y w.r.t. x,

$$\frac{dy}{dx}=(px+q)\frac{d}{dx}\bigg(\frac{r}{x}+s\bigg)+\\\bigg(\frac{r}{x}+s\bigg)\frac{d}{dx}(px+q)$$

(By product rule)

$$=(px+q)\bigg[-\frac{r}{x^2}+0\bigg]+\\\bigg(\frac{r}{x}+s\bigg)(p×1+0)\\=(px+q)\bigg(-\frac{r}{x^2}\bigg)+\bigg(\frac{r}{x}+s\bigg)p\\=-\frac{pxr}{x^2}-\frac{qr}{x^2}+\frac{pr}{x}+ps\\=-\frac{pr}{x}-\frac{qr}{x^2}+\frac{pr}{x}+ps\\\Rarr\space\frac{dy}{dx}=ps-\frac{qr}{x^2}$$

4. (ax + b) (cx + d)2.

Sol. Let y = (ax + b) (cx + d)2

$$\frac{dy}{dx}=(ax+b)\frac{d}{dx}(cx+d)^2+\\(cx+d)^2\frac{d}{dx}(ax+b)$$

(By product rule)

$$=(ax+b)\frac{d}{dx}(c^2x^2+d^2+2cxd)\\+(cx+d)^2(a×1+0)$$

(∵ (a + b)2 = a2 + 2ab + b2)

= (ax + b) [c2 (2x) + 0 + 2c × 1 × d] + (cx + d)2 × a

= (ax + b) (2c2x + 2cd) + a(cx + d)2

= (ax + b)2c (cx + d) + a(cx + d)2

= 2c(ax + b) (cx + d) + a(cx + d)2

$$\textbf{5.}\space\frac{ax+b}{cx+d}.\\\textbf{Sol.}\space\text{Let\space} y=\frac{ax+b}{cx+d}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\\\frac{(cx+d)\frac{d}{dx}(ax+b)-(ax+b)\frac{d}{dx}(cx+d)}{(cx+d)^2}$$

(By quotient formula)

$$=\frac{(cx+d)(a×1+0)-(ax+b)(c×1+0)}{(cx+d)^2}$$

$$=\frac{a(cx+d)-(ax+b)c}{(cx+d)^2}\\=\frac{acx+ad-acx-bc}{(cx+d)^2}\\\Rarr\space\frac{dy}{dx}=\frac{ad-bc}{(cx+d)^2}.$$

$$\textbf{6.}\space\frac{1+\frac{1}{x}}{1-\frac{1}{x}}.\\\textbf{Sol.}\space\text{Let y}=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\\=\frac{dy}{dx}=\\\frac{\bigg(1-\frac{1}{x}\bigg)\frac{d}{dx}\bigg(1+\frac{1}{x}\bigg)-\bigg(1+\frac{1}{x}\bigg)\frac{d}{dx}\bigg(1-\frac{1}{x}\bigg)}{\bigg(1-\frac{1}{x}\bigg)^2}$$

$$=\frac{\bigg(1-\frac{1}{x}\bigg)\bigg(0-\frac{1}{x^2}\bigg)-\bigg(1+\frac{1}{x}\bigg)\bigg(0+\frac{1}{x^2}\bigg)}{\bigg(1-\frac{1}{x}\bigg)^2}$$

$$=\frac{\frac{-1}{x^2}+\frac{1}{x^3}-\frac{1}{x^2}-\frac{1}{x^3}}{\bigg(1-\frac{1}{x}\bigg)^2}\\=\frac{\frac{-2}{x^2}}{\bigg(1-\frac{1}{x}\bigg)^2}\\=\frac{-2}{x^2\bigg(1-\frac{1}{x}\bigg)^2}\\=\frac{-2}{x^2\frac{(x-1)^2}{x^2}}\\=\frac{-2}{(x-1)^2}.$$

$$\textbf{7.}\space\frac{1}{(ax^2+bx+c)}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{1}{(ax^2+bx+c)}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\\\frac{(ax^2+bx+c)\frac{d}{dx}(1)-1×\frac{d}{dx}(ax^2+bx+c)}{(ax^2+bx+c)^2}\\\text{(By quotient formula)}\\=\frac{(ax^2+bx+c)×0-1×(2ax+b)}{(ax^2+bx+c)^2}$$

$$\Rarr\space\frac{dy}{dx}=\frac{2ax+b}{(ax^2+bx+c)^2}$$

$$\textbf{8.}\space\frac{ax+b}{px^2+qx+r}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{ax+b}{px^2+qx+r}$$

Differentiating y w.r.t. x, we get

$$\text{(px}^2+qx+r)\frac{d}{dx}(ax+b)\\\frac{dy}{dx}=\frac{-(ax+b)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2}$$

(By quotient formula)

$$=\frac{(px^2+qx+r)(a×1+0)-(ax+b)(2px+q)}{(px^2+qx+r)^2}\\=\frac{(px^2+qx+r)a-(ax+b)(2px+q)}{(px^2+qx+r)^2}\\=\\\frac{(apx^2+aqx+ra)+(2apx^2+aqx+2bpx+qb)}{(px^2+qx+r)^2}\\=\frac{apx^2+aqx+ra-2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}\\=\frac{-apx^2-2bpx+ar-bq}{(px^2+qx+r)^2}.$$

$$\textbf{9.}\space\frac{px^2+qx+r}{ax+b}.\\\text{Sol.}\space\text{Let}\space y=\frac{px^2+qx+c}{ax+b}$$

Differentiating y w.r.t. x, we get

$$(ax+b)\frac{d}{dx}(px^2+qx+r)\\\frac{dy}{dx}=\frac{-(px^2+qx+r)\frac{d}{dx}(ax+b)}{(ax+b)^2}$$

(By quotient formula)

$$=\\\frac{(ax+b)(2px+q+0)-(px^2+qx+r)(a×1+0)}{(ax+b)^2}\\=\frac{(ax+b)(2px+q)-(px^2+qx+r)a}{(ax+b)^2}\\=\\\frac{(2apx^2+aqx+2bpx+bq)-(apx^2+aqx+ar)}{(ax+b)^2}\\=\frac{2apx+aqx+2bpx+bq-apx^2-aqx-ar}{(ax+b)^2}\\=\frac{dy}{dx}=\frac{apx^2+2bpx+bq-ar}{(ax+b)^2}$$

$$\textbf{10.}\space\frac{a}{x^4}-\frac{b}{x^2}+\text{cos x}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{a}{x^4}-\frac{b}{x^2}+\text{cos x}\\\frac{dy}{dx}=\frac{d}{dx}\frac{a}{x^4}-\frac{d}{dx}\frac{b}{x^2}+\frac{d}{dx}\space\text{cos x}\\=a\frac{d}{dx}x^{-4}-b\frac{d}{dx}x^{-2}-\text{sin x}\\=\frac{-4a}{x^5}+\frac{2b}{x^3}-\text{sin x}.$$

$$\textbf{11.}\space 4\sqrt{x}-2.\\\textbf{Sol.}\space\text{Let}\space y=4\sqrt{x}-2\\\Rarr\space\frac{dy}{dx}=\frac{4}{2\sqrt{x}}-0\\\frac{dy}{dx}=\frac{2}{\sqrt{x}}.$$

12. (ax + b)n.

Sol. Let y = (ax + b)n

Differentiating y w.r.t. x,

$$\Rarr\space\frac{dy}{dx}=n(ax+b)^{n-1}\frac{d}{dx}(ax+b)\\=n(ax+b)^{n-1}×a\\\Rarr\space\frac{dy}{dx}=na(ax+b)^{n-1}.$$

13. (ax + b)n (cx + d)m.

Sol. Let y = (ax + b)n (cx + d)m

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(ax+b)^n\frac{d}{dx}(cx+d)^m+\\(cx+d)^m\frac{d}{dx}(ax+b)^n\\=(ax+b)^nm(cx+d)^{m-1}\frac{d}{dx}(cx+d)+\\(cx+d)^mn(ax+b)^{n-1}\frac{d}{dx}(ax+b)$$

= m(ax + b)n (cx + d)m – 1 (c × 1 + 0)+ n(cx + d)m (ax + b)n – 1 (a × 1 + 0)

= m(ax + b)n (cx + d)m – 1c + n(cx + d)m (ax + b)n – 1a

$$\Rarr\space\frac{dy}{dx}=(ax+b)^{n-1}(cx+d)^{m-1}$$

[mc(ax + b) + na(cx + d)]

$$\frac{dy}{dx}=(ax+b)^{n-1}(cx+d)^{m-1}$$

[macx + mbc + nacx + nad]

14. sin (x + a).

Sol. y = sin (x + a)

$$\frac{dy}{dx}=\frac{d}{dx}\text{sin}(x+a)\\\text{cos}(x+a)\frac{d}{dx}(x+a)$$

= cos (x + a) [1 + 0]

= cos (x + a).

15. cosec x cot x.

Sol. Let y = cosec x cot x

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\text{cosec x}\frac{d}{dx}(\text{cot x)}+\text{cot x}\frac{d}{dx}(\text{cosec x})$$

(by product rule)

= – cosec x cosec2 x + cot x (– cosec x cot x)

= – cosec3 x – cot2 x cosec x

= – cosec x (cosec2 x + cot2 x).

$$\textbf{16.}\space\frac{\text{cos x}}{\text{1+ sin x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{cos x}}{\text{1+sin x}}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\frac{(1+sin x)\frac{d}{dx}\text{(cos x) - cos x}\frac{d}{dx}(\text{1+sin x})}{(\text{1+ sin)}^2}$$

(By quotient formula)

$$=\frac{(1+ sin x)(- sin x)-cos x(0 + cos x)}{(1+ sin x)}$$

$$=\frac{-\space\text{sin x - sin}^2x- \text{cos}^2x}{(1+ sin \space x)^2}\\\frac{\text{- sin x}-(\text{sin}^2x+ \text{cos}^2 x)}{(1 + \text{sin x})^2}\\=\frac{\text{- sin x-1}}{(\text{1+ sin x})^2}\\=\frac{-(\text{sin x + 1})}{(\text{sin x + 1})^2}\\=\frac{-1}{(\text{sin x + 1})}.$$

$$\textbf{17.}\space\frac{\text{sin x + cos x}}{\text{sin x - cos x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{sin x + cos x}}{\text{sin x - cos x}}$$

Differentiating y w.r.t. x, we get

$$\text{(sin x - cos x)}\frac{d}{dx}(\text{sin x + cos x})\\\frac{dy}{dx}=\frac{-(\text{sin x + cos x})\frac{d}{dx}(\text{sin x - cos x})}{(\text{sin x - cos x })^2}$$

(by quotient formula)

$$=\\\frac{\text{(sin x - cos x)(cos x- sin x)}-(\text{sin x + cos x})(\text{cos x + sin x})}{(\text{sin x - cos x})^2}$$

$$=\\\frac{-(\text{cos x - sin x})(\text{cos x - sin x}) - (\text{cos x + sin x})^2}{(\text{sin x - cos x})^2}\\=\frac{-(\text{cos x - sin x})^2 - (\text{cos x + sin x})^2}{\text{(sin x - cos x)}^2}\\=\\\frac{-\lbrack(\text{cos}^2x + \text{sin}^2x - 2 \space\text{cos x sin x})+(\text{cos}^2x + \text{sin}^2 x + \text{2 cos x sin x})\rbrack}{\text{(sin x - cos x)}^2}\\\frac{dy}{dx}=\\\frac{- \text{cos}^2x-sin^2x+\text{2 cos x sin x - cos}^2 x -\text{sin}^2x-\text{2 cos x sin x}}{\text{(sin x - cos x)}^2}\\=\frac{-2\text{cos}^2 x - 2\text{sin}^2 x}{(\text{sin x - cos x})}\\=\frac{-2}{(\text{ sin x - cos x})^2}.$$

$$\textbf{18.}\space\frac{\text{sec x - 1}}{\text{sec x +1}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{sec x-1}}{\text{sec x+1}}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\\\frac{\text{(sec x+1)}\frac{d}{dx}(\text{sec x - 1})-(\text{sec x -1})\frac{d}{dx}(\text{sec x + 1})}{(\text{sec x + 1})^2}$$

(By quotient formula)

$$=\\\frac{(\text{sec x + 1})(\text{sec x tanx - 0 })-(sec\space x-1)(\text{sec x tan x+0})}{\text{(sec x + 1)}^2}\\=\\\frac{(\text{sec x+1})(\text{sec \space x \space tanx})-(\text{sec x-1})(\text{sec x tan x })}{(\text{sec x +1})^2}\\=\frac{\text{sec x tan x}\lbrack\text{sec x+1- sec x+1}\rbrack}{\text{(sec x + 1)}^2}\\=\frac{\text{2 sec x tan x}}{\text{(sec x +1)}^2}.$$

19. sinn x.

Sol. Let y = sinn x

⇒ y = (sin x)n

Differentiating y w.r.t x, we get

$$\frac{dy}{dx}=\text{n(sin x)}^{n-1}×\frac{d}{dx}\text{(sin x)}\\\Rarr\space \frac{dy}{dx}=\text{n sin}^{n-1}\text{x cos x.}$$

$$\textbf{20.}\space\frac{\text{a+b sin x}}{\text{c+d cos x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{a+b sin x}}{\text{c+d cos x}}$$

Differentaiting w.r.t. x, we get

$$\text{(c + d cos x)}\frac{d}{dx}(\text{a+b sin x})\\\frac{dy}{dx}=\frac{-(a+b \space \text{sin x})\frac{d}{dx}(\text{c+d cos x})}{(\text{c+ d cos x})^2}\\=\\\frac{(\text{c+d cos x})[0+b \space cos x]-(a+b \space sin x)(0-d \space sin x)}{(\text{c+d cos x})^2}\\=\\\frac{(\text{c+d cos x})(b \space cos x)-(a+b\space sin x)(-d \space sin x)}{(\text{c+d cos x})^2}\\=\frac{\text{bc cos x + bd cos}^2x +\text{ad\space sin x + bd\space sin}^ 2 x}{(c+d\space cos x)^2}\\=\frac{\text{bc cos x + ad sin x + bd}(\text{cos}^2x + \text{sin}^2 x)}{(\text{c+d cos x})^2}\\\Rarr\space\frac{dy}{dx}=\frac{bc \space cos x + ad \space sin x + bd}{\text{(c+d cos x)}^2}$$

$$\textbf{21.}\space\frac{\text{sin}\space(x+a)}{\text{cos x}}.\\\textbf{Sol.}\space\text{Let y}=\frac{\text{sin}(x+a)}{\text{cos x}}\\\frac{dy}{dx}=\frac{\text{cos x}\frac{d}{dx}\text{sin}(x+a)-\text{sin}(x+a)\frac{d}{dx}\text{cos x}}{\text{cos}^2x}\\=\frac{dy}{dx}=\frac{\text{cos x . cos(x+a) + \text{sin}(x+a) sin x}}{\text{cos}^2x}\\\frac{dy}{dx}=\frac{\text{cos}(x+a-x)}{\text{cos}^2x}$$

[ cos A cos B + sin A sin B = cos (A – B)]

$$\frac{dy}{dx}=\text{cos a sec}^2x.$$

22. x4 (5 sin x – 3 cos x).

Sol. Let y = x4 (5 sin x – 3 cos x)

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=x^4\frac{d}{dx}(\text{5 sin x - 3 cos x})+\\(5 \space\text{sin x}- 3\space\text{cos x})\frac{d}{dx}(x^4)$$

(By product formula)

= x4 (5 cos x + 3 sin x) + (5 sin x – 3 cos x) 4x3

= x3 [x(5 cos x + 3 sin x) + 4(5 sin x – 3 cos x)]

= x3 [5x cos x + 3x sin x + 20 sin x – 12 cos x]

23. (x2 + 1) cos x.

Sol. Let y = (x2 + 1) cos x

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(x^2+1)\frac{d}{dx}(\text{cos x}) \space +\\\text{cos x}\frac{d}{dx}(x^2+1)$$

(By product formula)

= (x2 + 1) (– sin x) + cos x (2x)

= – x2 sin x – sin x + 2x cos x.

24. (ax2 + sin x) (p + q cos x).

Sol. Let y = (ax2+ sin x) (p + q cos x)

(By product formula)

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(ax^2+sinx)\frac{d}{dx}(p+q\space cos x)+\\(\text{p + q cos x})\frac{d}{dx}(ax^2 + sinx )$$

= (ax2 + sin x) (0 – q sin x) + (p + q cos x) (2ax + cos x)

= – q sin x (ax2 + sin x) + (p + q cos x) (2ax + cos x).

25. (x + cos x) (x – tan x).

Sol. Let y = (x + cos x) (x – tan x)

(By product formula)

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(x + cos x)\frac{d}{dx}(x- tan x)+\\(x-tan x)\frac{d}{dx}(x+cos x)$$

= (x + cos x) (1 – sec2 x) + (x – tan x) (1 – sin x)

= – (x + cos x) (sec2 x – 1) + (x – tan x) (1 – sin x)

= – (x + cos x) tan2 x + (x – tan x) (1 – sin x)

(∵ sec2 x – tan2 x = 1)

$$\textbf{26.}\space\frac{4x+5 \space sin \space x}{3x + 7\space\text{cos x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{4x+5 \space sin x}{3x+7\space cos x}$$

Differentiating y w.r.t. x, we get

$$\text{(3x+7\space\text{cos x})}\frac{d}{dx}(4x+5\space sin \space x)\\\frac{dy}{dx}=\frac{-(4x+5\space sin x)\frac{d}{dx}(3x+7\space cos x)}{(3x+7 cos x)^2}\\\text{(By quotient formula)}\\=\\\frac{(3x+7 cosx )(4×1 + 5\space cos x)-(4x+5\space sin x)(3-7 sin x)}{(3x+7\space\text{cos x})^2}$$

$$=\\\frac{12x+15x\space\text{cos x}+28 \space cos x + 35\space cos^2 x- 12x + 28 x sin x- 15 sin x + 35\text{sin}^2 x}{(3x+7\space cos x)^2}$$

$$=\\\frac{35 (cos ^2x + sin^{2}x)+15 x\space cos x+ 28\space cos x + 28 \text{x sin x}- 15 \text{sin x}}{(3x+7\space\text{cos x})^2}\\=\frac{35+15x\space\text{cos x + 28 x sin x - 15 sin x}}{(3x+7 cos x)^2}\\\lbrack\because\space \text{sin}^{2}x + \text{cos}^2x=1\rbrack$$

$$\textbf{27.}\space\frac{\text{x}^2\text{cos}\bigg(\frac{\pi}{\text{4}}\bigg)}{\text{sin x}}\textbf{.}\\\textbf{Sol.}\space\text{Let}\space y=\frac{x^2\text{cos}\bigg(\frac{\pi}{4}\bigg)}{\text{sin x}}$$

Differentiating w.r.t. x, we get

$$\Rarr\space \frac{dy}{dx}=\text{cos}\frac{\pi}{4}×\frac{d}{dx}\bigg(\frac{x^2}{\text{sin x}}\bigg)\\(\because\space\text{cos}\frac{\pi}{4}\space\text{is constant})$$

$$\Rarr\space\frac{dy}{dx}=\bigg(\text{cos}\frac{\pi}{4}\bigg)×\\\frac{\text{sin x}\frac{d}{dx}(x^2)-x^2\bigg(\frac{d}{dx}\text{sin x }\bigg)}{\text{sin}^2x}$$

(By quotient formula)

$$=\text{cos}\frac{\pi}{4}×\frac{(sin \space x)(2x)-x^2(cos x)}{sin^2 x}\\=\frac{x \text{cos}\frac{\pi}{4}[2 \text{sin x-x \space cos x}]}{\text{sin}^2x}$$

$$\textbf{28.}\space\frac{\text{x}}{\text{1 + tan x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{x}}{\text{1+tan x}}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\frac{(\text{1+tan x})\frac{d}{dx}(x)-x\frac{d}{dx}(\text{1+tan x})}{\text{(1+tan x)}^2}$$

$$=\frac{\text{(1+ tan x)(1)-x(0 + sec}^2x)}{(\text{1+ tan x})^2}\\=\frac{\text{1+tan x - x sec}^2x}{(\text{1+ tan x})^2}.$$

29. (x + sec x) (x – tan x).

Sol. Let y = (x + sec x) (x –tan x)

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(x + sec x)\frac{d}{dx}(x- tan x)+\\(x-tan x)\frac{d}{dx}(x+sec x)$$

(By product formula)

$$\frac{dy}{dx}=(x + sec x)(1 - sec^2x)+\\(x-tan x)(1 + sec x\space tan x)$$

$$\textbf{30}.\space\frac{x}{\text{sin}^nx}.\\\textbf{Sol.}\space y=\frac{x}{\text{sin}^n x}$$

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=\frac{\text{sin}^nx\frac{d}{dx}(x)-x\frac{d}{dx}(\text{sin}^nx)}{(\text{sin}^nx)^2}\\\text{(By quotient formula)}\\=\frac{\text{sin}^nx×1-x\frac{d}{dx}(sin\space x)^n}{\text{sin}^{2n}x}$$

Differentiating (sin x)n by chain rule,

$$=\frac{\text{sin}^nx-nx(sin\space x)^{n-1}\frac{d}{dx}(sin\space x)}{\text{sin}^{2n}x}\\=\frac{\text{sin}^nx-nx\space \text{sin}^{n-1} x \space cos x}{\text{sin}^{2n}x}\\=\frac{\text{sin}^{n-1}x[\text{sin x - nx cosx }]}{(\text{sin x})^{2n}}\\=\frac{\text{sin x- nx cos x}}{(\text{sin x})^{2n-n+1}}\\=\frac{\text{sin x - nx cos x}}{\text{sin}^{n+1}x}.$$