Chapter 12: Limits and Derivatives

Ans: Let $f(x)=\dfrac{\sin (x+a)}{\cos x}$

By the quotient rule,

$f^{\prime}(x)=\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a) \dfrac{d}{d x} \cos x}{\cos ^{2} x}$

$f^{\prime}(x)=\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a) \dfrac{d}{d x}(-\sin x)}{\cos ^{2} x}$

Let $g(x)-\sin (x+a) .$ Accordingly,$g(x+h)=\sin (x+h+a)$

By first principle,

$g^{\prime}(x)=\lim _{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+a)-\sin (x+a)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+2 a+h}{h}\right)\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}\right]$

$=\lim _{n \rightarrow 0} \cos \left(\dfrac{2 x+2 a+h}{h}\right) \cdot \lim _{n \rightarrow 0}\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\} \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]$

$=\left(\cos \dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{n \rightarrow 0} \dfrac{\sin h}{h}=1\right]$

$=\cos (x+a) \quad \ldots$ (ii)

From (i) and (ii), we obtain $f^{\prime}(x)=\dfrac{\cos x \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x}$

$=\dfrac{\cos (x+a-x)}{\cos ^{2} x}$

$=\dfrac{\cos a}{\cos ^{2} x}$

Ans: Let $f(x)=x^{4}(5 \sin x-3 \cos x)$

By product rule.

$f^{\prime}(x)=x^{4} \dfrac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}\left[5 \dfrac{d}{d x}(\sin x)-3 \dfrac{d}{d x}(\cos x)\right]+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)\left(4 x^{3}\right)$

$=x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x]$

Ans: Let $f(x)=\left(x^{2}+1\right) \cos x$

By product rule.

$f^{\prime}(x)=\left(x^{2}+1\right) \dfrac{d}{d x}(\cos x)+\cos x \dfrac{d}{d x}\left(x^{2}+1\right)$

$=\left(x^{2}+1\right)(-\sin x)+\cos x(2 x)$

$=-x^{2} \sin x-\sin x+2 x \cos x$

Ans: Let $f(x)=\left(a x^{2}+\sin x\right)(p+q \cos x)$

By product rule.

$f^{\prime}(x)=\left(a x^{2}+\sin x\right) \dfrac{d}{d x}(p+q \cos x)+(p+q \cos x) \dfrac{d}{d x}\left(a x^{2}+\sin x\right)$

$=\left(a x^{2}+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x)$

$=-q \sin x\left(a x^{2}+\sin x\right)+(p+q \cos x)(2 a x+\cos x)$

Ans: Let $f(x)=(x+\cos x)(x-\tan x)$

By product rule,

$f^{\prime}(x)=(x+\cos x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\cos x)$

$=(x+\cos x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left[1-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

Let $g(x)=\tan x .$ Accordingly,$g(x+h)=\tan (x+h)$

By first principle,

$g^{\prime \prime}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\tan (x+h)-\tan (x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos x \cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{n \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{h \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x}\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h)}\right)$

$=\dfrac{1}{\cos x} \cdot \cdot\left(\dfrac{1}{\cos (x+0)}\right)$

$=\dfrac{1}{\cos ^{2} x}$

$=\sec ^{2} x \quad$... (ii)

Therefore, from (i) and (ii). We obtain

$f^{\prime}(x)=(x+\cos x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left(-\tan ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x)$

Ans: Let $f(x)=\dfrac{4 x+5 \sin x}{3 x+7 \cos x}$

Quotient rule,

$f^{\prime}(x)=\dfrac{(3 x+7 \cos x) \dfrac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \dfrac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)\left[4 \dfrac{d}{d x}(x)+5 \dfrac{d}{d x}(\sin x)\right]-(4 x+5 \sin x)\left[3 \dfrac{d}{d x}(x)+7 \dfrac{d}{d x}(\cos x)\right]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)[4 x+5 \cos x]-(4 x+5 \sin x)[3-7 \sin x]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{12 x+15 x \cos x+28 x \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}$

By quotient rule, $f^{\prime}(x)=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x \dfrac{d}{d x}\left(x^{2}\right)-x^{2} \dfrac{d}{d x}(\sin x)}{\sin ^{2} x}\right]$

$=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x(2 x)-x^{2}(\cos x)}{\sin ^{2} x}\right]$

$=\dfrac{x \cos \dfrac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}$

Ans: Let $f(x)=\dfrac{x}{1+\tan x}$

$f(x)=\dfrac{(1+\tan x) \dfrac{d}{d x}(x)-(x) \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

$f^{\prime}(x)=\dfrac{(1+\tan x)-x \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

Let $g(x)=1+\tan x .$ Accordingly $g(x+h)=1+\tan (x+h)$.

By first principle, $\dot{g}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$\lim _{h \rightarrow 0}\left[\dfrac{1+\tan (x+h)-1-\tan (x)}{h}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]$

$\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]$

$-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)$

$-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x$

$\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x$

From (i) and (ii), we obtain

$\dot{f}(x)=\dfrac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}}$

Ans: Let $f(x)=(x+\sec x)(x-\tan x)$

By product rule.

$f(x)=(x+\sec x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\sec x)$

$-(x+\sec x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \tan x\right]+(x-\tan x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \sec x\right]$

$\left.-f(x+\sec x)\left[1-\dfrac{d}{d x} \tan x\right)\right]+(x-\tan x)\left[1+\dfrac{d}{d x} \sec x\right]$

$\ldots(\mathrm{i})$

Let $f_{1}(x)=\tan x, f_{2}(x)=\sec x$

Accordingly, $f_{1}(x+h) \cdot \tan (x+h)$ and $f_{2}(x+h)-\sec (x+h)$

$f_{1}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{1}(x+h)-f_{1}(x)}{h}\right)$

$=\lim _{h \rightarrow 0}\left[\dfrac{\tan (x+h)-\tan (x)}{h}\right]$

$\begin{array}{l}-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]\\-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)\\-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x\\\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x\\f_{2}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{2}+(x+h)-f_{2}(x)}{h}\right)\\=\lim _{h \rightarrow 0}\left(\dfrac{\sec (x+h)-\sec (x)}{h}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}\right)\\=\dfrac{1}{\cos x^{\prime \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\cos (x+h)}\right]\\=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\cos (x+h)}\right]\end{array}$

$=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2}\right)\left\{\dfrac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\cos (x+h)}\right]$

$=\sec x \dfrac{\left\{\lim _{n \rightarrow 0} \sin \left(\dfrac{2 x+h}{2}\right)\right\}\left\{\lim _{\dfrac{h}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\lim _{n \rightarrow 0} \cos (x+h)}$

$=\sec x \cdot \dfrac{\sin x \cdot 1}{\cos x}$

$\Rightarrow \dfrac{d}{d x} \sec x=\sec x \tan x$

From (i). (ii), and (iii), we obtain

$f^{\prime}(x)=(x+\sec x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1+\sec x \tan x)$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x}{\sin ^{n} x}$

By quotient rule, $f^{\prime}(x)=\dfrac{\sin ^{n} x \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

It can be easily shown that $\dfrac{d}{d x} \sin ^{n} x=n \sin ^{n-1} x \cos x$

Therefore,

$f^{\prime}(x)=\dfrac{\sin ^{n} \times \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n} x \cdot 1-x\left(\operatorname{nin}^{n-1} x \cos x\right)}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x}$

$=\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x}$