Chapter 12: Limits and Derivatives

Ans: Let $f(x)=4 \sqrt{x}-2$

$f^{\prime}(x)=\dfrac{d}{d x}(4 \sqrt{x}-2)=\dfrac{d}{d x}(4 \sqrt{x})-\dfrac{d}{d x}(2)$

$=4 \dfrac{d}{d x}\left(x^{\dfrac{1}{2}}\right)-0=4\left(\dfrac{1}{2} x^{\dfrac{1}{2}}\right)$

$=\left(2 x^{-\dfrac{1}{2}}\right)=\dfrac{2}{\sqrt{x}}$

Ans: Let $f(x)=(a x+b)^{n} .$ Accordingly, $f(x+h)-\{a(x+h)+b\}^{n}-(a x+a h+b)^{n}$

By first principle,

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

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

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

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left\{1+n\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1)}{2}\left(\dfrac{a h}{a x+b}\right)^{2}+\cdots\right\}-1\right] \quad$ (using binomial theorem)

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1) a^{2} h^{2}}{2(a x+b)^{2}}+\cdots\right.$ (Terms containing higher degrees of $\left.\left.h\right)\right]$

$=(a x+b)^{n} \lim _{n \rightarrow 0}\left[\dfrac{n a}{(a x+b)}+\dfrac{n(n-1) \nexists^{7} h^{2}}{2(a x+b)^{2}}+\cdots\right]$

$=(a x+b)^{n}\left[\dfrac{n a}{(a x+b)}+0\right]$

$=n a \dfrac{(a x+b)^{n}}{a x+b}$

$-n a(a x+b)^{n-1}$

Ans: Let $f(x)=(a x+b)^{n}(c x+d)^{m}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b)^{n} \dfrac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \dfrac{d}{d x}(a x+b)^{n}$

Now let $f_{1}(x)=(c x+d)^{m}$

$f_{1}(x+h)=(c x+c h+d)^{m}$

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

$=\lim _{n \rightarrow 0} \dfrac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

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

$=(c x+d)^{m} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\left(1+\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1)}{2} \dfrac{c^{2} h^{2}}{(c x+d)^{2}}+\cdots\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right.$ (Terms containing higher degree oh $\left.\left.h\right)\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0}\left[\dfrac{m c}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right]$

$=(C x+a)^{m}\left[\dfrac{m c h}{(c x+d)}+0\right]$

$=\dfrac{m c(c x+d)^{m}}{(c x+d)}$

$=m c(c x+d)^{m-1}$

$\dfrac{d}{d x}(c x+d)^{m}=m d(x+d)^{m-1}$

Similarly, $\dfrac{d}{d x}(a x+b)^{n}=n a(a x+b)^{n-1}$... (3)

Therefore, from (1), (2), and (3), we obtain

$f^{\prime}(x)=(a x+b)^{n}\left\{m c(c x+d)^{m-1}\right\}+(c+d)^{m}\left\{n a(a x+b)^{n-1}\right\}$

$=(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]$

Ans: Let, $f(x)=\sin (x+a)$

$f(x+h)=\sin (x+h+a)$

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

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

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

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

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

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

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

$=\cos (x+a)$

Ans: Let $f(x)=\operatorname{cosec} x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=\operatorname{cosec} x(\cot x)^{\prime}+\cot x(\operatorname{cosec} x)^{\prime} \ldots .(1)$

Let $f_{1}(x)=\cot x .$ Accordingly, $f_{1}(x+h)=\cot (x+h)$

By first principle,

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

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

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

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

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

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

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

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

$=\dfrac{-1}{\sin ^{2} x}$

$=-\operatorname{cosec}^{2} x$

$\therefore(\cot x)^{\prime}=-\operatorname{cosec}^{2} x \quad \ldots$ (2)

Now, let $f_{2}(x)=\operatorname{cosec} x .$ Accordingly, $f_{2}(x+h)=\operatorname{cosec}(x+h)$

By first principle, $f_{2}(x)=\lim _{n \rightarrow 0} \dfrac{f_{2}(x+h)-f_{2}(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec}(x)]$

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

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

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

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

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

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

$=\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+0)}$

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

$=-\operatorname{cosec} x \cdot \cot x$

$\therefore(\operatorname{cosec} x)^{\prime}=-\operatorname{cosec} x \cdot \cot x$

From (1), (2), and (3), we obtain

$f^{\prime}(x)=\operatorname{cosec} x\left(-\operatorname{cosec}^{2} x\right)+\cot x(-\operatorname{cosec} x \cot x)$

$=-\operatorname{cosec}^{3} x-\cot ^{2} x \operatorname{cosec} x$

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

By the quotient rule,

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

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

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

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

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

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

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

Ans:17: Let $f(x)=\dfrac{\sin x+\cos x}{\sin x-\cos x}$

By the quotient rule,

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

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

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

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

$=\dfrac{-[1+1]}{(\sin x-\cos x)^{2}}$

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

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

$f(x)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}=\dfrac{1-\cos x}{1+\cos x}$

By the quotient rule,

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

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

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

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

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

$=\dfrac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}}$

$=\dfrac{\dfrac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}}$

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

Ans: Let $y=\sin ^{n} x$

Accordingly, for $n=1, y=\sin x$

$\therefore \dfrac{d y}{d x}=\cos x$, i.e., $\dfrac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$.

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}(\sin x \sin x)$

$=(\sin x)^{\prime}\left(\sin x+\sin x(\sin x)^{\prime} \quad\right.$ (By Leibnitz product rule)

$=\cos x \sin x+\sin x \cos x$

$=2 \sin x \cos x$

$\ldots .(1)$

For $n=3, y=\sin ^{3} x$

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}\left(\sin x \sin ^{2} x\right)$

$=(\sin x)^{\prime} \sin ^{2} x+\sin x(\sin x)^{\prime}$

(By Leibnitz product rule)

$-\cos x \sin ^{2} x+\sin x(2 \sin x \cos x) \quad[$ Using $(1)]$

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

$=3 \sin ^{2} x \cos x$

We assert that $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

Let our assertion be true for $n=k$.

i.e., $\dfrac{d}{d x}\left(\sin ^{k} x\right)=k \sin ^{(k-1)} x \cos x \quad \ldots .$ (2)

Consider,

$\dfrac{d}{d x}\left(\sin ^{k+1} x\right)=\dfrac{d}{d x}\left(\sin x \sin ^{(k)} x\right)$

$=(\sin x)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{n}$

(By Leibnitz product rule)

$=\cos x \sin ^{k} x+\sin x\left(k \sin ^{k-1} \cos x\right) \quad[$ Using $(2)]$

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

$-(k+1) \sin ^{k} x \cos x$

Thus, our assertion is true for $n=k+1$.

Hence, by mathematical induction, $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

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

By the quotient rule,

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

$=\dfrac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}}$

$=\dfrac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)}{(C+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}}$