# Differentiation (Part-1) | S N De

$~1.~~$ Find $~\frac{dy}{dx}~$ using the rule $~\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

$~(i)~~y=\sin\sqrt{x^2+a^2}$

Sol. let $~y=\sin u \Rightarrow \frac{dy}{du}=\cos u$

Again, let $~~u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

Here, $~~v=x^2+a^2 \Rightarrow \frac{dv}{dx}=2x$

So, $~~\frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}. \frac{dv}{dx}\\ \text{or,}~~ \frac{dy}{dx}=\cos u .\frac{1}{2\sqrt v}.~2x\\~~~~~~~~~~~=\cos\sqrt{x^2+a^2} \cdot \frac{x}{\sqrt{x^2+a^2}}\\~~~~~~~~~~~=\frac{x \cos{\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}~~\text{(ans.)}$

$~(ii)~~y=\log_2(\sin x^3)$

Sol. $~~~y=\log_2(\sin x^3)=\log_e(\sin x^3) \times \log_e2$

let $~~~~y=\log_eu \times \log_2e\\~~~[\text{where}~~u=\sin x^3] \\ \therefore \frac{dy}{du}=\log_2e \cdot \frac 1u$

Now, $~~~~u=\sin v~~~[\text{where}~~v=x^3]\\ \Rightarrow \frac{du}{dv}=\cos v$

$~~v=x^3 \Rightarrow \frac{dv}{dx}=3x^2 \\ \therefore \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \\ \text{or,}~~ \frac{dy}{dx}=\log_2e \cdot \frac 1u \cdot \cos v \cdot 3x^2 \\ \text{or,}~~ \frac{dy}{dx}=\log_2e \cdot \frac{\cos v}{\sin v}\cdot3x^2 \\ \therefore \frac{dy}{dx}=3x^2\cot(x^3)\log_2e~~\text{(ans.)}$

$~(iii)~~y=e^{\sqrt{\cos x}}$

Sol. $~~y=e^{\sqrt{\cos x}} \\ \text{or,}~~y=e^u ~~[~\text{where}~~u=\sqrt{\cos x}]\\ \text{or,}~~\frac{dy}{du}=e^u$

Again, $~u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

Now, $~v=\cos x \Rightarrow \frac{dv}{dx}=-\sin x$

So, $~~\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}\\ \text{or,}~~\frac{dy}{dx}=e^u \cdot \frac{1}{2\sqrt v} \cdot (-\sin x) \\ \text{or,}~~\frac{dy}{dx}=-\frac{e^{\sqrt{\cos x}}\cdot \sin x}{2\sqrt{\cos x}}~~\text{(ans.)}$

$~(iv)~~y=\frac{1}{\sqrt{\log \sec x}}$

Sol. $~~y=\frac{1}{\sqrt{\log \sec x}} \\ \text{or,}~~y=\frac 1u~~[~u=\sqrt{\log \sec x}] \\ \text{or,}~~\frac{dy}{du}=-\frac{1}{u^2}$

Again, $~~u=\sqrt v \\ \text{or,}~~\frac{du}{dv}=\frac{1}{2\sqrt v}$

$~~v=\log(\sec x) \\ \text{or,}~~\frac{dv}{dx}=\frac{1}{\sec x} \cdot \sec x \tan x \\ \text{or,}~~\frac{dv}{dx}=\tan x \\ \therefore \frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}.\frac{dv}{dx} \\ \text{or,}~~\frac{dy}{dx}=(-1/u^2)\cdot \frac{1}{2\sqrt v} \cdot \tan x \\ \text{or,}~~ \frac{dy}{dx}=-\frac{\tan x}{2\log(\sec x) \cdot\sqrt{\log(\sec x)}}\\ \text{or,}~~\frac{dy}{dx}=-\frac 12 \tan x[\log(\sec x)]^{-3/2}~~\text{(ans.)}$

$~(v)~~y=\cos^{-1}\sqrt{2x-3}$

Sol. let $~~y=\cos^{-1}u\Rightarrow \frac{dy}{du}=-\frac{1}{\sqrt{1-u^2}}\\~~~[~\text{where}~~u=\sqrt{2x-3}~~]$

Again, suppose $u=\sqrt v \Rightarrow \frac{du}{dv}=\frac{1}{2\sqrt v}$

$~~v=2x-3 \Rightarrow \frac{dv}{dx}=2 \\ \therefore \frac{dy}{dx}=\frac{dy}{du} .\frac{du}{dv}. \frac{dv}{dx}\\ \text{or,}~~ \frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2\sqrt v} \cdot 2 \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt{1-(2x-3)}} \cdot \frac{1}{\sqrt{2x-3}} \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt{4-2x}\sqrt{2x-3}} \\ \text{or,}~~\frac{dy}{dx}=-\frac{1}{\sqrt 2} \cdot \frac{1}{\sqrt{2-x}\sqrt{2x-3}}~~\text{(ans.)}$

$~2.~~$ If the inverse function of $~~y=f(x)~$ is $~~x=g(y)~$ and $~f'(x)=\frac{1}{1+x^2},~$ then prove that, $~~g'(x)=1+[g(x)]^2.$

Sol. $~f'(x)=\frac{1}{1+x^2}\\ \text{or,}~~ f(x)=\displaystyle\int{\frac{dx}{1+x^2}}=\tan^{-1}x \\ \therefore y=\tan^{-1}x~~[\because y=f(x)] \\ \Rightarrow x=\tan y \\ \Rightarrow g(y)=\tan y\rightarrow(1)\\~~[\because x=g(y)]$

Hence, from $\,(1)\,$ we can say,

$~g(x)=\tan x ~~[~\text{replacing}~ ~y~~\text{by}~~ x] \\ \therefore g'(x)=\frac{d}{dx}(\tan x)=\sec^2x\rightarrow(2)$

Again, $~~~~1+[g(x)]^2\\=1+\tan^2x\\=\sec^2x \rightarrow(3)$

So, from $\,(2)\,$ and $\,(3)\,$ we can conclude that

$~~g'(x)=1+[g(x)]^2~~~\text{(proved)}$