Differentiation (Part-38) | S N De

 

$~5.~$ If $~y=\left[(\tan x)^{\tan x}\right]^{\tan x}$

Sol. $~~~y=\left[(\tan x)^{\tan x}\right]^{\tan x} \rightarrow(1) \\ \therefore  \log y=\tan x \log(\tan x)^{\tan x} \\ \text{or,}~~  \log y=\tan^2x \log(\tan x) \\ \therefore \frac 1y \frac{dy}{dx}=2\tan x \sec^2x \log(\tan x)\\~~~~~~~~~~~+\tan x \sec^2x \\ \text{or,}~~  \frac{dy}{dx}\\= \left[(\tan x)^{\tan x}\right]^{\tan x}\left[2\tan x \sec^2x \log(\tan x)\\~~~~~~+\tan x \sec^2x \right]$

At $~x=\pi/4~,$ the value of $~y=1~~[\text{By (1)},~~\because ~\tan(\pi/4)=1]$

$~~\therefore ~~\left[\frac{dy}{dx}\right]_{x=\pi/4}\\=1 \times [2 \times 1 \times (\sqrt 2)^2 \times \log 1+1 \times (\sqrt 2)^2]\\~~[\because ~\sec(\pi/4)=\sqrt 2,~\tan(\pi/4)=1]\\=2~~~\text{(ans.)}$

$~6.~$ If $~~y=\frac{1+\sin\theta+\cos \theta}{1+\sin\theta-\cos \theta},~~$ show that 

$~~\frac{dy}{d\theta}+\frac{1}{1-\cos\theta}=0.$

Sol. $~~~~y=\frac{1+\sin\theta+\cos \theta}{1+\sin\theta-\cos \theta} \\ \text{or,}~~  y=\frac{\sin \theta+ (1+\cos\theta)}{\sin\theta+(1-\cos\theta)} \\ \text{or,}~~ y=\frac{2\sin(\theta/2)\cos(\theta/2)+2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)+2\sin^2(\theta/2)}  \\ \text{or,}~~  y=\frac{2\cos(\theta/2)[\sin(\theta/2)+\cos(\theta/2)]}{2\sin(\theta/2)[\sin(\theta/2)+\cos(\theta/2)]} \\ \therefore y=\cot(\theta/2) \\ \text{or,}~~\frac{dy}{d\theta}=-\frac 12 \csc^2(\theta/2)  \\ \text{or,}~~ \frac{dy}{d\theta}=-\frac 12 \cdot \frac{1}{\sin^2(\theta/2)}  \\ \text{or,}~~ \frac{dy}{d\theta}=-\frac{1}{1-\cos\theta}  \\ \text{or,}~~ \frac{dy}{d\theta}+\frac{1}{1-\cos \theta}=0~~\text{(proved)}$

$\,7.\,$ If $~\tan y=\frac{\tan x+\sec x-1}{\tan x-\sec x+1},~$ show that , $~\frac{dy}{dx}=\frac 12.$

Sol. $~~\tan y\\=\frac{\tan x+\sec x-1}{\tan x-\sec x+1}\\=\frac{\sin x+1-\cos x}{\sin x-1+\cos x}\\=\frac{2\sin(x/2) \cos (x/2)+2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)-2\sin^2(x/2)}\\=\frac{2\sin(x/2)[\cos(x/2)+\sin(x/2)]}{2\sin(x/2)[\cos(x/2)-\sin(x/2)]}\\=\frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}\\=\frac{1+\tan(x/2)}{1-\tan(x/2)}\\=\frac{\tan(\pi/4)+\tan(x/2)}{1-\tan(\pi/4)\tan(x/2)}\\=\tan(\frac{\pi}{4}+\frac x2) \\ \therefore y=\frac{\pi}{4}+\frac x2 \\ \text{or,}~~\frac{dy}{dx}=\frac 12~~\text{(ans.)}$

$~8.~$ If $~y=\tan^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)~~$ show that, $~~\frac{dy}{dx}=-1$

Sol.  $~y=\tan^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)  \\ \text{or,}~~ y=\tan^{-1} \left(\frac{\frac ab-\tan x}{1+\frac ab \tan x}\right) \\ \text{or,}~~ y=\tan^{-1}\left(\frac ab\right)-\tan^{-1}(\tan x)  \\ \text{or,}~~ y=\tan^{-1}(a/b)-x \\ \therefore \frac{dy}{dx}=0-\frac{d}{dx}(x)\\ \text{or,}~~\frac{dy}{dx}=-1~~\text{(ans.)}$

$~9.~$ If $~\cos y=x\cos(a+y),~$ prove that, $~\frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a},~$ where $\,a \neq 0\,$ is constant.

Sol. $~~\cos y=x\cos(a+y) \\ \therefore x=\frac{\cos y}{\cos(a+y)} \\ \text{or,}~~\frac{dx}{dy}=\frac{-\sin y\cos(a+y)+\sin(a+y)\cos y}{\cos^2(a+y)} \\ \text{or,}~~\frac{dx}{dy}=\frac{\sin(a+y-y)}{\cos^2(a+y)} \\ \therefore \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}~~\text{(proved)}$

$\,10.\,$ If $~y^x=e^{y-x}~$ prove that, $\,\frac{dy}{dx}=\frac{(\log ey)^2}{\log y}$

Sol. $~~y^x=e^{y-x} \\ \text{or,}~~  \log(y^x)=\log(e^{y-x}) \\ \text{or,}~~  x \log y=(y-x) \log e \\ \text{or,}~~  x \log y=y-x \\ \text{or,}~~  \log y=\frac yx-1 \\ \text{or,}~~  \log y+1=\frac yx \\ \text{or,}~~  \log y+\log e=\frac yx \\ \text{or,}~~  \log(ey)=\frac yx \\ \text{or,}~~  x=\frac{y}{\log(ey)}  \\ \therefore \frac{dx}{dy}= \frac{\log(ey).1-y\cdot~ \frac{1}{ey}.~ e}{(\log ey)^2} \\ \text{or,}~~  \frac{dx}{dy}=\frac{\log(ey)-1}{(\log ey)^2}\\ \text{or,}~~  \frac{dx}{dy}=\frac{\log (ey)-\log e}{(\log ey)^2} \\ \text{or,}~~  \frac{dx}{dy}=\frac{\log\left(\frac{ey}{e}\right)}{(\log ey)^2}=\frac{ \log y}{(\log ey)^2} \\ \therefore \frac{dy}{dx}=\frac{(\log ey)^2}{\log y}~~\text{(proved)}$

$~11(i)~$ If $~y=\frac x2\sqrt{x^2-a^2}-\frac{a^2}{2}\log(x+\sqrt{x^2-a^2}),~$ show that, $~~\frac{dy}{dx}=\sqrt{x^2-a^2}$

Sol. $~y=\frac x2\sqrt{x^2-a^2}-\frac{a^2}{2}\log(x+\sqrt{x^2-a^2})\\ \therefore 2~y =x\sqrt{x^2-a^2}-a^2 \log(x+\sqrt{x^2-a^2}) \\ \text{or,}~~ 2~ \frac{dy}{dx}=\sqrt{x^2-a^2}\\~~~+x \cdot \frac 12(x^2-a^2)^{-1/2}\cdot 2x \\~~~~-a^2 \cdot\frac{1}{x+\sqrt{x^2-a^2}} \cdot \frac{d}{dx}(x+\sqrt{x^2-a^2}) \\ \text{or,}~~ 2~\frac{dy}{dx}=\sqrt{x^2-a^2}+\frac{x^2}{\sqrt{x^2-a^2}}\\~~~~-\frac{a^2}{x+\sqrt{x^2-a^2}} \cdot \left[1+\frac 12 (x^2-a^2)^{-1/2}\cdot 2x\right]\\ \text{or,}~~ 2~\frac{dy}{dx}=\frac{x^2-a^2+x^2}{\sqrt{x^2-a^2}}\\ ~~~~~~~~~~~-\frac{a^2}{x+\sqrt{x^2-a^2}}\times\left[1+\frac{x}{\sqrt{x^2-a^2}}\right] \\ \text{or,}~~ 2~\frac{dy}{dx}=\frac{2x^2-a^2}{\sqrt{x^2-a^2}}-\frac{a^2}{x+\sqrt{x^2-a^2}} \cdot \frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}} \\ \text{or,}~~ 2~\frac{dy}{dx}=\frac{2x^2-a^2}{\sqrt{x^2-a^2}}-\frac{a^2}{\sqrt{x^2-a^2}} \\ \text{or,}~~ 2~\frac{dy}{dx}=\frac{ 2x^2-a^2-a^2}{\sqrt{x^2-a^2}} \\ \text{or,}~~2~\frac{dy}{dx}=\frac{2(x^2-a^2)}{\sqrt{x^2-a^2}} \\ \text{or,}~~ \frac{dy}{dx}=\sqrt{x^2-a^2}~~\text{(proved)} $ 

$~11(ii)~~$ If $~y=2\sin^{-1}~\frac{x-2}{\sqrt 6}-\sqrt{2+4x-x^2},~$ show that, $~~\frac{dy}{dx}~$ at $~x=2~$ is $~\frac{2}{\sqrt 6}.$

Sol. $~y=2\sin^{-1}~\frac{x-2}{\sqrt 6}-\sqrt{2+4x-x^2} \\ \therefore \frac{dy}{dx}=2 \cdot  \frac{1}{\sqrt{1-\left(\frac{x-2}{\sqrt 6}\right)^2}} \cdot \frac{1}{\sqrt 6}\\~~~~~~~~~~~~-\frac 12 \cdot \frac{4-2x}{\sqrt{2+4x-x^2}} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2}{\sqrt 6} \cdot  \frac{\sqrt 6}{\sqrt{6-(x^2-4x+4)}} \\~~~~~~~~~~~~-\frac{2-x}{\sqrt{2+4x-x^2}} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2}{\sqrt{2+4x-x^2}}-\frac{2-x}{\sqrt{2+4x-x^2}} \\ \text{or,}~~ \frac{dy}{dx}=\frac{2-(2-x)}{\sqrt{2+4x-x^2}} \\ \text{or,}~~ \frac{dy}{dx}=\frac{x}{\sqrt{2+4x-x^2}} \\ \therefore \left[\frac{dy}{dx}\right]_{x=2}=\frac{2}{\sqrt{2+4 \times 2-2^2}}=\frac{2}{\sqrt 6}~~\text{(proved)}$