Differentiation (Part-38) | S N De

Differentiation (Part-26) | S N De

 $~14(i)~~\log(ax)^x$

Solution. 

let $~~y=\log(ax)^x=x\log(ax) \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}\left[x\log(ax)\right]\\=\frac{d}{dx}(x)~\cdot \log(ax)+x ~\cdot \frac{d}{dx}[\log(ax)]\\=1~\cdot \log(ax)+x~\cdot\frac{1}{ax} \cdot \frac{d}{dx}(ax)\\=\log(ax)+\frac 1a~\cdot a~\frac{d}{dx}(x)\\=\log(ax)+1 \cdot 1\\=\log(ax)+1~~\text{(ans.)}$

$~(ii)~~x^5+\frac 6x-\tan(x^2)$

Solution. 

Let $~~y=x^5+\frac 6x-\tan(x^2) \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[x^5+6x^{-1}-\tan(x^2)]\\=\frac{d}{dx}(x^5)+6\frac{d}{dx}(x^{-1})-\frac{d}{dx}[\tan(x^2)]\\=5x^4+6(-1)\cdot x^{-1-1}\\~~~~~-\sec^2(x^2)\cdot \frac{d}{dx}(x^2)\\=5x^4-6x^{-2}-\sec^2(x^2)~\cdot 2x\\=5x^4-\frac{6}{x^2}-2x\sec^2(x^2)~~\text{(ans.)}$

$~(iii)~~\log \sin(\sqrt{x^2+1})$

Solution. 

Let $~~y=\log \sin(\sqrt{x^2+1}) \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[\log \sin(\sqrt{x^2+1})]\\=\frac{1}{\sin(\sqrt{x^2+1})}~\cdot \frac{d}{dx}[\sin(\sqrt{x^2+1})]\\=\frac{1}{\sin(\sqrt{x^2+1})} \cdot \cos(\sqrt{x^2+1})\\~~~~ \cdot \frac{d}{dx} (\sqrt{x^2+1})\\=\frac{1}{\sin(\sqrt{x^2+1})} \cdot \cos(\sqrt{x^2+1}) \\~~~~\cdot \frac{d}{dx}[(x^2+1)^{1/2}]\\=\frac{1}{\sin(\sqrt{x^2+1})} \cdot \cos(\sqrt{x^2+1}) \\~~~~\cdot \frac 12(x^2+1)^{\frac 12-1} \cdot \frac{d}{dx}(x^2+1)\\=\frac{1}{\sin(\sqrt{x^2+1})} \cdot \cos(\sqrt{x^2+1}) \\~~~~\cdot \frac 12(x^2+1)^{-\frac 12} \cdot (2x+0)\\=\frac{1}{\sin(\sqrt{x^2+1})} \cdot \cos(\sqrt{x^2+1}) \cdot \frac{2x}{2\sqrt{x^2+1}}\\=\frac{x}{\sqrt{x^2+1}}~\cdot \cot(\sqrt{x^2+1})~~\text{(ans.)}$

$~(iv)~~2x\tan^{-1}x-\log(1+x^2)$

Solution. 

Let $~~y=2x\tan^{-1}x-\log(1+x^2) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[2x\tan^{-1}x-\log(1+x^2)]\\=2~\frac{d}{dx}(x)\cdot \tan^{-1}x+2x~\cdot \frac{d}{dx}(\tan^{-1}x)\\~~~~-\frac{d}{dx}[\log(1+x^2)]\\=2 \cdot1\cdot\tan^{-1}x+2x~\cdot \frac{1}{1+x^2}\\~~~~-\frac{1}{1+x^2}\cdot \frac{d}{dx}(1+x^2)\\=2\tan^{-1}x+\frac{2x}{1+x^2}-\frac{1}{1+x^2}\cdot (0+2x)\\=2\tan^{-1}x+\frac{2x}{1+x^2}-\frac{2x}{1+x^2}\\=2\tan^{-1}x~~\text{(ans.)}$

$~(v)~~\cot^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}~~~(0<x<\pi/2)$

Solution. 

Let $~~y=\cot^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}} \rightarrow(1)$

Now, $~\cot^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}\\= \cot^{-1}\sqrt{\frac{\sin^2(x/2)+\cos^2(x/2)-2\sin x/2 \cos x/2}{\sin^2(x/2)+\cos^2(x/2)+2\sin x/2 \cos x/2}}\\~~~[~\because~\sin x\\=\sin(2 \cdot x/2)]\\=2\sin(x/2)\cos(x/2)~]\\=\cot^{-1}\sqrt{\frac{(\cos(x/2)-\sin(x/2))^2}{(\cos(x/2)+\sin(x/2))^2}}\\=\cot^{-1}\left[\frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)+\sin(x/2)}\right]\\=\cot^{-1}\left[\frac{\frac{\cos(x/2)}{\sin(x/2)}-\frac{\sin(x/2)}{\sin(x/2)}}{\frac{\cos(x/2)}{\sin(x/2)}+\frac{\sin(x/2)}{\sin(x/2)}}\right]\\=\cot^{-1}\left[\frac{\cot(x/2)-1}{\cot(x/2)+1}\right]\\=\cot^{-1}\left[\frac{\cot(\pi/4)\cdot\cot(x/2)-1}{\cot(\pi/4)+\cot(x/2)}\right]\\=\cot^{-1}[\cot(\pi/4+x/2)]\\=\frac{\pi}{4}+\frac x2\rightarrow(2)$

Hence, from $~(1),~(2)~$ we get,

$~y=\frac{\pi}{4}+\frac x2 \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}(\frac{\pi}{4}+\frac x2)\\=\frac{d}{dx}(\pi/4)+\frac{d}{dx}(x/2)\\=0+\frac 12~\cdot \frac{d}{dx}(x)\\=\frac 12 \cdot 1\\=\frac 12~~\text{(ans.)}$

$~(vi)~~\frac{2x+1}{x^3}-\log\tan x+x\sin^{-1}(x^2)$

Solution. 

Let $~~y=\frac{2x+1}{x^3}-\log\tan x+x\sin^{-1}(x^2) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}\left[\frac{2x}{x^3}+\frac{1}{x^3}-\log\tan x+x\sin^{-1}(x^2)\right]\\=2~\cdot\frac{d}{dx}(x^{-2})+\frac{d}{dx}(x^{-3})\\~~~~-\frac{d}{dx}(\log\tan x)+\frac{d}{dx}[x\sin^{-1}(x^2)]\\=2 \cdot(-2x^{-2-1})+(-3x^{-3-1})\\~~~~-\frac{1}{\tan x}\cdot \frac{d}{dx}(\tan x)+\frac{d}{dx}(x)~\cdot \sin^{-1}(x^2)\\~~~~+x~\cdot\frac{d}{dx}[\sin^{-1}(x^2)]\\=-4x^{-3}-3x^{-4}-\frac{1}{\tan x}\cdot \sec^2x\\~~~~+1\cdot \sin^{-1}(x^2)+x~\cdot \frac{1}{\sqrt{1-(x^2)^2}}~\cdot \frac{d}{dx}(x^2)\\=-\frac{4}{x^3}-\frac{3}{x^4}-\frac{\cos x}{\sin x} \cdot \sec^2 x\\~~~~+\sin^{-1}(x^2)+\frac{x}{\sqrt{1-x^4}} \cdot 2x\\=-\frac{4}{x^3}-\frac{3}{x^4}-\frac{1}{\sin x} \cdot \sec x\\~~~~+\sin^{-1}(x^2)+\frac{2x^2}{\sqrt{1-x^4}}\\=-\frac{4}{x^3}-\frac{3}{x^4}-\csc x \cdot \sec x\\~~~~+\sin^{-1}(x^2)+\frac{2x^2}{\sqrt{1-x^4}}~~\text{(ans.)}$