Differentiation (Part-38) | S N De

 

Find the derivative w.r.t. $~~x~~[7-19]:$

$~~7(i)~2\tan^{-1}~\frac xa$

Sol. $~~\frac{d}{dx}\left(2\tan^{-1}~\frac xa\right)\\=\frac{2}{1+(x/a)^2}~\cdot \frac{d}{dx}(x/a)\\=\frac{2a^2}{a^2+x^2}~\cdot \frac 1a\\=\frac{2a}{a^2+x^2}~~\text{(ans.)}$

$~~(ii)~~\log(\cot^{-1}x)$

Sol. $~~\frac{d}{dx}[\log(\cot^{-1}x)]\\=\frac{1}{\cot^{-1}x}~\cdot \frac{d}{dx}(\cot^{-1}x)\\=\frac{1}{\cot^{-1}x}\cdot \left(-\frac{1}{1+x^2}\right)\\=-\frac{1}{(1+x^2)\cot^{-1}x}~~\text{(ans.)}$

$~(iii)~\sec(\tan^{-1}x)$

Sol. $~~\frac{d}{dx}[\sec(\tan^{-1}x)]\\=\sec(\tan^{-1}x)\tan(\tan^{-1}x))\frac{d}{dx}(\tan^{-1}x)\\=\sec(\tan^{-1}x) \cdot x \cdot \frac{1}{1+x^2}\\=\frac{x\sec(\tan^{-1}x)}{1+x^2}~\rightarrow(1)$

let $~~\tan^{-1}x=\theta~~$ so that $~~\tan\theta=x$

$~~\therefore \sec\theta\\=\sqrt{1+\tan^2\theta}\\=\sqrt{1+x^2}~~[~\because \sec\theta >0~]\rightarrow(2)$

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

$~~\frac{d}{dx}[\sec(\tan^{-1}x)]\\=\frac{x~\sqrt{1+x^2}}{1+x^2}\\=\frac{x}{\sqrt{1+x^2}}~~\text{(ans.)}$

$~~8(i)~~2\sec^{-1}2x-3\sin^{-1} x+\cos^{-1}(x^2)$

Sol. let $~y=2\sec^{-1}2x-3\sin^{-1}x+\cos^{-1}(x^2) \\ \therefore~\frac{dy}{dx}\\=2\frac{d}{dx}(\sec^{-1}2x)-3\frac{d}{dx}(\sin ^{-1}x)\\~~~~~+\frac{d}{dx}[\cos^{-1}(x^2)]\\=2\cdot \frac{1}{2x\sqrt{(2x)^2-1}}\cdot \frac{d}{dx}(2x)-3\left(\frac{1}{\sqrt{1-x^2}}\right)\\~~~~~+\left(-\frac{1}{\sqrt{1-(x^2)^2}}\right)\cdot \frac{d}{dx}(x^2)\\=2\cdot \frac{1}{2x\sqrt{(2x)^2-1}}\cdot2-\frac{3}{\sqrt{1-x^2}}\\~~~~~-\frac{1}{\sqrt{1-x^4}}\cdot 2x\\=\frac{2}{x\sqrt{4x^2-1}}-\frac{3}{\sqrt{1-x^2}}\\~~~~~-\frac{2x}{\sqrt{1-x^4}}~~\text{(ans.)}$

$~~(ii)~~2\csc^{-1}(3x)+3\csc2x$

Sol. let $~~y=2\csc^{-1}(3x)+3\csc2x\\ \therefore~\frac{dy}{dx}\\=2\frac{d}{dx}(\csc^{-1}(3x))+3\frac{d}{dx}(\csc 2x)\\=2\cdot \left(-\frac{1}{3x\sqrt{(3x)^2-1}}\right)\cdot \frac{d}{dx}(3x)\\~~~~~~~+3(-\csc2x\cot2x)\cdot \frac{d}{dx}(2x)\\=\frac{-2}{3x\sqrt{9x^2-1}}\cdot 3-3\csc2x\cot2x \cdot 2\\=\frac{-2}{x\sqrt{9x^2-1}}-6\csc 2x \cot 2x~\text{(ans.)}$

$~(iii)~~\cos^{-1}x +\cos^{-1}\sqrt{1-x^2}$

Sol. let $~~y=\cos^{-1} x+\cos^{-1}\sqrt{1-x^2}\rightarrow(1) $

Also let $~~\cos^{-1} x=\theta \rightarrow(2)\\ \therefore x=\cos \theta$

Again, $~~\cos^{-1}\sqrt{1-x^2}\\=\cos^{-1}\sqrt{1-\cos^2\theta}\\=\cos^{-1}\sqrt{\sin^2\theta}\\=\cos^{-1}(\sin\theta)\\=\cos^{-1}\left[\cos(\pi/2-\theta)\right]\\=\pi/2-\theta\rightarrow(3)$

Hence, from $~~(1),~(2),~(3)~$ we get,

$~~~~y=\theta+\pi/2-\theta \\ \therefore y=\pi/2 \\ \text{so, }~~\frac{dy}{dx}=\frac{d}{dx}(\pi/2)=0~~\text{(ans.)}$