Differentiation (Part-38) | S N De

 $~11(i)~~\log(\csc x+\cot x)$

Sol. let $~~y=\log(\csc x+\cot x) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\log(\csc x+\cot x)]\\=\frac{1}{\csc x+\cot x} \cdot \frac{d}{dx}(\csc x+\cot x)\\=\frac{1}{\csc x+\cot x} \cdot [\frac{d}{dx}(\csc x)+\frac{d}{dx}(\cot x)]\\=\frac{1}{\csc x+\cot x} \cdot [-\csc x \cot x +(-\csc^2x)]\\=\frac{1}{\csc x+\cot x} \cdot [-\csc x(\csc x+\cot x)]\\=-\csc x~~\text{(ans.)}$

Note : Here, $~~\csc x \rightarrow~~\text{cosec}~x.$

$~(ii)~~\log(x+\sqrt{x^2+a^2})$

Sol. let $~~y=\log(x+\sqrt{x^2+a^2}) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}~[\log(x+\sqrt{x^2+a^2})]\\=\frac{1}{x+\sqrt{x^2+a^2}}\cdot \frac{d}{dx}~(x+\sqrt{x^2+a^2})\\=\frac{1}{x+\sqrt{x^2+a^2}}.[\frac{d}{dx}(x)+\frac{d}{dx}({x^2+a^2})^{1/2}]\\=\frac{1}{x+\sqrt{x^2+a^2}} \cdot [1+\frac 12(x^2+a^2)^{\frac 12-1}\\~~~~\cdot \frac{d}{dx}(x^2+a^2)]\\=\frac{1}{x+\sqrt{x^2+a^2}} \cdot[1+\frac 12(x^2+a^2)^{-\frac 12}\cdot (2x+0)]\\=\frac{1}{x+\sqrt{x^2+a^2}} \cdot \left[1+\frac{x}{\sqrt{x^2+a^2}}\right]\\=\frac{1}{x+\sqrt{x^2+a^2}} \cdot \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\\=\frac{1}{\sqrt{x^2+a^2}}~~\text{(ans.)}$

$~(iii)~~\log(\sqrt{x-2}+\sqrt{x-3})$

Sol. let $~~y=\log(\sqrt{x-2}+\sqrt{x-3}) \\ \therefore \frac{dy}{dx}\\=\frac{d}{dx}[\log(\sqrt{x-2}+\sqrt{x-3})]\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}} \cdot \frac{d}{dx}(\sqrt{x-2}+\sqrt{x-3})\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}}  \cdot [\frac{d}{dx}(x-2)^{1/2}\\~~~~~~~+\frac{d}{dx}(x-3)^{1/2}]\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}}  \cdot [\frac 12(x-2)^{\frac 12-1}\\~~~~~~~+\frac 12 (x-3)^{\frac 12-1}]\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}}  \cdot [\frac 12 (x-2)^{-1/2}\\~~~~~~~+\frac 12(x-3)^{-1/2}]\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}}  \cdot \frac 12 \left(\frac{1}{\sqrt{x-2}}+\frac{1}{\sqrt{x-3}}\right)\\=\frac{1}{\sqrt{x-2}+\sqrt{x-3}}  \cdot \frac{\sqrt{x-3}+\sqrt{x-2}}{2\sqrt{(x-2)(x-3)}}\\=\frac{1}{2\sqrt{(x-2)(x-3)}}~~\text{(ans.)}$

$~(iv)~~\log\tan\left(\frac{\pi}{4}+\frac x2\right)$

Sol. let $~~y=\log\tan\left(\frac{\pi}{4}+\frac x2\right) \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[\log\tan\left(\frac{\pi}{4}+\frac x2\right)]\\=\frac{1}{\tan(\pi/4+x/2)} \cdot \frac{d}{dx}[\tan(\pi/4+x/2)]\\=\cot(\pi/4+x/2) \cdot \sec^2(\pi/4+x/2)\\~~~\cdot \frac{d}{dx}(\pi/4+x/2)\\=\frac{\cos(\pi/4+x/2)}{\sin(\pi/4+x/2)} \cdot \sec^2(\pi/4+x/2)\cdot (0+\frac 12 \cdot 1)\\=\frac{1}{\sin(\pi/4+x/2)} \cdot \sec(\pi/4+x/2)\cdot \frac 12\\=\frac{1}{2\sin(\pi/4+x/2)\cos(\pi/4+x/2)}\\=\frac{1}{\sin[2(\pi/4+x/2)]}\\=\frac{1}{\sin(\pi/2+x)}\\=\frac{1}{\cos x}\\=\sec x~~\text{(ans.)}$

$~12(i)~~\sec^n x$

Sol. let $~~y=\sec^nx \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(\sec^nx)\\=n\sec^{n-1}x\cdot \frac{d}{dx}(\sec x)\\=n\sec^{n-1}x \cdot (\sec x \tan x)\\=n\sec^{n-1+1}x \tan x\\=n\sec^n x\tan x~~\text{(ans.)}$

$~(ii)~~\csc^6x$

Sol. let $~~y=\csc^6x \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(\csc^6x)\\=6\csc^{5}x\cdot \frac{d}{dx}(\csc x)\\=6\csc^5x(-\csc x\cot x)\\=-6\csc^6x\cot x~~\text{(ans.)}$

Note : $~~\csc x \rightarrow ~\text{cosec}~x$

$~(iii)~~\tan ax+\cot bx$

Sol. $~~y=\tan ax+\cot bx \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}~(\tan ax+\cot bx)\\=\frac{d}{dx}~(\tan ax)+\frac{d}{dx}~(\cot bx)\\=\sec^2ax\cdot \frac{d}{dx}~(ax)+(-\csc^2bx)\cdot \frac{d}{dx}~(bx)\\=\sec^2ax ~\cdot (a \cdot 1)-\csc^2bx~\cdot (b \cdot 1)\\=a\sec^2ax-b\csc^2bx~~\text{(ans.)} $

$~(iv)~~(\sin^{-1}x)^m$

Sol. let $~~y=(\sin^{-1}x)^m \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[(\sin^{-1}x)^m]\\=m(\sin^{-1}x)^{m-1} \cdot \frac{d}{dx}(\sin^{-1}x)\\=m(\sin^{-1}x)^{m-1} \cdot~\frac{1}{\sqrt{1-x^2}}\\=\frac{m(\sin^{-1}x)^{m-1}}{\sqrt{1-x^2}}~~\text{(ans.)}$

$~(v)~~(\cot^{-1}x)^4$

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

Mathematics by S.N.DE for class 12 (WBHS,WBJEE, JEE- Main, JEE- Advanced) Paperback

$~(vi)~~\left(\cos^{-1}\frac xa\right)^3$

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

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

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