The differential coefficient of {tan ^{ - 1}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = {	an ^{ - 1}}\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} $ and $z = {\cos ^{ - 1}}({x^2})$.Substitute ${x^2} = \cos 2	heta $,where $0 \le \

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(A) Let $y = {	an ^{ - 1}}\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} $ and $z = {\cos ^{ - 1}}({x^2})$.Substitute ${x^2} = \cos 2	heta $,where $0 \le 	heta \le \pi/2$.Then $y = {	an ^{ - 1}}\sqrt {\frac{{1 - \cos 2	heta }}{{1 + \cos 2	heta }}} = {	an ^{ - 1}}\sqrt {\frac{{2{{\sin }^2}	heta }}{{2{{\cos }^2}	heta }}} = {	an ^{ - 1}}(	an 	heta ) = 	heta $.Also,$z = {\cos ^{ - 1}}(\cos 2	heta ) = 2	heta $.We need to find $\frac{{dy}}{{dz}}$.Since $y = 	heta $ and $z = 2	heta $,we have $\frac{{dy}}{{d	heta }} = 1$ and $\frac{{dz}}{{d	heta }} = 2$.Therefore,$\frac{{dy}}{{dz}} = \frac{{dy/d	heta }}{{dz/d	heta }} = \frac{1}{2}$.

The differential coefficient of ${\tan ^{ - 1}}\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} $ with respect to ${\cos ^{ - 1}}({x^2})$ is

Similar Questions

Differentiate $\tan ^{-1} x$ with respect to $\cot ^{-1} x$ for $x \in R$.

If $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$,then $\frac{d y}{d x}$ is

$\frac{d}{dx}\left( \tan^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right)$ is equal to

If $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$,then $\left(\frac{d y}{d x}\right)$ at $x=0$ is

The derivative of $y = \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$ with respect to $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module