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

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Question

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

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $y = {	an ^{ - 1}}\left( {\frac{x}{{1 + \sqrt {1 - {x^2}} }}} ight)$ and $z = {\sin ^{ - 1}}x$. Substitute $x = \sin 	heta$,where $	heta = {\sin ^{ - 1}}x$. Then $y = {	an ^{ - 1}}\left( {\frac{{\sin 	heta }}{{1 + \cos 	heta }}} ight)$. Using the trigonometric identities $\sin 	heta = 2\sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2\cos^2 \frac{	heta}{2}$,we get: $y = {	an ^{ - 1}}\left( {\frac{{2\sin \frac{	heta}{2} \cos \frac{	heta}{2}}}{{2\cos^2 \frac{	heta}{2}}}} ight) = {	an ^{ - 1}}\left( {	an \frac{	heta}{2}} ight) = \frac{	heta}{2}$. Since $	heta = {\sin ^{ - 1}}x$,we have $y = \frac{1}{2}{\sin ^{ - 1}}x$. Now,we need to find the derivative of $y$ with respect to $z$: $\frac{{dy}}{{dz}} = \frac{d}{{dz}}\left( {\frac{1}{2}z} ight) = \frac{1}{2}$. Thus,the differential coefficient is $\frac{1}{2}$.

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

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