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

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Question

(A) Let ${y_1} = {	an ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} ight)$ and ${y_2} = {	an ^{ - 1}}x$. Substitute $x = 	an 	heta$,so $	h

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(A) Let ${y_1} = {	an ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} ight)$ and ${y_2} = {	an ^{ - 1}}x$. Substitute $x = 	an 	heta$,so $	heta = {	an ^{ - 1}}x$. Then ${y_1} = {	an ^{ - 1}}\left( {\frac{{\sqrt {1 + {{	an }^2}	heta } - 1}}{{	an 	heta }}} ight) = {	an ^{ - 1}}\left( {\frac{{\sec 	heta - 1}}{{	an 	heta }}} ight) = {	an ^{ - 1}}\left( {\frac{{1 - \cos 	heta }}{{\sin 	heta }}} ight) = {	an ^{ - 1}}\left( {	an \frac{	heta }{2}} ight) = \frac{	heta }{2} = \frac{1}{2}{	an ^{ - 1}}x$. Now,differentiate ${y_1}$ with respect to ${y_2} = {	an ^{ - 1}}x$: $\frac{{d{y_1}}}{{d{y_2}}} = \frac{d}{{d{y_2}}}\left( {\frac{1}{2}{y_2}} ight) = \frac{1}{2}$.

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

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