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

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Question

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

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(A) Let ${y_1} = {	an ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} ight)$ and ${y_2} = {\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} ight)$.Substitute $x = 	an 	heta$,where $	heta = {	an ^{ - 1}}x$.Then,${y_1} = {	an ^{ - 1}}\left( \frac{{2	an 	heta }}{{1 - {{	an }^2}	heta }} ight) = {	an ^{ - 1}}(	an 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.Similarly,${y_2} = {\sin ^{ - 1}}\left( \frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }} ight) = {\sin ^{ - 1}}(\sin 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.Now,we need to find the derivative $\frac{{d{y_1}}}{{d{y_2}}}$.Since ${y_1} = 2{	an ^{ - 1}}x$ and ${y_2} = 2{	an ^{ - 1}}x$,we have ${y_1} = {y_2}$.Therefore,$\frac{{d{y_1}}}{{d{y_2}}} = \frac{d}{{d{y_2}}}({y_2}) = 1$.

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

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