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

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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 \

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$1/2$

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(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

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