The derivative of {sin ^{ - 1}}left( {frac{{2 | vedclass

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(B) Let $p = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} ight)$.Using the substitution $x = 	an 	heta$,we get $p = {\sin ^{ - 1}}(\sin 2	heta

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$1$

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(B) Let $p = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} ight)$.Using the substitution $x = 	an 	heta$,we get $p = {\sin ^{ - 1}}(\sin 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.Now,let $q = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)$.Using the same substitution $x = 	an 	heta$,we get $q = {\cos ^{ - 1}}(\cos 2	heta) = 2	heta = 2{	an ^{ - 1}}x$.Since $p = q$,the derivative of $p$ with respect to $q$ is $\frac{{dp}}{{dq}} = \frac{d}{{dq}}(q) = 1$.

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

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