The derivative of tan^{-1} left( frac{x}{sqrt | vedclass

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(B) Let $u = 	an^{-1} \left( \frac{x}{\sqrt{1 - x^2}} ight)$ and $v = \sin^{-1}(x)$.Substitute $x = \sin(	heta)$,where $	heta = \sin^{-1}(x)$.The

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(B) Let $u = 	an^{-1} \left( \frac{x}{\sqrt{1 - x^2}} ight)$ and $v = \sin^{-1}(x)$.Substitute $x = \sin(	heta)$,where $	heta = \sin^{-1}(x)$.Then,$u = 	an^{-1} \left( \frac{\sin(	heta)}{\sqrt{1 - \sin^2(	heta)}} ight) = 	an^{-1} \left( \frac{\sin(	heta)}{\cos(	heta)} ight) = 	an^{-1}(	an(	heta)) = 	heta$.Since $v = \sin^{-1}(x) = 	heta$,we have $u = v$.Therefore,the derivative of $u$ with respect to $v$ is $\frac{du}{dv} = \frac{d}{dv}(v) = 1$.

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

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