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

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(D) Let $u = 	an ^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight)$ and $v = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,where $	heta = \s

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) Let $u = 	an ^{-1}\left(\frac{x}{\sqrt{1-x^2}}ight)$ and $v = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,where $	heta = \sin ^{-1} x$.Then $u = 	an ^{-1}\left(\frac{\sin 	heta}{\sqrt{1-\sin ^2 	heta}}ight) = 	an ^{-1}\left(\frac{\sin 	heta}{\cos 	heta}ight) = 	an ^{-1}(	an 	heta) = 	heta$.And $v = \sin ^{-1}(3 \sin 	heta - 4 \sin ^3 	heta) = \sin ^{-1}(\sin 3 	heta) = 3 	heta$.We need to find $\frac{du}{dv}$.Since $u = 	heta$ and $v = 3 	heta$,we have $\frac{du}{d	heta} = 1$ and $\frac{dv}{d	heta} = 3$.Therefore,$\frac{du}{dv} = \frac{du/d	heta}{dv/d	heta} = \frac{1}{3}$.

Differentiation of $\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is $....$

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