Derivative of tan ^{-1} sqrt{frac{1-x}{1+x}} | vedclass

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(D) Let $y = 	an ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.For $y$,substitute $x = \cos 2	heta$. Then $\sqrt{\frac{1-\cos 2	het

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$\frac{1}{6}$

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(D) Let $y = 	an ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.For $y$,substitute $x = \cos 2	heta$. Then $\sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} = \sqrt{\frac{2\sin^2 	heta}{2\cos^2 	heta}} = 	an 	heta$.So,$y = 	an^{-1}(	an 	heta) = 	heta = \frac{1}{2} \cos^{-1} x$.For $z$,substitute $x = \cos 	heta$. Then $z = \cos^{-1}(\cos 3	heta) = 3	heta = 3 \cos^{-1} x$.Now,differentiate $y$ and $z$ with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} ight) = \frac{-1}{2\sqrt{1-x^2}}$.$\frac{dz}{dx} = 3 \left( \frac{-1}{\sqrt{1-x^2}} ight) = \frac{-3}{\sqrt{1-x^2}}$.Finally,$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-1/(2\sqrt{1-x^2})}{-3/\sqrt{1-x^2}} = \frac{1}{6}$.

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\cos ^{-1}\left(4 x^3-3 x\right)$ is

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