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

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(B) Let $u = 	an^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} ight)$. Put $x = 	an 	heta$, then $	heta = 	an^{-1} x$.$u = 	an^{-1} \left( \frac{\sec

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$\frac{1}{4}$

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(B) Let $u = 	an^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} ight)$. Put $x = 	an 	heta$, then $	heta = 	an^{-1} x$.$u = 	an^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight) = 	an^{-1} \left( \frac{1 - \cos 	heta}{\sin 	heta} ight) = 	an^{-1} \left( 	an \frac{	heta}{2} ight) = \frac{	heta}{2} = \frac{1}{2} 	an^{-1} x$.Thus, $\frac{du}{dx} = \frac{1}{2(1+x^2)}$.Let $v = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$. For $|x| \le 1$, $v = 2 	an^{-1} x$.Thus, $\frac{dv}{dx} = \frac{2}{1+x^2}$.The derivative of $u$ with respect to $v$ is $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/(2(1+x^2))}{2/(1+x^2)} = \frac{1}{4}$.

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

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