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

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(B) Let $y = {	an ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} ight)$ and $z = {	an ^{ - 1}}\left( {\frac{{2x\sqrt {1 - {x^2}} }}{{1 - 2{x^2

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$\frac{1}{4}$

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(B) Let $y = {	an ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} ight)$ and $z = {	an ^{ - 1}}\left( {\frac{{2x\sqrt {1 - {x^2}} }}{{1 - 2{x^2}}}} ight)$.For $y$,put $x = 	an 	heta$,then $	heta = {	an ^{ - 1}}x$.$y = {	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{{dy}}{{dx}} = \frac{1}{2(1 + {x^2})}$.For $z$,put $x = \sin 	heta$,then $	heta = {\sin ^{ - 1}}x$.$z = {	an ^{ - 1}}\left( {\frac{{2\sin 	heta \cos 	heta }}{{1 - 2\sin^2 	heta }}} ight) = {	an ^{ - 1}}\left( {\frac{{\sin 2	heta }}{{\cos 2	heta }}} ight) = {	an ^{ - 1}}(	an 2	heta ) = 2	heta = 2{\sin ^{ - 1}}x$.Thus,$\frac{{dz}}{{dx}} = \frac{2}{{\sqrt {1 - {x^2}} }}$.Now,$\frac{{dy}}{{dz}} = \frac{{dy/dx}}{{dz/dx}} = \frac{1}{2(1 + {x^2})} 	imes \frac{{\sqrt {1 - {x^2}} }}{2} = \frac{{\sqrt {1 - {x^2}} }}{{4(1 + {x^2})}}$.At $x = 0$,$\frac{{dy}}{{dz}} = \frac{{\sqrt {1 - 0} }}{{4(1 + 0)}} = \frac{1}{4}$.

The derivative of ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\tan ^{ - 1}}\left( {\frac{{2x\sqrt {1 - {x^2}} }}{{1 - 2{x^2}}}} \right)$ at $x = 0$ is:

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