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

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

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$\frac{\sqrt{3}}{10}$

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(B) Let $f = 	an ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}ight)$.Put $x = 	an 	heta$,so $	heta = 	an ^{-1} x$.Then $f = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an ^{-1}\left(\frac{1 - \cos 	heta}{\sin 	heta}ight) = 	an ^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2} = \frac{1}{2} 	an ^{-1} x$.Thus,$\frac{df}{dx} = \frac{1}{2(1+x^{2})}$.Let $g = 	an ^{-1}\left(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}ight)$.Put $x = \sin 	heta$,so $	heta = \sin ^{-1} x$.Then $g = 	an ^{-1}\left(\frac{2 \sin 	heta \cos 	heta}{1-2 \sin ^{2} 	heta}ight) = 	an ^{-1}(	an 2	heta) = 2	heta = 2 \sin ^{-1} x$.Thus,$\frac{dg}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.Therefore,$\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{1}{2(1+x^{2})} \cdot \frac{\sqrt{1-x^{2}}}{2} = \frac{\sqrt{1-x^{2}}}{4(1+x^{2})}$.At $x = \frac{1}{2}$,$\frac{df}{dg} = \frac{\sqrt{1-(1/2)^{2}}}{4(1+(1/2)^{2})} = \frac{\sqrt{3/4}}{4(5/4)} = \frac{\sqrt{3}/2}{5} = \frac{\sqrt{3}}{10}$.

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

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