If u = tan^{-1} left( frac{sqrt{1 + x^2} - 1} | vedclass

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(C) Given $u = 	an^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} ight)$ and $v = 2 	an^{-1} x$.Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.Subst

Vedclass · Accepted Answer

$1/4$

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(C) Given $u = 	an^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} ight)$ and $v = 2 	an^{-1} x$.Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.Substituting $x = 	an 	heta$ in $u$:$u = 	an^{-1} \left( \frac{\sqrt{1 + 	an^2 	heta} - 1}{	an 	heta} ight) = 	an^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight)$$u = 	an^{-1} \left( \frac{1 - \cos 	heta}{\sin 	heta} ight) = 	an^{-1} \left( \frac{2 \sin^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)} ight)$$u = 	an^{-1} (	an(	heta/2)) = 	heta/2 = \frac{1}{2} 	an^{-1} x$.Now,$v = 2 	an^{-1} x$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.$\frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{1 + x^2}$.$\frac{dv}{dx} = 2 \cdot \frac{1}{1 + x^2}$.Therefore,$\frac{du}{dv} = \frac{\frac{1}{2(1 + x^2)}}{\frac{2}{1 + x^2}} = \frac{1}{4}$.

If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

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