If u = frac{tan^{-1} x}{tan^{-1} x + 1} and v | vedclass

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(B) Let $t = 	an^{-1} x$. Then $u = \frac{t}{t+1}$ and $v = 	an^{-1}(t)$.We need to find $\frac{du}{dv} = \frac{du/dt}{dv/dt}$.First,differentiate $

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$\frac{1 + (	an^{-1} x)^2}{(1 + 	an^{-1} x)^2}$

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(B) Let $t = 	an^{-1} x$. Then $u = \frac{t}{t+1}$ and $v = 	an^{-1}(t)$.We need to find $\frac{du}{dv} = \frac{du/dt}{dv/dt}$.First,differentiate $u$ with respect to $t$ using the quotient rule:$\frac{du}{dt} = \frac{(t+1)(1) - t(1)}{(t+1)^2} = \frac{1}{(t+1)^2}$.Next,differentiate $v$ with respect to $t$:$\frac{dv}{dt} = \frac{1}{1+t^2}$.Now,calculate $\frac{du}{dv}$:$\frac{du}{dv} = \frac{du/dt}{dv/dt} = \frac{1/(t+1)^2}{1/(1+t^2)} = \frac{1+t^2}{(t+1)^2}$.Substituting $t = 	an^{-1} x$ back into the expression:$\frac{du}{dv} = \frac{1 + (	an^{-1} x)^2}{(1 + 	an^{-1} x)^2}$.

If $u = \frac{\tan^{-1} x}{\tan^{-1} x + 1}$ and $v = \tan^{-1}(\tan^{-1} x)$,then $\frac{du}{dv} = \dots$

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