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

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(A) Let $y = {	an ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} ight)$.Dividing numerator and denominator by $\cos x$,we get:$y = {\

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$2$

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(A) Let $y = {	an ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} ight)$.Dividing numerator and denominator by $\cos x$,we get:$y = {	an ^{ - 1}}\left( {\frac{{	an x - 1}}{{	an x + 1}}} ight) = {	an ^{ - 1}}\left( {\frac{{	an x - 	an(\frac{\pi}{4})}}{{1 + 	an x \cdot 	an(\frac{\pi}{4})}}} ight)$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have:$y = {	an ^{ - 1}}\left( {	an \left( {x - \frac{\pi }{4}} ight)} ight)$.Since $x \in \left( {0, \frac{\pi }{2}} ight)$,then $x - \frac{\pi }{4} \in \left( {-\frac{\pi }{4}, \frac{\pi }{4}} ight)$,which is within the principal range of $	an^{-1}$.Thus,$y = x - \frac{\pi }{4}$.Now,we need to find the derivative of $y$ with respect to $u = \frac{x}{2}$.Since $u = \frac{x}{2}$,we have $x = 2u$.Substituting $x$ in $y$,we get $y = 2u - \frac{\pi }{4}$.Therefore,$\frac{dy}{du} = \frac{d}{du}(2u - \frac{\pi }{4}) = 2$.

The derivative of ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\frac{x}{2}$,where $x \in \left( {0, \frac{\pi }{2}} \right)$,is

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