Differentiation of tan ^{-1}left(frac{sqrt{1+ | vedclass

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(B) Let $u = 	an ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}ight]$ and $v = \cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}ight]$.Substitute

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(B) Let $u = 	an ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}ight]$ and $v = \cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}ight]$.Substitute $x = 	an 	heta$,then $	heta = 	an ^{-1} x$.For $u$:$u = 	an ^{-1}\left[\frac{\sec 	heta - 1}{	an 	heta}ight] = 	an ^{-1}\left[\frac{1-\cos 	heta}{\sin 	heta}ight] = 	an ^{-1}\left[\frac{2 \sin ^2 (	heta/2)}{2 \sin (	heta/2) \cos (	heta/2)}ight] = 	an ^{-1}(	an (	heta/2)) = \frac{	heta}{2} = \frac{	an ^{-1} x}{2}$.For $v$:$v = \cos ^{-1}\left[\sqrt{\frac{1+\sec 	heta}{2 \sec 	heta}}ight] = \cos ^{-1}\left[\sqrt{\frac{\cos 	heta + 1}{2}}ight] = \cos ^{-1}\left[\sqrt{\frac{2 \cos ^2 (	heta/2)}{2}}ight] = \cos ^{-1}(\cos (	heta/2)) = \frac{	heta}{2} = \frac{	an ^{-1} x}{2}$.Since $u = v$,we have $\frac{du}{dv} = \frac{d}{dv}(v) = 1$.

Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right)$ is

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