Derivative of tan ^{-1}left(frac{sqrt{1+x^2}- | vedclass

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(A) Let $y = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}ight)$ and $z = \cos ^{-1}(x^2)$.Substitute $x^2 = \cos 2	h

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Let $y = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}ight)$ and $z = \cos ^{-1}(x^2)$.Substitute $x^2 = \cos 2	heta$,which implies $	heta = \frac{1}{2} \cos ^{-1}(x^2)$.Then,$y = 	an ^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta}+\sqrt{1-\cos 2	heta}}ight)$.Using the identities $1+\cos 2	heta = 2\cos^2 	heta$ and $1-\cos 2	heta = 2\sin^2 	heta$,we get:$y = 	an ^{-1}\left(\frac{\sqrt{2}\cos 	heta - \sqrt{2}\sin 	heta}{\sqrt{2}\cos 	heta + \sqrt{2}\sin 	heta}ight) = 	an ^{-1}\left(\frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta}ight)$.Dividing numerator and denominator by $\cos 	heta$,we have:$y = 	an ^{-1}\left(\frac{1 - 	an 	heta}{1 + 	an 	heta}ight) = 	an ^{-1}\left(	an\left(\frac{\pi}{4} - 	hetaight)ight) = \frac{\pi}{4} - 	heta$.Substituting $	heta = \frac{1}{2} \cos ^{-1}(x^2)$ back into the equation:$y = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1}(x^2) = \frac{\pi}{4} - \frac{1}{2} z$.Now,differentiating $y$ with respect to $z$:$\frac{dy}{dz} = \frac{d}{dz}\left(\frac{\pi}{4} - \frac{1}{2} zight) = -\frac{1}{2}$.

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

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