If y=cos ^{-1}left(frac{a^2-x^2}{a^2+x^2}righ | vedclass

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(C) Given $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}ight)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}ight)$.Let $x=a 	an 	heta$,then $	heta=	an ^{-1

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$\frac{4 a}{x^2+a^2}$

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(C) Given $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}ight)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}ight)$.Let $x=a 	an 	heta$,then $	heta=	an ^{-1}\left(\frac{x}{a}ight)$.Substituting $x$ in the expression:$y=\cos ^{-1}\left(\frac{a^2-a^2 	an ^2 	heta}{a^2+a^2 	an ^2 	heta}ight) + \sin ^{-1}\left(\frac{2 a^2 	an 	heta}{a^2+a^2 	an ^2 	heta}ight)$$y=\cos ^{-1}\left(\frac{1-	an ^2 	heta}{1+	an ^2 	heta}ight) + \sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^2 	heta}ight)$Using trigonometric identities $\cos 2	heta = \frac{1-	an^2	heta}{1+	an^2	heta}$ and $\sin 2	heta = \frac{2	an	heta}{1+	an^2	heta}$:$y=\cos ^{-1}(\cos 2 	heta) + \sin ^{-1}(\sin 2 	heta)$$y=2 	heta + 2 	heta = 4 	heta$.Substituting back $	heta = 	an ^{-1}\left(\frac{x}{a}ight)$:$y=4 	an ^{-1}\left(\frac{x}{a}ight)$.Now,differentiating with respect to $x$:$\frac{d y}{d x} = 4 \cdot \frac{1}{1+(\frac{x}{a})^2} \cdot \frac{1}{a}$$\frac{d y}{d x} = 4 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a} = \frac{4 a}{a^2+x^2}$.

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to

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