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

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

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

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(C) Given,$y = \cos^{-1}\left(\frac{a^2-x^2}{a^2+x^2}ight) + \sin^{-1}\left(\frac{2ax}{a^2+x^2}ight)$.Substitute $x = a 	an 	heta$,which implies $	heta = 	an^{-1}\left(\frac{x}{a}ight)$.Then,$y = \cos^{-1}\left(\frac{a^2 - a^2 	an^2 	heta}{a^2 + a^2 	an^2 	heta}ight) + \sin^{-1}\left(\frac{2a^2 	an 	heta}{a^2 + a^2 	an^2 	heta}ight)$.Simplifying the terms inside the brackets:$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 $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we get:$y = \cos^{-1}(\cos 2	heta) + \sin^{-1}(\sin 2	heta)$.$y = 2	heta + 2	heta = 4	heta$.Substituting back $	heta = 	an^{-1}\left(\frac{x}{a}ight)$,we have $y = 4 	an^{-1}\left(\frac{x}{a}ight)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 4 \cdot \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a} = 4 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{4a}{a^2 + x^2}$.

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

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