If y=cos ^{-1}left(frac{a^2}{sqrt{x^4+a^4}}ri | vedclass

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(A) Given $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}ight)$.Let $x^2=a^2 	an 	heta$,then $	heta=	an ^{-1}\left(\frac{x^2}{a^2}ight)$.Substit

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

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(A) Given $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}ight)$.Let $x^2=a^2 	an 	heta$,then $	heta=	an ^{-1}\left(\frac{x^2}{a^2}ight)$.Substituting $x^2$ in the expression for $y$:$y=\cos ^{-1}\left(\frac{a^2}{\sqrt{(a^2 	an 	heta)^2+a^4}}ight) = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^4 	an ^2 	heta+a^4}}ight)$$y=\cos ^{-1}\left(\frac{a^2}{a^2 \sqrt{	an ^2 	heta+1}}ight) = \cos ^{-1}\left(\frac{1}{\sec 	heta}ight)$$y=\cos ^{-1}(\cos 	heta) = 	heta$Thus,$y=	an ^{-1}\left(\frac{x^2}{a^2}ight)$.Differentiating with respect to $x$ using the chain rule:$\frac{d y}{d x} = \frac{1}{1+\left(\frac{x^2}{a^2}ight)^2} \cdot \frac{d}{d x}\left(\frac{x^2}{a^2}ight)$$\frac{d y}{d x} = \frac{1}{1+\frac{x^4}{a^4}} \cdot \frac{2x}{a^2} = \frac{a^4}{a^4+x^4} \cdot \frac{2x}{a^2}$$\frac{d y}{d x} = \frac{2 a^2 x}{x^4+a^4}$.

If $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$,then $\frac{d y}{d x}$ is

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