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

\(y=\cos ^{-1}\left(\frac{\mathrm{a}^2}{\sqrt{x^4+\mathrm{a}^4}}\right)\)
Put \(x^2=\mathrm{a}^2 \tan \theta\)
\(\begin{aligned}
& \therefore \quad \theta=\tan ^{-1}\left(\frac{x^2}{\mathrm{a}^2}\right) \\
& \therefore \quad y=\cos ^{-1}\left(\frac{\mathrm{a}^2}{\sqrt{\mathrm{a}^4 \tan ^2 \theta+\mathrm{a}^4}}\right) \\
& \Rightarrow y=\cos ^{-1}\left(\frac{\mathrm{a}^2}{\sqrt{\mathrm{a}^4\left(1+\tan ^2 \theta\right)}}\right) \\
& \Rightarrow y=\cos ^{-1}\left(\frac{1}{\sec \theta}\right) \\
& \Rightarrow y=\cos ^{-1}(\cos \theta) \\
& \Rightarrow y=\theta \\
& \Rightarrow y=\tan ^{-1}\left(\frac{x^2}{\mathrm{a}^2}\right)
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{1}{1+\left(\frac{x^2}{\mathrm{a}^2}\right)^2} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^2}{\mathrm{a}^2}\right) \\
& =\frac{\mathrm{a}^4}{\mathrm{a}^4+x^4} \cdot \frac{2 x}{\mathrm{a}^2} \\
& =\frac{2 \mathrm{a}^2 x}{x^4+\mathrm{a}^4}
\end{aligned}\)