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

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
  1. $\frac{2 \mathrm{a}^2 x}{x^4+\mathrm{a}^4}$
  2. $\frac{2 a^2 x^2}{\sqrt{x^4+a^4}}$
  3. $\frac{a^4 x^4}{x^4+a^4}$
  4. $\frac{a^4 x^2}{2 \sqrt{x^4+a^4}}$

Solution

$y=\cos ^{-1}\left(\frac{\mathrm{a}^2}{\sqrt{x^4+\mathrm{a}^4}}\right)$ 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 $\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{1}{1+\left(\frac{x^2}{\mathrm{a}^2}\right)^2} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^2}{\mathrm{a}^2}\right) \\ & =\frac{\mathrm{a}^4}{\mathrm{a}^4+x^4} \cdot \frac{2 x}{\mathrm{a}^2} \\ & =\frac{2 \mathrm{a}^2 x}{x^4+\mathrm{a}^4} \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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