Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ w.r.t. $\cos ^{-1}\left(4 x^3-3 x\right)$ is

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ w.r.t. $\cos ^{-1}\left(4 x^3-3 x\right)$ is
  1. $\frac{-1}{6}$
  2. $\frac{2}{3}$
  3. $\frac{3}{2}$
  4. $\frac{1}{6}$

Solution

Let $y=\tan ^{-1} \sqrt{\frac{1-x}{1+x}}, \mathrm{z}=\cos ^{-1}\left(4 x^3-3 x\right)$ for $y$, substitute $x=\cos 2 \theta_1$ and for z , substitute $x=\cos \theta$ $\begin{aligned} & \therefore \quad y=\tan ^{-1} \sqrt{\frac{1-\cos 2 \theta_1}{1+\cos 2 \theta_1}}=\tan ^{-1}\left(\tan \theta_1\right)=\theta_1 \text { and } \\ & \mathrm{z}=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right) \\ &=\cos ^{-1}(\cos 3 \theta)=3 \theta \\\ herefore \quad y & =\frac{\cos ^{-1} x}{2} \text { and } \mathrm{z}=3 \cos ^{-1} x \\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x} =\frac{-1}{2 \sqrt{1-x^2}} \text { and } \frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{-3}{\sqrt{1-x^2}} \\ & \therefore \quad \frac{\mathrm{~d} y}{\mathrm{dz}} =\frac{\frac{\mathrm{d} y}{\mathrm{~d} x}}{\frac{\mathrm{~d} z}{\mathrm{~d} x}}=\frac{1}{6} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

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