Question

Differentiate ${\cos }^{-1}\left(4x^3-3x\right)$ with respect to ${\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right),\mathrm{if}\frac{1}{2}<x<1$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\cos }^{-1}\left(4x^3-3x\right) \\ \mathrm{Put},x=\cos \theta \\ \Rightarrow \theta ={\cos }^{-1}x \\ \mathrm{Now},u={\cos }^{-1}\left(4{\cos }^3\theta -3\cos \theta \right) \\ \Rightarrow u={\cos }^{-1}\left(\cos 3\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Let},v={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \\ \Rightarrow v={\tan }^{-1}\left(\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\right) \\ \Rightarrow v={\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right) \\ \Rightarrow v={\tan }^{-1}\left(\tan \theta \right)...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ \frac{1}{2}<x<1 \\ \Rightarrow \frac{1}{2}<\cos \theta <1 \\ \Rightarrow 0<\theta <\frac{\mathrm{\pi }}{3} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=3\theta \left[\mathrm{Since},{\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\right] \\ \Rightarrow u=3{\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=\frac{-3}{\sqrt{1-x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{From}\mathrm{equation}\left(\mathrm{ii}\right), \\ v=\theta \left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,\mathrm{if}\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow v={\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{dv}{dx}=\frac{-1}{\sqrt{1-x^2}}...\left(\mathrm{iv}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{iv}\right), \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\left(\frac{-3}{\sqrt{1-x^2}}\right)\left(-\frac{\sqrt{1-x^2}}{1}\right) \\ \therefore \frac{du}{dv}=3 \end{gathered}$$