Question

The derivative of ${\cos }^{-1}\left(2x^2-1\right)$ with respect to ${\cos }^{-1}x$ is
(a) 2

(b) $\frac{1}{2\sqrt{1-x^2}}$

(c) $2/x$

(d) $1-x^2$

Answer 1

(a) 2

$$\begin{gathered} \mathrm{Let}u={\cos }^{-1}\left(2x^2-1\right) \\ \mathrm{Put}x=\cos \theta \\ \Rightarrow \theta ={\cos }^{-1}x \\ \frac{d\theta }{dx}=\frac{-1}{\sqrt{1-x^2}} \\ \mathrm{Now},u={\cos }^{-1}\left(\cos 2\theta \right) \\ \Rightarrow u=2\theta \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{du}{dx}=2\frac{d\theta }{dx} \\ \Rightarrow \frac{du}{dx}=\frac{-2}{\sqrt{1-x^2}}...\left(i\right) \\ \mathrm{and}, \\ v={\cos }^{-1}x \\ \Rightarrow v={\cos }^{-1}\left(\cos \theta \right) \\ \Rightarrow v=\theta \end{gathered}$$

$$\begin{gathered} \frac{dv}{dx}=\frac{d\theta }{dx} \\ \Rightarrow \frac{dv}{dx}=\frac{-1}{\sqrt{1-x^2}}...\left(ii\right) \\ \mathrm{Dividing}\left(\mathrm{i}\right)\mathrm{by}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}, \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{-2}{\sqrt{1-x^2}}\times \frac{\sqrt{1-x^2}}{-1} \\ \Rightarrow \frac{du}{dv}=2 \end{gathered}$$