Browse Questions

# The derivative of $\cos^{-1}(2x^2-1)$ w.r.t $\cos^{-1}x$ is

$\begin{array}{1 1}(A)\;2 & (B)\;\frac{-1}{2\sqrt{1-x^2}}\\(C)\;\frac{2}{x} & (D)\;1-x^2\end{array}$

Toolbox:
• Let $u=f(x)$ and $v=g(x)$ be two functions of $x$.The derivative of $f(x)$ w.r.t $g(x)$ (i.e) $\large\frac{du}{dv}=\large\frac{\Large\frac{du}{dx}}{\Large\frac{dv}{dx}}$
Step 1:
Let $u=\cos^{-1}(2x^2-1)$ and $v=\cos^{-1}x$
Put $x=\cos \theta\Rightarrow \theta=\cos^{-1}x$
$u=\cos^{-1}(2\cos^2\theta-1)$
But $2\cos^2\theta-1=\cos 2\theta$
Therefore $u=\cos^{-1}(\cos 2\theta)$
$\Rightarrow u=2\theta$
$\qquad=2.\cos^{-1}x$
$\large\frac{du}{dx}=$$-2\big(\large\frac{-1}{\sqrt{1-x^2}}\big)$
Step 2:
Consider $v=\cos^{-1}x$
$\large\frac{dv}{dx}=\large\frac{-1}{\sqrt{1-x^2}}$
Therefore $\large\frac{\Large\frac{du}{dx}}{\Large\frac{dv}{dx}}=\frac{\Large\frac{-2}{\sqrt{1-x^2}}}{\Large\frac{-1}{\sqrt{1-x^2}}}$
$\Rightarrow 2$
Hence the correct option is $A$