Question

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

• A. $2$
• B. $\frac{-1}{2\sqrt{1-x^2}}$
• C. $\frac{2}{x}$
• D. $1-x^2$

Answer 1

Answer: (A)

Let $u={\cos }^{-1}\left(2x^2-1\right)$ and $v={\cos }^{-1}x$
Now, $u={\cos }^{-1}\left(2x^2-1\right)$
Put $x=\cos \theta$
$\therefore u={\cos }^{-1}\left(2{\cos }^2\theta -1\right)$
$={\cos }^{-1}(\cos 2\theta )$
$=2\theta =2{\cos }^{-1}x$
Again, $\frac{du}{dv}=\frac{\left(\frac{du}{dx}\right)}{\left(\frac{dv}{dx}\right)}$
$=\frac{\frac{d}{dx}\left(2{\cos }^{-1}x\right)}{\frac{d}{dx}\left({\cos }^{-1}x\right)}$
$=\frac{\left(-\frac{2}{\sqrt{1-x^2}}\right)}{\left(\frac{-1}{\sqrt{1-x^2}}\right)}=2$