Question

The derivative of ${\sin }^{-1}(2x\sqrt{1-x^2})$ with respect to ${\sin }^{-1}(3x-4x^3)$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $1$
• E. $0$

Answer 1

Answer: (A)

Let $y={\sin }^{-1}(2x\sqrt{1-x^2})$ ?. (i) and $z={\sin }^{-1}(3x-4x^3)$ ...(ii)
Now, $x=cos\theta$ putting in Eq. (i), we get $y={\sin }^{-1}(2\cos \theta \sqrt{1-{\cos }^2\theta })$
$={\sin }^{-1}(2\cos \theta \sin \theta )$
$={\sin }^{-1}(\sin 2\theta )$
$\Rightarrow$ $y=2\theta$
$\Rightarrow$ $y=2{\cos }^{-1}x$
Differentiating it w.r.t. $\theta$ , we get $\frac{dz}{d\theta }=3$ ...(iii)
Also, putting $x=sin\theta$ in Eq. (ii), we get $z={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )={\sin }^{-1}(\sin 3\theta )$
$\therefore$ $z=3\theta$
Differentiating it w.r.t. $\theta$ , we get $\frac{dz}{d\theta }=3$ ...(iv) Now, $\frac{dy}{dz}=\frac{dy}{d\theta }.\frac{d\theta }{dz}$
$=2.\frac{1}{3}=\frac{2}{3}$
$\therefore$ $\frac{d({\sin }^{-1}2x\sqrt{1-x^2})}{d({\sin }^{-1}(3x-4x^3))}=\frac{2}{3}$