Question

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

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

Answer 1

Answer: (A)

$\frac{2}{3}$

Answer 2

Let $y={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})$ ?. (i) and $z={{\sin }^{-1}}(3x-4{{x}^{3}})$ ...(ii)
Now, $x=cos\theta$ putting in E (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 E (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, $\frac{dy}{dz}=\frac{dy}{d\theta }.\frac{d\theta }{dz}$
$=2.\frac{1}{3}=\frac{2}{3}$
$\therefore$ $\frac{d({{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}})}{d({{\sin }^{-1}}(3x-4{{x}^{3}}))}=\frac{2}{3}$