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 $
• E. $ 0 $

Answer 1

Answer: (A)

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 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-4{{x}^{3}}))}=\frac{2}{3} $