Differentiation Question 195

Question: $ \frac{d}{dx}{{\sin }^{-1}}(3x-4x^{3})= $

[RPET 2003]

Options:

A) $ \frac{3}{\sqrt{1-x^{2}}} $

B) $ \frac{-3}{\sqrt{1-x^{2}}} $

C) $ \frac{1}{\sqrt{1-x^{2}}} $

D) $ \frac{-1}{\sqrt{1-x^{2}}} $

Show Answer

Answer:

Correct Answer: A

Solution:

Put $ x=\sin \theta , $ we get $ \frac{d}{dx}{{\sin }^{-1}}(3x-4x^{3}) $

$ =\frac{d}{dx}{{\sin }^{-1}}(\sin 3\theta )=\frac{3}{\sqrt{1-x^{2}}} $ .