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}}} $ .