Differentiation Question 59
Question: If $ y={{\tan }^{-1}}( \frac{x}{\sqrt{1-x^{2}}} ) $ , then $ \frac{dy}{dx}= $
[MP PET 1999]
Options:
A) $ -\frac{1}{\sqrt{1-x^{2}}} $
B) $ \frac{x}{\sqrt{1-x^{2}}} $
C) $ \frac{1}{\sqrt{1-x^{2}}} $
D) $ \frac{\sqrt{1-x^{2}}}{x} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{\tan }^{-1}}( \frac{x}{\sqrt{1-x^{2}}} ) $
Put $ x=\sin \theta $ , $ \therefore $ $ dx=\cos \theta d\theta $ , $ \frac{d\theta }{dx}=\frac{1}{\cos \theta } $
$ \therefore y={{\tan }^{-1}}( \frac{\sin \theta }{\cos \theta } ) $
$ \Rightarrow $ $ y=\theta $
$ \therefore $ $ dy=d\theta $
Now $ \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx} $ = $ 1.\frac{1}{\cos \theta }=\sec \theta =\frac{1}{\sqrt{1-x^{2}}} $ .
BETA
