Differentiation Question 66
Question: The differential coefficient of $ {{\tan }^{-1}}( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} ) $ is
[MP PET 2003]
Options:
A) $ \sqrt{1-x^{2}} $
B) $ \frac{1}{\sqrt{1-x^{2}}} $
C) $ \frac{1}{2\sqrt{1-x^{2}}} $
D) x
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y={{\tan }^{-1}}( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} ) $
Put $ x=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{{\cos }^{-1}}x $
$ y={{\tan }^{-1}}( \frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }} ) $
$ y={{\tan }^{-1}}( \frac{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }} ) $
$ y={{\tan }^{-1}}( \frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta } )={{\tan }^{-1}}( \frac{1-\tan \theta }{1+\tan \theta } ) $
$ y={{\tan }^{-1}}(\tan (\pi /4-\theta )) $
Therefore $ y=\frac{\pi }{4}-\theta $
$ y=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x $
Therefore $ \frac{dy}{dx}=\frac{1}{2}( \frac{1}{\sqrt{1-x^{2}}} ) $ .
BETA
