Differentiation Question 274
Question: Differential coefficient of $ {{\tan }^{-1}}( \frac{x}{1+\sqrt{1-x^{2}}} ) $ w.r.t $ {{\sin }^{-1}}x, $ is
[Kurukshetra CEE 2002]
Options:
A) $ \frac{1}{2} $
B) 1
C) 2
D) $ \frac{3}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\tan }^{-1}}[ \frac{x}{1+\sqrt{1-x^{2}}} ] $
Put $ x=\sin \theta $
Therefore $ y={{\tan }^{-1}}[ \frac{\sin \theta }{1+\cos \theta } ]={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2} $
Therefore $ y=\frac{1}{2}{{\sin }^{-1}}x $ and let $ z={{\sin }^{-1}}x $
Hence $ \frac{dy}{dz}=\frac{( \frac{dy}{dx} )}{( \frac{dz}{dx} )} $ = $ \frac{\frac{1}{2}\frac{d}{dx}{{\sin }^{-1}}x}{\frac{d}{dx}{{\sin }^{-1}}x} $ = $ f’’(e^{x}).e^{2x}+f’(e^{x}).e^{x} $ .
BETA
