Applications Of Derivatives Question 205
Question: The derivative of $ {{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} ) $ with respect to $ {{\tan }^{-1}}( \frac{2x\sqrt{1-x^{2}}}{1-2x^{2}} ) $ at $ x=0 $ , is
Options:
A) $ \frac{1}{8} $
B) $ \frac{1}{4} $
C) $ \frac{1}{2} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y={{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} ) $ , $ z={{\tan }^{-1}}( \frac{2x,\sqrt{1-x^{2}}}{1-2x^{2}} ) $
Putting $ x=\tan \theta , $ we get
$ y={{\tan }^{-1}}( \frac{\sec \theta -1}{\tan \theta } )={{\tan }^{-1}}( \tan \frac{\theta }{2} )=\frac{1}{2}{{\tan }^{-1}}x $
$ \therefore \frac{dy}{dx}=\frac{1}{2(1+x^{2})} $
Putting $ x=\sin \theta $ , we get
$ z={{\tan }^{-1}}( \frac{2\sin \theta \cos \theta }{\cos 2\theta } )={{\tan }^{-1}}(\tan 2\theta )=2\theta $
$ \therefore z=2{{\sin }^{-1}}x\Rightarrow \frac{dz}{dx}=\frac{2}{\sqrt{1-x^{2}}} $
Thus $ \frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{1}{4(1+x^{2})}\sqrt{1-x^{2}}\Rightarrow {{( \frac{dy}{dx} )}_{x=0}}=\frac{1}{4} $ .
BETA
