Differentiation Question 139
Question: The differential coefficient of $ {{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} ) $ with respect to $ {{\tan }^{-1}} $ x is
[Kurukshetra CEE 1998; RPET 1999]
Options:
A) $ \frac{1}{2} $
B) $ -\frac{1}{2} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y_1={{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} ) $ and $ y_2={{\tan }^{-1}}x $
Now $ \frac{dy_1}{dx}=\frac{d}{dx}[ {{\tan }^{-1}}\tan \frac{\theta }{2} ], $ [By putting $ x=\tan \theta ] $
Therefore $ \frac{dy_1}{dx}=\frac{d}{dx}[ {{\tan }^{-1}}\tan \frac{\theta }{2} ]=\frac{1}{2(1+x^{2})} $ & $ \frac{dy_2}{dx}=\frac{1}{1+x^{2}} $
Hence $ \frac{dy_1}{dy_2}=\frac{1}{2} $ .
BETA
