Differentiation Question 70
Question: $ \frac{d}{dx}( {{\tan }^{-1}}\frac{\sqrt{1+x^{2}}-1}{x} ) $ is equal to
[MP PET 2004]
Options:
A) $ \frac{1}{1+x^{2}} $
B) $ \frac{1}{2(1+x^{2})} $
C) $ \frac{x^{2}}{2\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-1)} $
D) $ \frac{2}{1+x^{2}} $
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Answer:
Correct Answer: B
Solution:
Let $ y={{\tan }^{-1}}\frac{\sqrt{1+x^{2}}-1}{x} $
Put $ x=\tan \theta , $ then $ y={{\tan }^{-1}}( \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } ) $
$ y={{\tan }^{-1}}( \frac{\sec \theta -1}{\tan \theta } )={{\tan }^{-1}}( \frac{1-\cos \theta }{\sin \theta } ) $
$ y={{\tan }^{-1}}( \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} )={{\tan }^{-1}}\tan \frac{\theta }{2} $
$ y=\frac{\theta }{2}=\frac{1}{2}{{\tan }^{-1}}x $ , $ (\because \theta ={{\tan }^{-1}}x) $ . Hence $ \frac{dy}{dx}=\frac{1}{2(1+x^{2})} $ .
BETA
