Applications Of Derivatives Question 202
Question: $ \frac{d}{dx}[ {{\tan }^{-1}}\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} ]= $
[Roorkee 1980; Karnataka CET 2005]
Options:
A) $ \frac{-x}{\sqrt{1-x^{4}}} $
B) $ \frac{x}{\sqrt{1-x^{4}}} $
C) $ \frac{-1}{2\sqrt{1-x^{4}}} $
D) $ \frac{1}{2\sqrt{1-x^{4}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=A $
Putting $ x^{2}=\cos 2\theta , $ we get
$ A=\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } $
$ A=\frac{1+\tan \theta }{1-\tan \theta }=\tan ( \theta +\frac{\pi }{4} ) $
Now $ y={{\tan }^{-1}}\tan ( \theta +\frac{\pi }{4} )=\theta +\frac{\pi }{4}=\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}x^{2} $
$ \therefore \frac{dA}{dx}=\frac{1}{2}\frac{-1}{\sqrt{1-x^{4}}}2x=\frac{-x}{\sqrt{1-x^{4}}} $
BETA
