Differentiation Question 336
Question: $ \frac{d}{dx}{{\tan }^{-1}}\frac{x}{\sqrt{a^{2}-x^{2}}}= $
Options:
A) $ \frac{a}{a^{2}+x^{2}} $
B) $ \frac{-a}{a^{2}+x^{2}} $
C) $ \frac{1}{a\sqrt{a^{2}-x^{2}}} $
D) $ \frac{1}{\sqrt{a^{2}-x^{2}}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d}{dx}{{\tan }^{-1}}\frac{x}{\sqrt{a^{2}-x^{2}}} $
Putting $ x=a\sin \theta , $ we get $ =\frac{d}{dx}[ {{\tan }^{-1}}\frac{a\sin \theta }{a\cos \theta } ]=\frac{d}{dx}[ {{\tan }^{-1}}\tan \theta ]=\frac{d}{dx}[\theta ] $
Substituting value of $ \theta $ , so $ |x| $ .
BETA
