Differentiation Question 358
Question: $ \frac{d}{dx}{{\cos }^{-1}}\sqrt{\frac{1+x^{2}}{2}}= $
[AI CBSE 1988]
Options:
A) $ \frac{-1}{2\sqrt{1-x^{4}}} $
B) $ \frac{1}{2\sqrt{1-x^{4}}} $
C) $ \frac{-x}{\sqrt{1-x^{4}}} $
D) $ \frac{x}{\sqrt{1-x^{4}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ x^{2}=\cos 2\theta $ , we have $ \frac{d}{dx}[ {{\cos }^{-1}}\sqrt{\frac{1+x^{2}}{2}} ]=\frac{d}{dx}[{{\cos }^{-1}}\cos \theta ] $
$ =\frac{d}{dx}[\theta ]=\frac{d}{dx}[ \frac{1}{2}{{\cos }^{-1}}x^{2} ]=\frac{-x}{\sqrt{1-x^{4}}} $ .
BETA
