Differentiation Question 270
Question: If $ f(x)={{\cos }^{-1}}
[ \frac{1-{{(\log x)}^{2}}}{1+{{(\log x)}^{2}}} ], $ then the value of $ f’(e)= $
[Karnataka CET 1999; Pb. CET 2000]
Options:
A) 1
B) 1/e
C) 2/e
D) $ \frac{2}{e^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)={{\cos }^{-1}}[ \frac{1-{{(\log x)}^{2}}}{1+{{(\log x)}^{2}}} ] $
$ =2{{\tan }^{-1}}(\log x) $
Therefore $ {f}’(x)=2.\frac{1}{1+{{(\log x)}^{2}}}.\frac{1}{x}.\text{Therefore }{f}’(e)=\frac{1}{e} $ .
BETA
