Question

If $f\left(x\right) = \cos^{-1} \left[\frac{1-\left(\log x\right)^{2}}{1+\left(\log x\right)^{2}}\right]$ then the value of $f'(e)$ is equal to

• A. 1
• B. $\frac{1}{e}$
• C. $\frac{2}{e}$
• D. $\frac{2}{e^2}$

Answer 1

Answer: (B)

$\frac{1}{e}$

Answer 2

Rewriting the equation as $\cos [ f ( x )]=\frac{1-(\log x )^{2}}{1+(\log x)^{2}}$.
Now taking derivatives we get,
$-\sin [ f ( x )] f '( x )=\frac{\left[1+(\log x)^{2}\right][-2(\log x ) x ]-\left[1-(\log x )^{2}\right][2(\log x ) x ]}{\left[1+(\log x)^{2}\right]^{2}}$
Evaluating at $x = e$ we get,
$-f'(x)=\frac{(2)(-2 / e)-(0)(2 / 2)}{2^{2}}$
This implies that $f '( x )=1 / e$.

Therefore, the correct option is (B): $\frac 1e$