Question

If $f\left(x\right)={\cos }^{-1}\left(\frac{1-{\left(\log x\right)}^2}{1+{\left(\log x\right)}^2}\right)$, then $f\text{'}\left(e\right)$ is equal to

• A. $\frac{2}{e}$
• B. $\frac{1}{e}$
• C. $\frac{3}{e}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{e}$

Explanation for the correct option.

Step 1. Simplify the given function.

Let $\log x=\tan \theta$, then $\theta ={\tan }^{-1}\left(\log x\right)$.

Now using the substitution $\log x=\tan \theta$, the equation $f\left(x\right)={\cos }^{-1}\left(\frac{1-{\left(\log x\right)}^2}{1+{\left(\log x\right)}^2}\right)$ becomes:

$\begin{array}{rcl}f\left(x\right)& =& {\cos }^{-1}\left(\frac{1-{\left(\tan \theta \right)}^2}{1+{\left(\tan \theta \right)}^2}\right)\\ & =& {\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\\ & =& {\cos }^{-1}\cos 2\theta \left[\because cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right]\\ & =& 2\theta \\ & =& 2{\tan }^{-1}\left(\log x\right)\left[\theta ={\tan }^{-1}\left(\mathrm{logx}\right)\right]\end{array}$

Step 2. Find the value of $f\text{'}\left(e\right)$.

Differentiate both sides of the equation .

$f\text{'}\left(x\right)=2\frac{1}{1+{\left(\log x\right)}^2}\frac{1}{x}\left[\because \frac{d}{\mathrm{dx}}\mathrm{logx}=\frac{1}{x},\frac{d}{\mathrm{dx}}{\tan }^{-1}x=\frac{1}{1+x^2}\right]$

Now substitute $x=e$ and find the value of $f\text{'}\left(e\right)$.

$\begin{array}{rcl}f\text{'}\left(e\right)& =& 2\frac{1}{1+{\left(\log e\right)}^2}\frac{1}{e}\\ & =& 2\frac{1}{1+1^2}\frac{1}{e}\\ & =& 2\frac{1}{2}\frac{1}{e}\\ & =& \frac{1}{e}\end{array}$

Hence, the correct option is B.