Question

If $y={\cos }^{-1}\left(\frac{1-\log \left(x\right)}{1+\log \left(x\right)}\right)$ then $\frac{dy}{dx}$at $x=e$ is

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

Answer 1

The correct option is C

$\frac{1}{2e}$

Explanation for the correct option.

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate $y={\cos }^{-1}\left(\frac{1-\log \left(x\right)}{1+\log \left(x\right)}\right)$ with respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left({\cos }^{-1}\left(\frac{1-\log \left(x\right)}{1+\log \left(x\right)}\right)\right)\\ & =& \frac{-1}{\sqrt{1-{\left({\displaystyle \frac{1-\log \left(x\right)}{1+\log \left(x\right)}}\right)}^2}}\left(\frac{(1+\log \left(x\right))\left({\displaystyle \frac{-1}{x}}\right)-\left(1-\log \left(x\right)\right)\left(\frac{1}{x}\right)}{{\left(1+\log \left(x\right)\right)}^2}\right)\end{array}$

Step 2. Find the value of $\frac{dy}{dx}$at $x=e$.

Put $x=e$in the above equation.

$\begin{array}{rcl}{\overline{)\frac{dy}{dx}}}_{x=e}& =& \frac{-1}{\sqrt{1-{\left({\displaystyle \frac{1-\log \left(e\right)}{1+\log \left(e\right)}}\right)}^2}}\left(\frac{(1+\log \left(e\right))\left({\displaystyle \frac{-1}{e}}\right)-\left(1-\log \left(e\right)\right)\left(\frac{1}{e}\right)}{{\left(1+\log \left(e\right)\right)}^2}\right)\\ & =& \frac{-1}{\sqrt{1-{\left({\displaystyle \frac{1-1}{1+1}}\right)}^2}}\left(\frac{(1+1)\left({\displaystyle \frac{-1}{e}}\right)-\left(1-1\right)\left(\frac{1}{e}\right)}{{\left(1+1\right)}^2}\right)\\ & =& \frac{-1}{\sqrt{1-0}}\left(\frac{{\displaystyle \frac{-2}{e}}-0}{2^2}\right)\\ & =& -1\left(\frac{-1}{2e}\right)\\ & =& \frac{1}{2e}\end{array}$

Hence, the correct option is C.