Question

If $x^y=e^{2(x–y)}$, then $\frac{dy}{dx}$ is equal to:

• A. $\frac{2(1+\log x)}{{(2+\log x)}^2}$
• B. $\frac{(1+\log x)}{{(2+\log x)}^2}$
• C. $\frac{2}{2+\log x}$
• D. $\frac{2(1-\log x)}{{(2+\log x)}^2}$

Answer 1

The correct option is A

$\frac{2(1+\log x)}{{(2+\log x)}^2}$

Explanation for the correct option.

$x^y=e^{2(x–y)}$

By taking log on both sides we get,

$\begin{array}{rcl}y\log x& =& 2(x-y)\log e\\ & \Rightarrow & y\log x+2y=2x\\ \Rightarrow y& =& \frac{2x}{2+\log x}\end{array}$

Differentiating with respect to $x$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\left(2+\log x\right)\left(2\right)-2x\left(0+{\displaystyle \frac{1}{x}}\right)}{{\left(2+\log x\right)}^2}\left[\mathrm{by}\mathrm{quotient}\mathrm{rule}\right]\\ & =& \frac{\left(2+\log x\right)\left(2\right)-2}{{\left(2+\log x\right)}^2}\\ & =& \frac{2\left(2+\log x-1\right)}{{\left(2+\log x\right)}^2}\\ & =& \frac{2\left(1+\log x\right)}{{\left(2+\log x\right)}^2}\end{array}$

Hence, option A is correct.