Question

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

• A. $y(y–x\log y)$
• B. $y(y+x\log y)$
• C. $x(x+y\log x)$
• D. $x(y–x\log y)$

Answer 1

The correct option is A

$y(y–x\log y)$

Explanation for the correct option.

The given equation is $x^y=y^x$.

Taking log on the both sides, we get

$\begin{array}{rcl}\log \left(x^y\right)& =& \log \left(y^x\right)\\ & \Rightarrow & y\log x=x\log y\end{array}$

By differentiating both sides w.r.t $x$, we get

$\begin{array}{rcl}\log x\frac{dy}{dx}+y\left(\frac{1}{x}\right)& =& \log y\left(1\right)+x\left(\frac{1}{y}\right)\frac{dy}{dx}\\ \Rightarrow \log x\frac{dy}{dx}+\frac{y}{x}& =& \log y+\left(\frac{x}{y}\right)\frac{dy}{dx}\\ \Rightarrow \left(\frac{x}{y}-\log x\right)\frac{dy}{dx}& =& \frac{y}{x}-\log y\\ & \Rightarrow & \left(\frac{x-y\log x}{y}\right)\frac{dy}{dx}=\frac{y-x\log y}{x}\\ & \Rightarrow & x\left(x-y\log x\right)\frac{dy}{dx}=y\left(y-x\log y\right)\end{array}$

Hence, option A is correct.