Question

If $x^y{y}^x=1$, then $\frac{dy}{dx}$ is:

• A. $\frac{-y\left(y+x\log y\right)}{x\left(y\log x+x\right)}$
• B. $\frac{y\left(x+y\log x\right)}{x\left(x\log y+y\right)}$
• C. $\frac{y\left(y+x\log y\right)}{x\left(y\log x+x\right)}$
• D. None of these

Answer 1

The correct option is A

$\frac{-y\left(y+x\log y\right)}{x\left(y\log x+x\right)}$

Explanation for the correct option.

Step 1: Simplify by taking log.

The given equation is $x^y{y}^x=1$.

By taking log on both sides we get,

$\begin{array}{rcl}\log \left(x^y{y}^x\right)& =& \log \left(1\right)\\ & \Rightarrow & y\log x+x\log y=0\end{array}$

Step 2: Differentiate with respect to $x$.

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

$\begin{array}{rcl}\log x\frac{dy}{dx}+y\left(\frac{1}{x}\right)+\log y\left(1\right)+x\left(\frac{1}{y}\times \frac{dy}{dx}\right)& =& 0\left[\mathrm{by}\mathrm{product}\mathrm{rule}\right]\\ \Rightarrow \left(\log x\right)\frac{dy}{dx}+\left(\frac{x}{y}\right)\frac{dy}{dx}& =& -\left[\frac{y}{x}+\log y\right]\\ \Rightarrow \left(\log x+\frac{x}{y}\right)\frac{dy}{dx}& =& -\left[\frac{y+x\log y}{x}\right]\\ \Rightarrow \left(\frac{y\log x+x}{y}\right)\frac{dy}{dx}& =& -\left[\frac{y+x\log y}{x}\right]\\ \Rightarrow \frac{dy}{dx}& =& \frac{-y\left(y+x\log y\right)}{x\left(y\log x+x\right)}\end{array}$

Hence, option A is correct.