Question

If $x^y=e^{x-y},x>0$, then the value of $\frac{dy}{dx}$ at $(1,1)$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1: Find $\frac{dy}{dx}$

The given equation is $x^y=e^{x-y}$

By taking log on the both sides, we get

$\begin{array}{rcl}y\log x& =& \left(x-y\right)\log e\\ & \Rightarrow & y\log x=\left(x-y\right)\left(as\log e=1\right)\end{array}$

Differentiate with respect to $x$

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

Step 2: Find $\frac{dy}{dx}$at $(1,1)$.

$\begin{array}{rcl}{\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}& =& \frac{1-1}{1}\times \frac{1}{\left(\log 1+1\right)}\\ & =& 0\end{array}$

Hence, option A is correct.