Question

If $x^y=e^{x-y}$ then
$\frac{dy}{dx}=\frac{logx}{(1-logx{)}^2}$.

• A. True
• B. False

Answer 1

The correct option is B False

$x^y=e^{x-y}$

taking log to the both sides,

$\log x^y=\log e^{x-y}$

As,

$\log a^b=b\log a$

So,

$y\log x=\left(x-y\right)\log e$

$y\log x=x-y$

differentiating with respect to x,

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

Thus, the equation is false