Question

if x^y= e ^x-y then Dy/dx =​

Answer 1

Step-by-step explanation:

We have,

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

Taking log both sides,

$y ln(x) = x - y$

$\implies \: y( ln(x) + 1) = x$

Differentiating both sides, we have,

$\implies( ln(x) + 1) \frac{dy}{dx} + y( \frac{1}{x} ) = 1$

$\implies \frac{dy}{dx} = (1 - \frac{y}{x} ) \times \frac{1}{ ln(x) + 1}$

$\implies \frac{dy}{dx} = \frac{(x - y)}{x ln(x) + x}$