Question

Find dy /DX for y^x= e^( x+ log y)

Answer 1

Step-by-step explanation:

${y}^{x} = {e}^{x + logy} \\ taking \: log \: on \: both \: sides \\ x logy = x + logy \: \\ differentiating \: on \: both \: sides \\ \frac{x}{y} \frac{dy}{dx} + logy = 1 + \frac{1}{y} \frac{dy}{dx} \\ \frac{x - 1}{y} \frac{dy}{dx} = 1 - logy \\ \frac{dy}{dx} = \frac{y(1 - logy)}{x - 1} \\ \\ hope \: it \: will \: help \: you \:$