Question

If x^y=y^x find dy/dx.... ​

Answer 1

Answer:

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

Step-by-step explanation:

x^y=y^x

y ln(x) = x ln(y)

y' ln(x) + y (1/x) = 1 ln(y) + x (1/y) y'

y' ln(x) + y/x = ln(y) + x/y y'

y' ( ln(x) - x/y ) = ln(y) - y/x

y' =

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