That was hard :O

Hello Max!

Find $\frac{dy}{dx}$ when $y= x^{y} +y^{x}$  

$lny=y\times lnx+ x\times lny$

$lny-x\times lny=y\times lnx$

$lny\times \left(1-x\right)= y\times lnx$

$\frac{y}{lny} = \frac{\left(1-x\right) }{lnx}$

$y=\frac{lny}{lnx} \times \left(1-x\right)$

From here I have to capitulate.
To differentiate, I need
y isolated on one side of the equation.
I'm excited about the solution!

Greetings asinus :- )  !

Not quite asinus!   ln(y) is not equal to y*ln(x) + x*ln(y)

Do dy/dx = d(x^y)/dx  +  d(y^x)/dx  first, then treat the terms on the right seperately.

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As follows:

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