Quick derivative question

[fffbone]

How do you find the derivative of y=x^y ?

[master_coda]

\begin{align*}
y&=x^y \\
\ln y&=y\ln x \\
\frac{y^\prime}{y}&=y^\prime\ln x+\frac{y}{x} \\
y^\prime\left(\frac{1}{y}-\ln x\right)&=\frac{y}{x} \\
y^\prime&=\frac{y^2}{x-xy\ln x}
\end{align*}

[fffbone]

Not quite the answer I was looking for, but thanks any how. I already know how the problem is solved.

[master_coda]

You asked me how to find the derivative. I showed you a way to do it. What were you looking for?

[fffbone]

I was looking for something more like this: x'=y^(1/y)*(1/y^2-ln(y)/y^2).

I solved it though, guess just got stuck for a minute.

[master_coda]

Usually people want the derivative of y wrt x. Plus, you gave an equation for y in terms of x and y.

[fffbone]

Since y does not equal f(x), y can not be expressed in terms of x. We would have to express x in terms of y.

Of course, it would be much better if the equation was x=y^x, then it can be expressed as y=x^(1/x).