Quick derivative question

1. Jan 10, 2004

fffbone

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

2. Jan 10, 2004

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*}

3. Jan 10, 2004

fffbone

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

4. Jan 10, 2004

master_coda

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

5. Jan 11, 2004

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.

6. Jan 11, 2004

master_coda

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

7. Jan 11, 2004

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).