# Differentiation of x^(1/x)

1. Nov 18, 2008

I was looking at results of different numbers in the equation $$\sqrt[x]{x}$$ and found out that the biggest result came when it was $$\sqrt[e]{e}$$. I know this can be re-written as $$x^{1/x}$$ and that the gradient would be 0 at x = e. How would you differentiate y = $$x^{1/x}$$, I can't seem to do it using any of the laws I know.

2. Nov 18, 2008

### D H

Staff Emeritus
Find a function $f(x)$ that lets you rewrite $y=x^{1/x}$ in the form $y=\exp(f(x))$.

3. Nov 18, 2008

I don't see how that works but I'll give it a go.

4. Nov 18, 2008

### lurflurf

use the power rule
[u^v]'=v*[u^(v-1)]*u'+[u^v]*[log(u)]*v'

5. Nov 18, 2008

### D H

Staff Emeritus
Rather than committing the power rule to memory, I find it much easier to remember a couple simple facts which happen to be of use in a lot of other places,
1. $$\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}\,f'(x)$$

2. $$f(x) = e^{\ln f(x)}\;\Rightarrow\;u(x)^{v(x)} = e^{v(x)\,\ln u(x)}$$
and thus

\aligned \frac{d}{dx}\left(u(x)^{v(x)}\right) &= \frac{d}{dx}\left(e^{v(x)\,\ln u(x)}\right) \\ &= e^{v(x)\,\ln u(x)}\frac d {dx}\left(v(x)\,\ln u(x)\right) \\ &= u(x)^{v(x)}\left(v'(x)\,\ln u(x) + \frac{v(x)\,u'(x)}{u(x)}\right)

6. Nov 18, 2008

### HallsofIvy

Equivalently, if y= x1/x, then ln(y)= (1/x)ln(x). Use "implicit differentiation".

7. Nov 18, 2008

### lurflurf

or just use it on
y^x=x

8. Nov 18, 2008