# Derivative of e^(x^x) with respect to x

1. Aug 5, 2011

### labinojha

1. The problem statement, all variables and given/known data

Derivative of e^(x^x) with respect to x

2. Relevant equations

3. The attempt at a solution

Computed using wolframalpha. I have attached the image . Would anyone explain to me the part I have highlighted with the blue box.

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2. Aug 5, 2011

### BruceW

imagine that at first, you don't realise that both of the variables are x, so then you have an equation like this: $u^v$ So then, he calculates the total derivative of this equation in a general way, then at the end he replaces u by x and v by x.

Edit: ha, I'm talking about wolfram as a 'he'. Also, there is another way to solve this.

Last edited: Aug 5, 2011
3. Aug 5, 2011

### tiny-tim

hi labinojha!

this is the partial derivative version of the https://www.physicsforums.com/library.php?do=view_item&itemid=353"

if $a$ depends on $b_1,\cdots b_n$, and $b_1,\cdots b_n$ depend only on $c$, then:

$$\frac{da}{dc}\ =\ \frac{\partial a}{\partial b_1}\frac{db_1}{dc}\ +\ \cdots \frac{\partial a}{\partial b_n}\frac{db_n}{dc}\ =\ (\mathbf{\nabla_b}\,a)\cdot \frac{d\mathbf{b}}{dc}$$​

in your case, a is xx, b1 and b2 are u and v,

and so a is a function a(u,v) of two variables, and we need to apply the chain rule to each variable separately

(btw, easier would be to say xx = exln(x) )

Last edited by a moderator: Apr 26, 2017
4. Aug 5, 2011

### labinojha

$\frac{\partial}{\partial x}$$u^{v}$=$\frac{\partial}{\partial x}$$u^{v}$.$\frac{\partial}{\partial x}$$u+\frac{\partial}{\partial x}$$u^{v}$.$\frac{\partial}{\partial x}$v

Can i get any reason or possibly a derivation for this ?

Hi tiny-tim!

http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd10.html

Last edited: Aug 5, 2011
5. Aug 5, 2011

### labinojha

Hi Bruce!
The other way i used to do it was to take the natural logarithm of both the sides(one side of the equation being y to suppose it as the function) two times in a row and then differentiating them both sides .

Is this what you were talking about ? :)

Last edited: Aug 5, 2011
6. Aug 5, 2011

### BruceW

I was talking about the thing tiny-tim said:
$$x^x = e^{xln(x)}$$
But yes, the other way you do it is right as well. I guess there are several equivalent ways to do this problem.