# Derivative of e^x^x^2?

1. Oct 25, 2007

### erjkism

can someone explain to me how this could be solved?

so far i have:
f(x)=e^x^x^2
lnf(x)= x^2lne^x)

(e and ln cancel?)

f'(x)/f(x)= (x^2)x
f'(x)= f(x) x^3

= (e^x^x^2)(x^3)??

is that right? or do i need to use the power rule or something?

2. Oct 26, 2007

### mathwonk

get a clue. use the chain rule.

3. Oct 26, 2007

### TMM

Check to see that you're substituting the natural log derivative correctly.

4. Oct 26, 2007

### ice109

mathwonk you're mean

how do i take the derivative of $f(g(h(x)))$ with respect to x erjkism?

5. Oct 26, 2007

### Zurtex

Looking at how you've tackled the problem, I'm going to assume you mean:

(e^x)^(x^2), now because ^ isn't an associative binary operator this is quite a bit different from e^(x^(x^2)) and in turn it quite a bit different from ((e^x)^x)^2.

Think about what the actual brackets are and apply the chain rule carefully. I'm fairly confident you've calculated wrong no matter how you look it at, because you haven't really considered what you're differentiating.

6. Oct 26, 2007

### CRGreathouse

Which do you mean?

1. $$e^{x^{x^2}}=\exp(x^{x^2})$$

2. $$(e^x)^{x^2}=\exp(x^3)$$

3. $$((e^x)^x)^2=\exp(2x^2)$$

Last edited: Oct 26, 2007
7. Oct 26, 2007

### HallsofIvy

I had assumed that you meant
$$e^{x^{x^2}}=\exp(x^{x^2})$$ (sometimes called an "exponential stack")
since, as CRGreathouse pointed out, the other possibilities can be written more simply.
If that is the case, then
$$ln(f)= ln(e^{x^{x^2}})= x^{x^2}$$
NOT $x^2 ln(e^x)$.

8. Oct 26, 2007

### TMM

Why can't
$$e^{x^{x^2}}$$
be equal to
$x^2 ln(e^x)$?

Isn't that a property of logarithms?

9. Oct 26, 2007

### CRGreathouse

No. The property is

$$\log b^x=x\cdot\log b$$

but

$$e^{x^{x^2}}\neq (e^x)^{x^2}$$

in general.

10. Oct 26, 2007

### coomast

Hello erjkism.

You're question is one that can easily be solved wrongly. So here the method. Work out the details and prove the formula's used, it's important to learn the methods of proving them in case you can't exactly remember the end results.

Under the assumption that the equation is indeed $$exp(x^{x^{2}})$$, you could start with the following derivative:

$$\frac{d}{dx}\left[exp(f(x))\right]=\frac{df(x)}{dx}\cdot exp(f(x))$$

This will leave you with $$\frac{d}{dx}\left[x^{x^{2}}\right]$$ to solve.

Using the following formula:

$$\frac{d}{dx}\left([f(x)]^{g(x)}\right)=[f(x)]^{g(x)}\cdot ln(f(x)) \cdot\frac{dg(x)}{dx} +g(x) \cdot [f(x)]^{g(x)-1}\cdot \frac{df(x)}{dx}$$

with $$f(x)=x$$ and $$g(x)=x^2$$, gives you as the result:

$$exp(x^{x^{2}}) \cdot x^{x^2+1} \cdot \left(1+ln(x^2)\right)$$

I hope I didn't make a mistake....