# Derviate x^x^x

1. Oct 23, 2007

### erjkism

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

differentiate f(x)= x^x^x

2. Relevant equations
chain rule
product rule

3. The attempt at a solution

x^x (lnx)

i dont know what to do after this

2. Oct 23, 2007

### quasar987

A function raised to another function is an exponential:

In general, $$f(x)^{g(x)} = \exp(\ln(f(x)^{g(x)}))=\exp(g(x)\ln(f(x)))$$

And you know how to differentiate an exponential.

So, can you use what I wrote to write x^x^x as an exponential?

3. Oct 23, 2007

### arildno

Would that be:
1. $$x^{(x^{x})}=x^{x^{x}}$$
2. $$(x^{x})^{x}=x^{x^{2}}$$

Learn to use parentheses..

4. Oct 24, 2007

### HallsofIvy

Don't just leave x^x(ln x) by itself! If f= x^x^x, then ln(f)= x^x ln(x). Now DO IT AGAIN! ln(ln(f))= ln(x^x ln(x))= ln(x^x)+ ln(ln(x))= xln(x)+ ln(ln(x)).

Use the chain rule to differentiate both ln(ln(f(x)) and ln(ln(x)).

5. Oct 25, 2007

### Gib Z

I sometimes get annoyed with exponential notation for exactly that reason. My opinion is if the exponent is any larger than 1 term, write it is terms of exp(...).

6. Oct 26, 2007

### nizi

Neglecting the given attempt, I put
$$y = x^{x^{x}}$$
$$z = x^{x}$$
and develop as follows.
$$\ln y = \ln x^{x^{x}} = z \ln x$$
$$\frac{y'}{y} = z' \ln x + z \frac{1}{x}$$
here I calculate the differentiation of $$z$$
$$z = x^{x}$$
$$\ln z = x \ln x$$
$$\frac{z'}{z} = \ln x + x \frac{1}{x}$$
$$z' = z \left( { \ln x + 1 } \right) = x^{x} \left( { \ln x + 1 } \right)$$
Accordingly
$$y' = y \left( { x^{x} \left( { \ln x + 1 } \right) \ln x + x^{x} \frac{1}{x} } \right) = x^{x^{x}+x-1} \left( { x \left( { \ln x + 1 } \right) \ln x + 1 } \right)$$