Differentiating f(x) = x^x^x: A Step-By-Step Guide

[erjkism]

Homework Statement

differentiate f(x)= x^x^x

Homework Equations

chain rule
product rule

The Attempt at a Solution

x^x (lnx)

i don't know what to do after this

 

[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?

 

[arildno]

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

Learn to use parentheses..

 

[HallsofIvy]

Homework Statement

differentiate f(x)= x^x^x

Homework Equations

chain rule
product rule

The Attempt at a Solution

x^x (lnx)

i don't know what to do after this

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)).

 

[Gib Z]

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

Learn to use parentheses..

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(...).

 

[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
$$y' = y \left( { x^{x} \left( { \ln x + 1 } \right) \ln x + x^{x} \frac{1}{x} } \right) = x^{x^{x}+x-1} \left( { x \left( { \ln x + 1 } \right) \ln x + 1 } \right)$$