the derivative of y=x^x^x

• Jan 28th 2010, 05:45 PM
Evan.Kimia
the derivative of y=x^x^x
Find derivative of y=x^x^x.

Whoa. Not sure how to attack this but heres what i think:

use logarithmic diff, so it would be xln(x^x)
Then product rule, so x*derivative of ln(x^x)+ln(x^x), or can i do the same law of logs again to make it x*derivative of ln(xln(x))+ln(x)?
• Jan 28th 2010, 05:50 PM
General
Quote:

Originally Posted by Evan.Kimia
Find derivative of y=x^x^x.

Whoa. Not sure how to attack this but heres what i think:

use logarithmic diff, so it would be xln(x^x)
Then product rule, so x*derivative of ln(x^x)+ln(x^x), or can i do the same law of logs again to make it x*derivative of ln(xln(x))+ln(x)?

First, You should specify it:
Is it $\displaystyle y=(x^x)^x$ or $\displaystyle y=x^{(x^x)}$ ?

Surely, for both cases you will use the logarithmic differentiation.
But there is a big difference between them when you do the Log.D.
• Jan 28th 2010, 05:51 PM
Evan.Kimia
its y=x^(x^x). I was about to edit this post: Can i just make it x^(x^2)?
• Jan 28th 2010, 05:54 PM
pickslides
Quote:

Originally Posted by Evan.Kimia
its y=x^(x^x). I was about to edit this post: Can i just make it x^(x^2)?

$\displaystyle x\times x = x^2 \neq x^x$
• Jan 28th 2010, 05:54 PM
General
Quote:

Originally Posted by Evan.Kimia
its y=x^(x^x). I was about to edit this post: Can i just make it x^(x^2)?

No !
$\displaystyle y=(x^x)^x=x^{ (x^2) }$
But $\displaystyle y=x^{(x^x)} \neq x^{ x^2 }$
• Jan 29th 2010, 02:00 AM
HallsofIvy
Be careful with your notation! $\displaystyle y= (x^x)^x= x^{x^2}$ but $\displaystyle x^{(x^x)}\ne x^{x^2}$.

If $\displaystyle y= x^{(x^x)}$, then $\displaystyle ln(y)= x^x ln(x)$ and, taking the logarithm again, $\displaystyle ln(ln(y))= x ln(x)+ ln(x)$.
• Jan 29th 2010, 05:29 AM
General
Quote:

Originally Posted by HallsofIvy
Be careful with your notation! $\displaystyle y= (x^x)^x= x^{x^2}$ but $\displaystyle x^{(x^x)}\ne x^{x^2}$.

If $\displaystyle y= x^{(x^x)}$, then $\displaystyle ln(y)= x^x ln(x)$ and, taking the logarithm again, $\displaystyle ln(ln(y))= x ln(x)+ {\color{red}ln(x)}$.

It should be $\displaystyle ln(ln(x))$.
• Jan 29th 2010, 05:37 AM
Evan.Kimia
I figure it out. I tried applying ln again, but it got wayy to sloppy. Since i solved the derivative for x^x before separately while trying to do this problem i was able to just plug in the derivative when doing the product rule on ln(y)=x^xln(x). My final answer was x^x^x[((x^x)*1/x)+(ln(x)*x^x(ln(x)+1)]. I then found this video on youtube and the guy did exactly what i did :). Thanks for all the help!

If you want to check out the video, its at YouTube - Calculus: Derivative of x^(x^x)