# differentiating

• Jun 29th 2008, 02:13 AM
javax
differentiating
hello.
I'm I bit confused on differentiating this function

$\displaystyle y = x^{x^x}$
• Jun 29th 2008, 02:18 AM
Moo
Hello,

Is it $\displaystyle (x^x)^x$ or $\displaystyle x^{x^x}$?
(see the latex code ;))

In both cases, you will need that :

$\displaystyle x^x=e^{x \ln x}$

Using the chain rule :

$\displaystyle (x^x)'=(x \ln x)' e^{x \ln x}=(x \ln x)' x^x$

$\displaystyle (x \ln x)'=1+\ln x$ (product rule)

Therefore $\displaystyle \boxed{(x^x)'=(1+\ln x)x^x}$

Does it help ?
• Jun 29th 2008, 02:25 AM
javax
y
It was $\displaystyle x^{x^x}$. Yes Moo it helped. Thanks a lot!
• Jun 29th 2008, 08:12 AM
Mathstud28
Quote:

Originally Posted by javax
hello.
I'm I bit confused on differentiating this function

$\displaystyle y = x^{x^x}$

$\displaystyle y=x^{x^{x}}$

So $\displaystyle \ln(y)=\ln(x)\cdot{x^x}$

Differentiating we get

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

Now as moo showed above

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

Now seeing that y is our original equation and solving for y' we get

$\displaystyle y'=x^{x^{x}}\bigg[\ln(x)(\ln(x+1)x^x+x^{x-1}\bigg]$