hello.
I'm I bit confused on differentiating this function
$\displaystyle y = x^{x^x}$
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 ?
$\displaystyle y=x^{x^{x}}$
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
$\displaystyle y'=x^{x^{x}}\bigg[\ln(x)(\ln(x+1)x^x+x^{x-1}\bigg]$