Taking the derivative of $f(x)=x^{e^{e^x}}$ How can I take the derivative of $f(x)=x^{e^{e^x}}$?
How do I apply the chain rule?  Thanks for the help!
 A: $f(x)= e^{e^{e^x}\ln x}$ (smooth on $(0,\infty)$, so $$
f^\prime(x)= \left(\frac{d}{dx}e^{e^x}\ln x\right)\cdot e^{e^{e^x}\ln x}
$$
Only the first term is somehow tricky, but using the rules for the derivative of a product of functions and using at some point the chain rule to get $\frac{d}{dx}e^{e^x} = e^xe^{e^x}$, you'll be fine.
A: The function is
$$f(x)=x^{e^{e^x}}$$
The derivative is:
$$x^{e^{e^x}} \left[ e^{e^x} \left(\dfrac{1}{x} + e^x \ log_e{x} \right) \right]$$
This is found using a method called "implicit differentiation". So you start off with your formula:
$$y=x^{e^{e^x}}$$
Then you take the logarithm of each side...we'll use the natural logarithm:
$$ln(y)=ln \left( x^{e^{e^x}} \right)$$
$$ln(y)=e^{e^x}[ln(x)]$$
Now we differentiate both sides. You use the chain rule for the left side and the product rule for the right side
$$\left( \dfrac{1}{y} \right) y'=e^{e^x} \left( \dfrac{1}{x} \right) +(e^{e^x})'[ln(x)]$$
Now you use the chain rule to take care of $(e^{e^x})'$:
$$\left( \dfrac{1}{y} \right) y'=e^{e^x} \left( \dfrac{1}{x} \right) +e^{e^x}e^x[ln(x)]$$
Now you are very close. Multiply both sides by $y$ to get the derivative, $y'$ isolated.
$$y'=y \left[ e^{e^x} \left( \dfrac{1}{x} \right) +e^{e^x}e^x[ln(x)] \right]$$
Essentially, we are done but we can simplify it a little more by substituting $y$ in:
$$y'=x^{e^{e^x}} \left[ e^{e^x} \left( \dfrac{1}{x} \right) +e^{e^x}e^x[ln(x)] \right]$$
We can also factor a little bit to obtain:
$$y'=x^{e^{e^x}} \left[ e^{e^x} \left( \dfrac{1}{x} +e^x ln(x) \right) \right]$$
And were done!
A: One way to deal with functions of the form $y = x^{g(x)}$ is by taking a logarithm. So if
$$
y = x^{e^{e^x}},
$$
then 
$$
\ln(y) = \ln(x)e^{e^{x}}.
$$
Now take a $\frac{d}{dx}$ (and use the product rule on the right hand side):
$$
\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}e^{e^{x}} + \ln(x)\frac{d}{dx}e^{e^{x}}.
$$
Now all you need to do is to find
$$
\frac{d}{dx} e^{e^{x}}
$$
and solve for $\frac{dy}{dx}$.
