Why can't I use the chain rule for the derivative of $x^{x^2}$? I realize you can do this with implicit differentiation, but I thought you could also take the derivative by using the chain rule. Hence, it should be $2x\cdot x^{x^2-1}$
What's wrong with this?
 A: You can, but not in the way you are thinking.  To apply the chain rule, you clearly have to specify your inner and outer functions:
$$\frac d{dx}f(g(x))=f'(g(x))g'(x)$$
What are your $f(x)$ and $g(x)$?  You seem to pick $f(x)=x^x$ and $g(x)=x^2$, but then,
$$f(g(x))=(x^2)^{(x^2)}=x^{2x^2}\ne x^{x^2}$$
(if you are thinking power rule, then what about the exponential rule where you take the derivative of $a^x$?)

If you are insistent on the chain rule, you could apply the multivariable chain rule.  Here, I will let $y=u^v$, $u=x$, and $v=x^2$.  Thus,
$$\frac{dy}{dx}=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}=vu^{v-1}+(\ln(u)u^v)(2x)$$

$$\frac{d}{dx}x^{x^2}=x^{x^2}(2x\ln(x)+x)$$


If you think this is too much, you could try rewriting to make it work for the regular chain rule:
$$x^{x^2}=(e^{\ln(x)})^{x^2}=e^{x^2\ln(x)}$$
Now you can set $f(x)=e^x$ and $g(x)=x^2\ln(x)$ to get

$$\frac d{dx}x^{x^2}=e^{x^2\ln(x)}(2x\ln(x)+x)=x^{x^2}(2x\ln(x)+x)$$


As a general tip, when dealing with exponential functions, you may wish to apply logarithmic differentiation:
$$y=x^{x^2}\implies\ln(y)=\ln(x^{x^2})=x^2\ln(x)$$
$$\frac{y'}y=2x\ln(x)+x$$

$$y'=y(2x\ln(x)+x)=x^{x^2}(2x\ln(x)+x)$$

A: In order to use the chain rule, you need to write $f(x) = x^{x^2}$ as a composition $f(x) = g(h(x))$ of two functions whose derivative you can compute and then use the formula
$$ f'(x) = g'(h(x)) h'(x). $$
In your case, we can write 
$$ f(x) = x^{x^2} = e^{x^2 \ln x} $$
and then use the chain rule for $g(x) = e^x$ and $h(x) = x^2 \ln x$ resulting in
$$ f'(x) = e^{x^2 \ln x} (2x \ln x + x) = x^{x^2 + 1} (2 \ln x + 1).$$
The reason of rewriting $x^{x^2}$ as $e^{x^2 \ln x}$ (other than the fact it is very often the definition of $x^{x^2}$!) lies precisely in that in enables you to use the chain rule with the functions I wrote above.
A: $$
\begin{align}
\lim_{h \to 0} \frac1h \bigg((x+h)^{(x+h)^2}-x^{x^2}\bigg) 
& = \lim_{h \to 0} \frac{x^{x^2+2hx}}h \bigg((1+\frac{h}x)^{x^2+2hx}-x^{-2hx} \bigg) \\[10pt]
& =  \lim_{h \to 0} \frac{x^{x^2+2hx}}h\bigg((1-x^{-2hx}+\frac{h}x(x^2+2hx) \bigg) \\[10pt]
& =  x^{x^2}\lim_{h \to 0}\frac{x^{2hx}-1}{h} +  \lim_{h \to 0}x^{x^2+2hx}(x+2h) \\[10pt] 
& =  x^{x^2}\lim_{h \to 0}\frac{(x^{2x})^h-1}{h} +  x^{x^2}x \\[10pt]
& = x^{x^2}(\log x^{2x} +x) \\[10pt]
& = x^{x^2}(2\log x +1)x
\end{align}
$$
A: As you know, we can prove that:
$$(x^n)'=nx^{n-1},$$
For any $n\in\mathbb{N}$ ou even any $n\in\mathbb{R}$. But when we prove this we can't tell anything about the following derivative:
$$(x^{f(x)})'.$$
In fact, we have proven the identity only when the exponent is a number, and not a function. This is a very common mistake amongst my calculus students, and the calculations in the answer above are the correct way to get such derivative.
