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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?

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    Write it as $e^{x^2\ln x}$. How you apply the chain rule to $x^x$, by example? – 2017-02-19
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    Write down explicitly the functions that (you think) are part of the "*chain*", and you'll see what's missing. – 2017-02-19
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    I agree with dxiv. Try to write down the two functions that you think your function is the composite of. For example, for $y = \sin(x^2)$, you can break it down into two steps, $u = x^2$ and $y = \sin u$. In your case, you'll see that this isn't so simple to achieve. – 2017-02-19

4 Answers 4

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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.

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    I'm not sure why you need to write it as e^(x^2.... – 2017-02-19
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    @Goldname because the power rule only applies for constant exponents, so you can't do $x^x\to x(x^{x-1})$ the way you can for $x^c$. If you know the derivative of $x^x$ then you don't need to rewrite it, but it's a safe guess that you dont, since hat problem and this problem are more or less solved the same way. – 2017-02-19
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    @Goldname because **it is the definition** of any power of functions, that is: $$f(x)^{g(x)}:=e^{g(x)\ln f(x)}$$ In other words: writing $x^x$ is a short way to write $e^{x\ln x}$. – 2017-02-19
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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:

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

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

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

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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.

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$$ \begin{align} \lim_{h \to 0} \frac1h \bigg((x+h)^{(x+h)^2}-x^{x^2}\bigg) & = \lim_{h \to 0} \frac{x^{x^2+2hx}}h \bigg((1+\frac{h}x)^{x^2+2hx}-x^{-2hx} \bigg) \\[10pt] & = \lim_{h \to 0} \frac{x^{x^2+2hx}}h\bigg((1-x^{-2hx}+\frac{h}x(x^2+2hx) \bigg) \\[10pt] & = x^{x^2}\lim_{h \to 0}\frac{x^{2hx}-1}{h} + \lim_{h \to 0}x^{x^2+2hx}(x+2h) \\[10pt] & = x^{x^2}\lim_{h \to 0}\frac{(x^{2x})^h-1}{h} + x^{x^2}x \\[10pt] & = x^{x^2}(\log x^{2x} +x) \\[10pt] & = x^{x^2}(2\log x +1)x \end{align} $$