Studying past exam problems for my exam in ~$4$ weeks, and I came across this derivative as one of the questions. I actually have no idea how to solve it.
$$\frac{d}{dx} (x^{x^2})$$
Using the chain rule on it letting $x^2 = u$ led to me getting $2x^{x^2-2}$, which isn't right. The function acts like $e^x$ so I am thinking I have to convert it to this form.
So I took it to be:
$$\frac{d}{dx} (e^{{x^2}log(x)})$$
Not really sure where to go from here, or if I am going in the right direction.
Best Answer
Yes, you've done it correctly so far, and could proceed by the chain rule.
On the other hand, set
$$y = x^{x^2}$$
Then $\ln{y} = x^2 \ln{x}$; taking a derivative on both sides and using the chain rule for the left leads to
$$\frac{y'}{y} = 2x \ln{x} + \frac{x^2}{x}$$