Practicing the chain rule I'm trying to understand the answer to the problem in this screenshot
The goal is to use the chain rule to find the derivative of
$$e^{\sin(x^2)}$$
As seen in the screenshot, the correct answer is
$$2x e^{\sin(x^2)}\cos(x^2)$$
From my understanding the chain goes:
$$f'(g'(h(x)))  h'(x)  g'(x) $$
but I am not too sure how it gets $\cos(x^2)  $

Your statement of the chain rule is incorrect. The derivative of $f(g(h(x)))$ with respect to $x$ is $f'(g(h(x)))g'(h(x))h'(x)$.
In your case, $f(x) = e^x$, $g(x) = \sin(x)$, and $h(x) = x^2$, so
$$f(g(h(x))) = f(g(x^2)) = f(\sin(x^2)) = e^{\sin(x^2)}$$
To apply the chain rule, we need the derivatives of $f$, $g$, and $h$, which are:
$$\begin{aligned}
f'(x) &= e^x \\
g'(x) &= \cos(x) \\
h'(x) &= 2x \\
\end{aligned}$$
Then the three factors of the chain rule are:
$$\begin{aligned}
f'(g(h(x)) &= f'(\sin(x^2)) = e^{\sin(x^2)} \\
g'(h(x)) &= \cos(x^2) \\
h'(x) &= 2x \\
\end{aligned}
$$
Multiplying these together, we get the final answer:
$$f'(g(h(x)))g'(h(x))h'(x) = 2x e^{\sin(x^2)}\cos(x^2)$$
which matches the correct answer at your link.
