How do find the fourth derivative of $e^{x^2}$ So using chain rule for the first derivative:
$$f' = 2x\cdot e^{x^2}$$
then using product rule for the $$f'' = 2x(2xe^{x^2}) + e^{x^2}\cdot 2$$
Is there an easy way to get the third? Does the second deriative simplify to:
$$4x^2e^{x^2} + 2e^{x^2} = (4x^2 + 2)(e^{x^2})$$
Can I then take chain rule again to get $f^{(3)}$?

Use Maclaurin expansion: $y=e^{x^2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$:
$$\begin{align}y'&=\sum_{n=1}^{\infty} \frac{2nx^{2n-1}}{n!}\\
y'' &=\sum_{n=1}^{\infty} \frac{2n(2n-1)x^{2n-2}}{n!}\\
y''' &=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)x^{2n-3}}{n!}\\
y^{(4)}&=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)(2n-3)x^{2n-4}}{n!}=\\
&=4\sum_{n=2}^{\infty} \frac{(2n-1)(2n-3)x^{2(n-2)}}{(n-2)!}=\\
&=4\sum_{n=0}^{\infty} \frac{(2n+3)(2n+1)x^{2n}}{n!}=\\
&=16\sum_{n=1}^{\infty} \frac{(n-1+1)x^{2n}}{(n-1)!}+32\sum_{n=1}^{\infty} \frac{x^{2n}}{(n-1)!}+12\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}=\\
&=(16x^4+48x^2+12)e^{x^2}.\end{align}$$
