Question:
Let $f: \mathbb R \to \mathbb R$ be infinitely differentiable and let $P_n(t)$ be it's Taylor Series of order $n$ about $x=0$.
Let $g(x)=f(x^3)$.
Show that $Q(t)=P_n(t^3)$ is the Taylor Series of order $3n$ for $g$ about $x=0$.
As for what I've tried:
I followed my instincts and wrote out the series of differentials for $g(x)$ to see what comes out, but I got a lot of terms and can't find a way to connect it to it's Taylor Series:
$g^{(n)}(x)=((3x^2)f'(x^3))^{(n-1)}=((6x)f'(x^3)+(9x^4)f''(x^3))^{(n-2)}=((6)f'(x^3)+(54x^3)f''(x^3)+(27x^6)f'''(x^3))^{(n-3)}$
Not sure how to go about this problem. Any ideas?