Let $f'(0)=f''(0)=1$, $f^{(12)}(x)$ is differentiable and $g=f(x^{10})$ What's the value of $g^{(11)}(0)$? 
Let $f'(0)=f''(0)=1$,  $f^{(12)}(x)$ is differentiable and $g=f(x^{10})$
What's the value of $g^{(11)}(0)$?

I think it is important to use the fact that $f^{(12)}(x)$ is differentiable. However, I don't know hot to use it.
Can anyone help me solve the question?
 A: From Taylor's Theorem we have 
$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+h(x)x^2=f(0)+x+\frac{1}{2}x^2+h(x)x^2, \ x \in \mathbb{R},$$ 
where $h$ is a function s.t. $\displaystyle{\lim_{x \to 0}h(x)=0}.$
The polynomial $f(0)+x+\frac{1}{2}x^2$  is unique with the above property.
Since $g(x)=f\left(x^{10}\right)$ and $f$ is $13$ times differentiable $\Rightarrow \ \ g$ is $13$ times differentiable. Now $$g(x)=f\left(x^{10}\right)=f(0)+x^{10}+\frac{1}{2}x^{20}+h\left(x^{10}\right)x^{20}=\\
f(0)+x^{10}+x^{13}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^{7}\right), \ x \in \mathbb{R}.$$
Since $\displaystyle{\lim_{x \to 0}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^7\right) =0}$ from uniqueness in Taylor's theorem we conclude that $f(0)+x^{10}=g(0)+g'(0)x+\frac{g''(0)}{2}x^2+\ldots +\frac{g^{(13)}(0)}{13!}x^{13}$.
Therefore $g^{(11)}(0)=0.$
Actually $g^{(k)}(0)=0, \ \ \ \forall k \in \{1,2,\ldots,9,11,12,13\}$  and 
$g(0)=f(0), \ g^{(10)}(0)=10!.$
The fact that $f^{(12)}$ is differentiable is important because to apply Taylor's theorem to $g$ we need to know how many times $g$ is differentiable. Of course he could have said that $f^{(10)}$ is differentiable.
A: Taylor's Theorem tells us that a $k$-times differentiable function can be written in the form: 
$$f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_kx^k + h(x)x^{k}$$
where $h(x) \to 0$ as $x \to 0$. If $f'(0) = f''(0) = 0$ then $a_1 = a_2 = 0.$ Thus:
$$f(x) = a_0 + a_3x^3 + a_4x^4 + \cdots + a_kx^k + h(x)x^{k}.$$
You define $g(x) := f(x^{10})$ and so we have:
$$g(x) = a_0 + a_3x^{30} + a_4x^{40} + \cdots + a_kx^{10k} + h(x^{10})x^{10k}.$$
Clearly $g^{(p)}(0) = 0$ for all $1 \le p \le 29.$ In particular $g^{(11)}(0) = 0.$ Notice we only really needed $f$ to be twice differentiable. Anything else was a bonus. 
Without Taylor's Theorem you could apply the chain rule ten times.
