Given that $f'(0) = f''(0) = 1$, $f^{(12)}$ exists, and $g(x)= f(x^{10})$. Find $g^{(11)}(0)$. Given $f'(0) = f''(0) = 1$, $f^{(12)}$ exists, and $g\colon x \mapsto f(x^{10})$. Find $g^{(11)}(0)$. 
 A: Hint:
By chain rule, we have 
$$g'(x) = 10f'(x^{10})x^9$$
and by product rule and chain rule 
$$g''(x) = 10\times 10f''(x^{10})x^9x^9+10\times 9f'(x^{10})x^9 = 100x^{18}f''(x^{10}) + 10\times9f'(x^{10})x^9$$
Since $g^{(11)}(x)$ will be evaluated at $x=0$, the first term can be discarded directly considering the exponent grows for each differentiation. 
Continuing, you will get
$$g^{(10)}(x) = \text{loads of discarded terms} + 10\times9 \times \dots\times 2 f'(x^{10})$$
Now calculate $g^{(11)}(x)$ and set $x=0$ to get the result. 
(Side note: We have used the fact that $f^{(12)}(x)$ exists in the discarded terms above, since this implies that all derivatives $f^{(n)}(x)$ for $n < 12$ also exist.)
A: Given any $t\mapsto f(t)$ defined in a neighborhood $\ ]{-h},h[\ $ of $t=0$ the function
$$g(x):=f\bigl(x^{10}\bigr)\qquad\bigl(|x|<h'\bigr)\tag{1}$$
is even, and if $f$ is sufficiently differentiable at $0$ one necessarily has $g^{(11)}(0)=0$: Since $g(x)=g(-x)$ for all $x$, by the chain rule one has $g^{(k)}(x)=(-1)^k g^{(k)}(-x)$ for all $k\geq0$, which implies $g^{(k)}(0)=0$ for odd $k$.
For an approach that does not make use of the special situation at hand assume for simplicity that $f\in C^\infty$. Then by Taylor's theorem we may write
$$f(t)=f(0)+t+{t^2\over2}+t^3 f_1(t) \qquad\bigl(|t|<h\bigr)$$
with $f_1\in C^\infty$. It follows from $(1)$ that
$$g(x)=f(0)+x^{10}+{x^{20}\over 2}+x^{30}f_1\bigl(x^{10}\bigr)\ .$$
Here one can read off that all $g^{(k)}(0)$ for $k\in[1,29]\setminus\{10,20\}$ are $=0$.
A: By Faa di Bruno we have
$$
 g^{(11)}(x) = \sum_{\substack{m_j \ge 0\\ m_1 + 2m_2 + \cdots + 11m_{11} = 11}} \frac{11!}{m_1!1!^{m_1} \cdots m_{11}!11!^{m_{11}}}f^{(m_1 + \cdots + m_{11})}(x^{10})\prod_{j=1}^{11} \bigl((x^{10})^{(j)}\bigr)^{m_j}
$$
Noting that $(x^{10})^{(j)}|_{x=0} = 0$ when $j \ne 10$, only those summands with $m_j = 0$ for $j \ne 10$ are $\ne 0$ at $x=0$. But for any non-negative $m_j$ with $m_1 + \cdots + 11m_{11} = 11$ we must have $m_j > 0$ for some $j \ne 10$. Hence $g^{(11)}(0) = 0$.
