Find $g^{10}(x)$ and $g^{11}(x)$ when $x=0$.

Let $f$ be a smooth function such that $f'(0) = f''(0) = 1$. Let $g(x) = f(x^{10})$. Find $g^{(10)}(x)$ and $g^{(11)}(x)$ when $x=0$.

I tried applying chain rule multiple times:

$$g'(x) = f'(x^{10})(10x^9)$$

$$g''(x) = \color{red}{f'(x^{10})(90x^8)}+\color{blue}{(10x^9)f''(x^{10})}$$

$$g^{(3)}(x)=\color{red}{f'(x^{10})(720x^7) + (90x^8)f''(x^{10})(10x^9)}+\color{blue}{(10x^9)(10x^9)f^{(3)}(x^{10})+f''(x^{10})(90x^8)}$$

The observation here is that, each time we take derivative, one "term" becomes two terms $A$ and $B$, where $A$ has power of $x$ decreases and $B$ has power of $x$ increases. $A$ parts will become zero when evaluated at zero, but what about $B$ parts?

We have that $f(x)=f(0)+x+x^2/2+o(x^2)$. Therefore the expansion of $g$ at $0$ is $$g(x)=f(x^{10})=f(0)+x^{10}+x^{20}/2+o(x^{20}).$$ Hence $g^{(10)}(0)/10!$, which is the coefficient of $x^{10}$, is equal to $1$, and we conclude that $g^{(10)}(0)=10!$. Are you able now to find $g^{(11)}(0)$?
• So in fact $g^{(n)}(0) = 0$ for $10 <n < 20$? – 3x89g2 Aug 28 '16 at 18:01
• @Misakov Yes it is. We are using the fact that the Taylor expansion is unique. So if $f$ is regular and $f(x)=\sum_{k=0}^n a_k x^k+o(x^n)$ then $a_k=f^{(k)}(0)/k!$ for $k=0,\dots,n$. – Robert Z Aug 29 '16 at 10:16
\begin{align*} g'(x) &= 10x^9f'(x^{10})\\ g''(x) &= 10x^9f''(x^{10})+10.9x^8f'(x^{10})\\ & \vdots\\ g^{(9)}(x)&= p(x)+10.9.\cdots .3x^2f''(x^{10})+10.9.\cdots .2x^1f'(x^{10})\\ g^{(10)}(x)&= q(x)+10.9.\cdots .2x^1f''(x^{10})+10!x^0f'(x^{10})\\ g^{(11)}(x)&= r(x)+10!f''(x^{10}) \end{align*} where $p(x)$, $q(x)$ and $r(x)$ are real polynomials which haven't constant term, so $p(0)=q(0)=r(0)=0$. Consequently $g^{(11)}(0)=g^{(10)}(0)=10!$
From there, you should be able to identify which terms in the sum go to zero when $x$ is evaluated at $0$.