The 32nd derivative of a real function 
Let  $$ f(x) :=\exp{[{\sin{(x^{10}+x^{11})}}]}, \qquad x \in \mathbb R. $$
  Calculate  $$ \frac{d^{32}}{dx^{32}}f(x)\mid_{x=0} .  $$

I'm really puzzled. What should I do? Should I use some power series? But the question is: how can I use them?
Well, since $\exp(\cdot)$ and $\sin(\cdot)$ are analytic, I can write 
$$
f(x)= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty} \left[(-1)^k\frac{(x^{10}+x^{11})^{2k+1}}{(2k+1)!}\right]^n\frac{1}{n!}
$$
But now what do I have to do? 
Thanks in advance.  
 A: Using the power series expansion is not so complicated if you notice that differentiating 32 times will eliminate any power of $x$ below 32, and setting $x=0$ will eliminate any term which has a power of $x$ greater than 32. There are high powers of $x$ involved so we do not need to look at many terms, and we can cut off the expansion once we get to $x^{32}$. It seems to me that just writing it out explicitly, leaving out irrelevant terms as you go, is as easy as any other method.
Do two stages using $R$ for an irrelevant remainder:
$$\sin(x^{10}+x^{11}) = x^{10}+x^{11}-\frac {(x^{10}+x^{11})^3}{3!}+R = x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2+R$$
Now note that when this is exponentiated, the terms in $x^{30}$ and higher will become irrelevant when the whole expression is squared or cubed, so (ignoring irrelevant terms):
$$e^{x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2} = 1+\left(x^{10}+x^{11}-\frac {x^{30}} 6-\frac{x^{31}} 2-\frac{x^{32}} 2\right)+\frac 1 2 (x^{10}+x^{11})^2+\frac 1 6(x^{10}+x^{11})^3+R$$
The only ultimately relevant term, as noted above, is the one in $x^{32}$, and picking this out from the terms we have we see that: $$f(x)=-\frac{x^{32}}2+\frac{x^{32}}2+R$$
The coefficient we are looking for vanishes, and the answer to the question is therefore 0.
A: HINT: Suppose that you’ve written out the power series as $f(x)=\sum_{n\ge 0}a_nx^n$. When you take the $32$-nd derivative, the terms in $x^n$ with $n<32$ will disappear. Then when you evaluate $f^{(32)}(x)$ at $x=0$, all of the terms except the constant term will disappear. Thus, you need only find $a_{32}$. To do this, you need to examine the summation
$$f(x)= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty} \left[\frac{(x^{10}+x^{11})^{2k+1}}{(2k+1)!}\right]^n\frac{1}{n!}$$
closely to see which combinations of $k$ and $n$ give you $x^{32}$ terms.
(Actually, first you need to correct it: the Taylor series for the sine alternates, and you’re missing a factor of $(-1)^k$. And make sure that you realize that the $\frac1{n!}$ is not inside the inner summation.)
A: The function $f$ is analytic hence $f(x)=\sum\limits_{n\geqslant0}a_nx^n$ for some coefficients $(a_n)_{n\geqslant0}$ and $f^{(n)}(0)=n!a_n$ for every $n\geqslant0$. Furthermore, for every $n\geqslant0$, $f(x)=\sum\limits_{k=0}^na_kx^k+o(x^{n+1})$. Hence the task is to compute the coefficient $a_{32}$ of $x^{32}$ in the expansion of $f(x)$ at $x=0$.
To do so, note that $$\sin(u)=u-\frac16u^3+O(u^5)$$ when $u\to0$ and consider $u=x^{10}+x^{11}=x^{10}(1+x)$. Then $$\sin(u)=x^{10}(1+x)-\frac16x^{30}(1+x)^3+O(x^{50}),$$ that is, $\sin(u)=x^{10}v+o(x^{32})$ with $v=1+x-\frac16x^{20}-\frac12x^{21}-\frac12x^{22}$. 
Likewise, $$\mathrm e^w=1+w+\frac12w^2+\frac16w^3+O(w^4)$$ when $w\to0$ hence $$f(x)=1+x^{10}v+\frac12x^{20}v^2+\frac16x^{30}v^3+O(x^{40}).$$ The $x^{32}$ terms in the RHS are $x^{10}(-\frac12x^{22})$ in $x^{10}v$, $0$ in $x^{20}v^2$ since $v^2$ has no $x^{12}$ term, and $3x^{32}$ in $x^{30}v^3$ since $v^3=1+3x+3x^2+o(x^2)$.
Finally, $a_{32}=-\frac12+\frac163=0$ hence $f^{(32)}(0)=0$.
