Here is the question: Let f(x) = x sin (x^N) . Calculate F^k(0) for the smallest integer k such that F^k(0) does not equal 0. (Note that the integer k determines the kth derivative).
In attempting this question, I wrote the function as a Taylor Series, giving me the result of
x Sin[x^n] = x Sum[((-1)^k (x^n)^(1 + 2 k))/(1 + 2 k)!, {k, 0, Infinity}]
However, I am not exactly sure as to where I continue from here. I do know that the answer depends on N, thus the answer being in some form of N.
We have
$$ x \sin(x^N) = x (x^N - \frac{x^{3N}}{3!} + \frac{x^{5N}}{5!} - \dots) = x^{N+1} - \frac{x^{3N + 1}}{3!} + \frac{x^{5N + 1}}{5!} - \dots $$
and so the first non-vanishing derivative of $f(x) = x\sin(x^N)$ (when $N > 0$) is the $(N + 1)$-th derivative and since the coefficient of $x^{N+1}$ in the Taylor series of $f$ is $\frac{f^{(N+1)}}{(N+1)!}$ we have
$$ 1 = \frac{f^{(N+1)}(0)}{(N + 1)!} \implies f^{(N+1)}(0) = (N+1)! $$