Calculus determine whether a function is differentiable (hard) 
I know that i need to prove 
$$\lim_{ h\to0^+}\dfrac{ g(x+h) - g(x) }{h}$$
exists but i cannot find a way to simplify the expression ...
 A: This is a cute, although way over engineered question. Let us denote the function we want to investigate by $g_n(x)$ (because it depends on $n \in \mathbb{N}$). Note that the function $f$ is monotonic increasing so is $k$ is even we have $f(x^k) \geq f(0) = e - 1 > 0$. If $k$ is odd then then $f(x^k) > 0$ if $x > -1$ and $f(x^k) < 0$ if $x < -1$ and $f(x^k) = 0$ if $x = -1$. Hence, we have
$$ g_n(x) = \begin{cases} 100 f(x) - \sum_{k=1}^n f(x^k) & x > -1 \\
100 f(x) - \sum_{k=1}^n (-1)^k f(x^k) & x < -1.
\end{cases}$$
From this formula, it is clear that $g_n$ is differentiable at all $x \neq -1$. To see what happens at $x = -1$, note that
$$ g_n'(x) = \begin{cases}
100 f'(x) - \sum_{k=1}^n k x^{k-1} f'(x^k) & x > -1, \\
-100f'(x) - \sum_{k=1}^n (-1)^k k x^{k-1} f'(x^k) & x < -1. 
\end{cases} $$
Since $g_n(x)$ is the restriction of a continuously differentiable function on $(-1,\infty)$ and $(-\infty, -1)$, we can calculate the one-sided derivatives of $g_n$ at $x = -1$ by taking the one-sided limits of $g_n'(x)$. Hence,
$$ (g_n)_{+}'(-1) = \lim_{x \to -1^{+}} g_n'(x) = 100 f'(-1) - \sum_{k=1}^n k (-1)^{k-1} f'((-1)^k), \\
(g_n)_{-}'(-1) = \lim_{x \to -1^{-}} g_n'(x) = -100 f'(-1) + \sum_{k=1}^n k f'((-1)^k).$$
In order for $g_n$ to be differentiable at $x = -1$, we must have $(g_n)_{+}'(-1) = (g_n)_{-}'(-1)$ so we get the equation
$$ 200 f'(-1) = \sum_{k=1}^n k(1 + (-1)^{k-1}) f'((-1)^k) = \sum_{k \leq n \text{ odd}} 2k f'(-1) \iff \\
100 = \sum_{k \leq n \text{ odd}} k.
$$
Hence, the function is differentiable iff $n = 19$ or $n = 20$ and so the final answer is $39$. 

It is also nice to observe this behavior graphically. For the original function, there are some resolution related issues involved in plotting the function but if we consider instead the function
$$ g_n(x) = 4|f(x)| - \sum_{k=1}^n |f(x^k)| $$
then the same calculation as above shows that $g_n$ should be differentiable only when $n = 3$ or $n = 4$ which is consistent with the plots of the graphs of $g_n$ for $n=1,\dots,5$:

