I have function: $f(x) = \operatorname{sgn}(x^3-4x)\sin(x^3-2x^2)$
which can be represented as:
$ f(x) = -\sin(x^3-2x^2) ;\quad x \in (-\infty, -2) \cup (0,2)$,
$ f(x) = \sin(x^3 - 2x^2); \quad x \in (-2, 0) \cup (2,\infty) $,
$f(x) = 0\quad$ if $x=-2\:$ or $\:x=0\:$ or $\:x=2 $.
Therefore I have derivatives:
$ f(x)' = -\cos(x^3-2x^2)(3x^2-4) ;\quad x \in (-\infty, -2) \cup (0,2)$,
$ f(x)' = \cos (x^3 - 2x^2)(3x^2-4);\quad x \in (-2, 0) \cup (2,\infty) $,
$f(x)' = 0\quad$ if $x=-2\:$ or $\:x=0\:$ or $\:x=2 $.
Is that correct? How can I find one sided derivatives, are there any?
$\sin$ is continuous everywhere and $sgn$ is continuous everywhere except $x = 0$. So $sgn(x^3-4x)\sin(x^3-2x^2)$ will be continuous everywhere except maybe where $x^3-4x = 0$. i.e. everywhere except $x=-2,0,2$.
And yes, your calculations for $f'(x)$ are correct on all the points $x \ne [-2|0|2]$.
At $x= 0,2$ we have $x^3 - 2x^2 = 0$ so $\sin(x^3-2x^2) = 0$ so $\lim_{x\to [0|2]}sgn(x^3-4x)\sin(x^3-2x^2) = 0$ so $sgn(x^3-4x)\sin(x^3-2x^2)$ is continuous at $x = 0,2$.
But at $x = -2$ , $x^3 - 2x^2 = -16$ and $\sin(x^3-2x^2) = \sin(-16)=k \ne 0$. $\lim_{x\to -2^+} sgn(x^3-4x)\sin(x^3-2x^2) = k$ and $\lim_{x\to -2^-} sgn(x^3-4x)\sin(x^3-2x^2)=-k$ so $f$ is not continuous at $x = -2$ and $f$ is not differentiable at $x = -2$.
So your result is incorrect in that $f'(-2)$ does not exist.
$f$ will be differentiable at $x = [0|2]$ if the left and right side limits of $f'$ are equal.
But $\cos(x^3-2x^2)(3x^2-4) = 1*(3x^2 - 4) = 8$ if $x = 2$ and $\cos(x^3-2x^2)(3x^2-4) = 1*(3x^2 - 4) = -4$ if $x = 0$.
So $\lim_{x\to 0^-} \cos(x^3-2x^2)(3x^2-4)= -4\ne 4 = \lim_{x \to 0^+} \cos(x^3-2x^2)(3x^2-4)$ and $\lim_{x\to 2^-} \cos(x^3-2x^2)(3x^2-4)=8\ne -8 = \lim_{x\to 2^+} \cos(x^3-2x^2)(3x^2-4)$.
so $f$ is not differentiable at $x = 0,2$.
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You error is in assuming that as $f(x_1) = c$ a constant then $f'(x_1) = 0$.
By this reasoning I could assume if $f(x) = x^4$ then as $f(2) = 16$ then $f'(2) = 0$ which is obviously not correct.
For $f(x_1) = c$ to be a constant function and therefore $f'(x_1)= 0$ we must have $f(x) = c$ for all $x$ in an interval around $x_1$.