Why is the derivative of $f(x) = x-(x^2-2x)$ not defined at $x= 0$?
For a function $f(x) = x-|x^2-2x|$, the differentiation is possible when is broken into a piece wise function. i.e.
$$f(x) =\begin{cases}x-(x^2-2x)&x^2-2x\ge0\\ x+(x^2-2x)&x^2-2x < 0\end{cases}$$
$$f'(x)=\begin{cases}1-(2x-2)&x^2-2x > 0\\ 1+(2x-2)&x^2-2x < 0\end{cases}$$
The derivative at $x=0$ is not defined. Why? Please explain.
Let $x$ be small positive. Then $x^2-2x$ is negative, so $x-|x^2-2x|=x-(2x-x^2)=-x+x^2$.
Let $x$ be negative. Then $x^2-2x$ is positive, so $x-|x^2-2x|=x-(x^2-2x)=3x-x^2$.
So if $h$ is small positive, then $\frac{f(0+h)-f(0)}{h}=-1+h$.
If $h$ is negative, then $\frac{f(0+h)-f(0)}{h}=3-h$.
Thus the limit of $\frac{f(0+h)-f(0)}{h}$ as $h$ approaches $0$ from the right is $-1$.
The limit of $\frac{f(0+h)-f(0)}{h}$ as $h$ approaches $0$ from the left is $3$.
It follows that $\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ does not exist.
Remark: It might be useful to have software graph the function, so that you can see the "kink" at $x=0$. This example is an more complicated version of the probably familiar fact that the function $g(x)=|x|$ is not differentiable at $x=0$.
$$f'(0)=\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}\frac{x-|x^2-2x|}{x}=\lim_{x \to 0}\frac{x-|x(x-2)|}{x} $$ If $x \to 0^+$ then $x(x-2)<0$: $$\lim_{x \to 0^+}\frac{x-|x(x-2)|}{x}=\lim_{x \to 0^{+}}\frac{x+x(x-2)}{x}=\lim_{x \to 0^{+}}1+(x-2)=-1$$
If $x \to 0^-$ then $x(x-2)>0$: $$\lim_{x \to 0^+}\frac{x-|x(x-2)|}{x}=\lim_{x \to 0^{+}}\frac{x-x(x-2)}{x}=\lim_{x \to 0^{+}}1-(x-2)=3$$ Then $$\lim_{x \to 0}\frac{x-|x(x-2)|}{x}=f'(0)$$ does not exist.
I think the problem is the way you were taught derivatives. There is a theorem which states that the derivative of $f(g(x))$ is $f'(g(x)) g'(x)$ in every point where both $f$ and $g$ are differentiable. In your case you have $g(x)=x^2-2x$ and $f(y) = |y|$ which is differentiable for $y\neq 0$ and from the theorem you find that your function is differentiable for $x\neq 0$ and $x\neq 2$. But the theorem tells you nothing in the case that the assumptions are not satisfied. When $g(x)=0$ i.e. for $x=0$ and $x=2$ you should use the definition of derivative and check if the limits exists and are finite: $$ \lim_{x\to 0} \frac{f(g(x+h))-f(g(x))}{h},\qquad \lim_{x\to 2} \frac{f(g(x+h))-f(g(x))}{h} $$
If you look at the function $|x^2-2x|$ as a function defined piecewise you are using another theorem which tells you that the derivative of a function in a point $x$ only depend on the values of $f$ in a neighbourhood of $x$. So if you have a function defined piecewise as: $$ f(x) = \begin{cases} g(x)\text{ if } x \in I\\ h(x) \text{ if } x \in J \end{cases} $$ and if $g$ and $h$ are differentiable, you can conclude that $f$ is differentiable in the interior of $I$ and in the interior of $J$. Again this tells you nothing about the points in the boundary of $I$ or $J$. In such points you need, again, to apply the definition of derivative.
Actually there is a positive result which you can use to avoid to check the limit of the incremental ratio. By l'Hopital theorem one can show that if a function $f$ is continuous in a point $x_0$ and if it is differentiable in a neighbourhood of $x_0$ and if $f'(x)\to \ell$ as $x\to x_0$, then $f$ is differentiable in $x_0$ and $f'(x_0)=\ell$.
