Derivative of max function

It might be of help to sketch the function or write it without the $\max$. We get $$f(x) = \begin{cases} (1-x)^2 & \text{if } x \leq 1\\ 0 & \text{if }x \geq 1\end{cases}$$ It is easy to work out the derivative everywhere except at $x=1$. At $x=1$, work out explicitly from definition. $$\lim_{h \to 0^+} \dfrac{f(1+h) - f(1)}{h} = 0$$ $$\lim_{h \to 0^-} \dfrac{f(1+h) - f(1)}{h} = \lim_{h\to 0^-} \dfrac{h^2}{h} = 0$$ Hence, we have $$f'(x) = \begin{cases} 2(x-1) & \text{if } x \leq 1\\ 0 & \text{if }x \geq 1\end{cases}$$

HINT : Analyze this function on different intervals.

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?