Derivative of max function

Question

I was just wondering what the derivative of $f(x) = \max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

since $(1-x)^2 \ge 0 $ (for x a real number), $\max(0,(1-x)^2)$ is equal to $(1-x)^2$ — Andre Holzner, Jun 08 '16 at 12:35

score 60 · Accepted Answer · answered Apr 21 '13 at 18:32

60

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 $$f'(x) = \begin{cases} 2(x-1) & \text{if } x \leq 1\\ 0 & \text{if }x \geq 1\end{cases}$$

answered Apr 21 '13 at 18:32

2

in your function definition for all x >= 1, you are returning 0. But let's take x = 2, then (1 - 2) ^ 2 will be (-1) ^2 which is nothing but 1 and according to op's max function, 1 should be returned. But since you gave the condition of x >= 1, we always return 0 even when x is something like 2. I think in comments what Andre Holzner said is correct. since (1−x)2≥0 (for x a real number), max(0,(1−x)2) is equal to (1−x)2 – theprogrammer May 29 '21 at 00:21

score 10 · Answer 2 · answered Apr 21 '13 at 18:29

10

HINT : Analyze this function on different intervals.

answered Apr 21 '13 at 18:29

Kalissar

283

1

If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)? – phil12 Apr 21 '13 at 18:31
1

Now what happens when x = 1 ? Does f'(1) exist ? – Kalissar Apr 21 '13 at 18:33

score 5 · Answer 3 · answered Jul 22 '17 at 06:14

5

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?

answered Jul 22 '17 at 06:14

khan

213

X0-user-0X · Answer 4 · 2023-09-27T17:09:06.000

I know that this is an old question but here an other way that can help.

It may be easier to apply the basic properties of derivatives on your problem knowing the following result $max\left \{ a;b \right \}=\frac{a+b+|a-b|}{2}$ and that $\frac{\partial |x|}{\partial x}=\frac{x}{|x|}=sign(x)$.

After all you need to do is to derivate: $(\frac{1-x+|1-x|}{2})^2$.
$\frac{\partial }{\partial x}(\frac{1-x+|1-x|}{2})^2=\frac{1}{4}2(1-x+|1-x|)\frac{\partial }{\partial x}(1-x+|1-x|)=\frac{1}{2}(1-x+|1-x|)(-1+\frac{\partial }{\partial x}(|1-x|))=\frac{1}{2}(1-x+|1-x|)(-1-sign(1-x))=...=(x-1)-|1-x|$ (different expression of the same function given above).

After you need to carefully check what happen at $x=1$ as allready explained above.

Derivative of max function

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