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Assume

$$f(x)=\left \{\begin{matrix} x-1 & x \leq 0\\ x+1 & x > 0 \end{matrix} \right.$$

Obviously, $df(x)/dx$ do not exists at $x=0$ but the one-sided differential exists.

What's the notation of one-sided differential?

$$d_-f(x)/d_-x$$, $$d_-f(x)/dx$$, or $$df(x_-)/dx_-$$?

Update:

The previous example is really bad, and I was confusing deritive with differential. Thank you very much for the anwsers. Maybe the following example will make the problem clearer:

$$z=\left \{\begin{matrix} y\times x & x \leq 0\\ 2\times y \times x & x > 0 \end{matrix} \right.$$

and both a and x is variable.

How do I represent $\partial z / \partial x$ at $x=0^+$ and $x=0^-$?

It seems $\lim_{x\to 0^+}\partial z / \partial x$ and $\lim_{x\to 0^-}\partial y / \partial x$ is what I want.

iuradz
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    Seems this is not a differential, it's a derivative. You can try to google for "left right derivative". – peter.petrov May 17 '16 at 12:48
  • I am a bit confused by your question. Whether you are asking about $\lim_{x\to0^+}f'(x)$ and $\lim_{x\to0^-}f'(x)$ or one-sided derivatives. In your example, $\lim_{x\to0^+}f'(x)=\lim_{x\to0^-}f'(x)=1$. However, the left-sided derivative is $1$ all right. But the right-sided derivative does not exist. – velut luna May 17 '16 at 13:11

2 Answers2

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Seems the question is about left/right derivative.
The notation is normally: $f'(x^+)$ and $f'(x^-)$.
In your case it would be: $f'(0^+)$ and $f'(0^-)$.

peter.petrov
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  • Will $f'(0^+)$ be mistaken as $\lim_{x\to 0^+}f'(x)$? – velut luna May 17 '16 at 12:54
  • If $f$ is piecewise smooth, there's no harm in the misinterpretation. In general the only way this goes bad is if you have a function which has a derivative in a half-neighborhood, and has a one-sided derivative, but these approach different values at the center. This seems to be excluded almost by definition. – Eric Stucky May 17 '16 at 13:00
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    @EricStucky In fact I am confused by the OP question. Whether he is asking about $\lim_{x\to0^+}f'(x)$ and $\lim_{x\to0^-}f'(x)$ or one-sided derivatives. In his example, $\lim_{x\to0^+}f'(x)=\lim_{x\to0^-}f'(x)=1$. However, the left-sided derivative is $1$ all right. But the right-sided derivative does not exist. – velut luna May 17 '16 at 13:09
  • @Mathaholic That's quite possible actually... but that's the notation I know of. – peter.petrov May 17 '16 at 13:12
  • ${f}'_{-}(0)$ is less ambigiuous, in my opinion. – Hans Lundmark May 17 '16 at 13:22
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I am not sure what you mean by one-sided derivatives especially when you say that they exist.

According to wikipedia, the notation for left and right derivative of $f(x)$ at $x=a$ is

$$\partial_+f(a)=\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$$

$$\partial_-f(a)=\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}$$

According to that definition, for your function

$$\partial_-f(0)=1$$

But

$$\partial_+f(0)$$ does not exists.

I agree with Hans Lundmark that alternative and more common notations may be $f'_-(a)$ and $f'_+(a)$.

As for $$\lim_{x\to a^+}f'(x)$$ and $$\lim_{x\to a^-}f'(x)$$ which need not be equal to the left- right- derivatives. I don't know any common notation for them. May be we have to put up with this and write $$\lim_{x\to a^+}f'(x)$$ and $$\lim_{x\to a^-}f'(x)$$ to avoid ambiguity.

In your example $$\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)=1$$

velut luna
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