0

Let $f$ be a continuous function.

We know $f$ is differentiable at $x = a$ if the limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ exists. (Normal derivative)

And, if $f$ is differentiable at $x = a$ then we can write $f'(a)$ as $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}$$

But the limit (symmetric derivative) $$\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$$ can exist when $f$ is not differntiable at $x = a$

For example, a function $x \to |x|$ is not differentiable at $x = 0$ but the limit exists : $$\lim_{h\to 0}\frac{f(0+h)-f(0-h)}{2h} = \lim_{h\to 0}\frac{|h|-|-h|}{2h} = 0$$

I checked some examples ($f$ is not differntiable but the symmetric derivative exists) and I got zero for all $f$.

So my question is :

  1. If $f$ is not differentiable at $x=a$ and symmetric derivative exists at $x=a$, then the symmetric derivative of function $f$ at $x=a$ must be zero? (How to prove? or I want to know counterexample.)

  2. Is there any nice approach to understand symmetric derivative? (I understood this as a slope of two points.)

Thanks for your help.

Just_A_User
  • 2,126
bFur4list
  • 2,604

1 Answers1

0

The symmetric derivative is basically the arithmetic mean of the left-hand and right-hand slopes. So if they're equal in magnitude but of opposite signs, you will get $0.$ For instance, if we're evaluating the derivative at $a=0$, then any even function such as $y=|x|$ will yield a symmetric derivative of $0.$

However, any piece-wise (or non-differentiable) function with "uneven" slopes will have a non-zero symmetric derivative. A simple example is $f(x)=x \text{ for }x>0$ and $f(x)=-2x \text{ for } x≤0$ as pointed out by @stephenkk in the comments.

Here is a YouTube video on symmetric derivatives with some examples if you wish to learn more.

devam_04
  • 748