# Definition of second derivative as a limit

I found a statement that the second derivative can be defined as:

$$\lim_{x \to a} \frac{f '(x)-f '(a)}{x-a}$$.

Does this definion follow from the definition of the first derivative as:

$$f ' (x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

If so, how? If not, where does it come from?

Edit: Mistake corrected, sorry.

• Your "$n$" should be an "$x$". Feb 25, 2014 at 17:07
• Your first expression makes no sense at all: first, there is $\;n\to a\;$ (perhaps you meant $\;x\to a\;$?) , and even then that's the definition of the first derivative at $\;a\;$ ... Feb 25, 2014 at 17:08
• Perhap you meant the "Second definition" rather than "second derivative"? Feb 25, 2014 at 17:09
• Where did you find this statement?
– Did
Feb 25, 2014 at 17:18

Assuming that $$f$$ is $$C^2$$ then $$\displaystyle\lim_{h\rightarrow 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2}=^{\text{L'Hospital's}}\displaystyle\lim_{h\rightarrow 0} \frac{f'(x+h)-f'(x-h)}{2h}$$

$$=^{\text{L'Hospital's}}\displaystyle\lim_{h\rightarrow 0} \frac{f''(x+h)+f''(x-h)}{2}=f''(x)$$

Yes, since

$$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h},$$

and the second derivative is the derivative of the derivative, we get

$$f''(x) = \lim_{h\rightarrow 0} \frac{f'(x+h)-f'(x)}{h}.$$

There are also difference quotients for the second derivative defined immediately in terms of $f$. The most commonly seen is

$$f''(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$$

This is commonly derived using Taylor expansions.

• Silly question but, How are the two versions of of $f''(x)$ the same Nov 4, 2021 at 21:34

The first formula does not define you the second derivative of $f$ at $a$, only the first derivative. These two definitions are equivalent. If you put $h=a-x$, you get one definition from the other.

And also, in the first formula under the limit sign you should have $a\to x$

• I did this, but I am getting $\frac{f(2x-a)-f(x)}{x-a}$. How is this equivalent to the desired fraction in the limit? Feb 25, 2014 at 17:17
• @kiwifruit There should be $h=a-x$. I have corrected it. Feb 25, 2014 at 17:49
• Ok, but then I get $\frac{f(a)-f(x)}{a-x}$. Is that equivalent to reversing both subtractions? Feb 25, 2014 at 18:12

The second derivative is defined applying the definition of derivative to the first derivative, i.e.: $$f''(x)=\lim_{h\to0}\frac{f'(x+h)-f'(x)}{h},$$ where: $$f'(y)=\lim_{h\to0}\frac{f(y+h)-f(y)}{h}.$$ I do not think the first expression you wrote makes any sense. What is $$n$$? If you meant $$x$$, that is the definition of the first derivative.

This is not a definition for the second derivative. This is an alternative definition for the first derivative.