Definition of second derivative as a limit

Question

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.

Answer 1

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) $

Answer 2

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.

Answer 3

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 $a\to x$

Answer 4

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.

Answer 5

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