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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.

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  • $\begingroup$ Your "$n$" should be an "$x$". $\endgroup$ Feb 25, 2014 at 17:07
  • $\begingroup$ 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 $\;a\;$ ... $\endgroup$
    – DonAntonio
    Feb 25, 2014 at 17:08
  • $\begingroup$ Perhap you meant the "Second definition" rather than "second derivative"? $\endgroup$ Feb 25, 2014 at 17:09
  • $\begingroup$ Where did you find this statement? $\endgroup$
    – Did
    Feb 25, 2014 at 17:18

5 Answers 5

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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) $

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  • $\begingroup$ With this derivation $-2f(x)$ can be replaced by anything not dependent on $h$. So it works only in one direction. You still need to prove the other direction. $\endgroup$ Apr 18 at 2:43
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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.

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    $\begingroup$ Silly question but, How are the two versions of of $f''(x)$ the same $\endgroup$
    – homosapien
    Nov 4, 2021 at 21:34
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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.

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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 $a\to x$

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  • $\begingroup$ 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? $\endgroup$
    – kiwifruit
    Feb 25, 2014 at 17:17
  • $\begingroup$ @kiwifruit There should be $h=a-x$. I have corrected it. $\endgroup$ Feb 25, 2014 at 17:49
  • $\begingroup$ Ok, but then I get $\frac{f(a)-f(x)}{a-x}$. Is that equivalent to reversing both subtractions? $\endgroup$
    – kiwifruit
    Feb 25, 2014 at 18:12
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This is not a definition for the second derivative. This is an alternative definition for the first derivative.

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