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Just making sure I understood:

$$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$

At a first glance I didn't understand why the above is true. It's because (in the case above) we can say that $x=x_0+\Delta x$; right?

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This is what I use as the definition of the derivative. However, if you define the derivative as $$f'(x_0):=\lim\limits_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$$ then your statement is correct.

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  • $\begingroup$ That is what I meant (That if x is what I stated above, we get exactly the defintion of the derivative.). Thank you =] $\endgroup$ – Py42 Sep 9 '12 at 21:09
  • $\begingroup$ @Py42 No problem. BTW, since you're new here I suggest you look at the faq. $\endgroup$ – Alex Becker Sep 9 '12 at 21:19
  • $\begingroup$ Is there a guideline/rule I've missed? (I'll delete this comment after you answer as it's off-topic.) $\endgroup$ – Py42 Sep 10 '12 at 6:26
  • $\begingroup$ @Py42 Not in particular, but it always helps to familiarize yourself with the rules and practices. For example, your question was perfectly fine but it would have been helpful if you had included your definition of derivative, since different courses define it differently (but equivalently, of course). Also I'm being a bit selfish: the faq tells you to "accept" the most helpful answer, and if you accepted mine it would give me points. :p $\endgroup$ – Alex Becker Sep 10 '12 at 17:51
  • $\begingroup$ I actually did try to accept yours, somewhy the system told me to wait (at the time, now I can and did). Anyway this question/answer now seems too obvious to even ask it (apparently I was very tired yesterday). $\endgroup$ – Py42 Sep 10 '12 at 19:13
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I've seen the definition written as $$ \frac{dy}{dx} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x}, $$ which is also equivalent to the forms discussed above.

If the definition you've seen is $$ f'(x_0) = \lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x_0} $$ (or you might call it $h$ instead of $\Delta x$) then you can say $$ \begin{align} x & = x_0+\Delta x, \\[12pt] \text{so }\Delta x & = x - x_0, \\[12pt] \text{and as }\Delta x & \to0,\text{ then }x \to x_0, \\ \end{align} $$ and one form becomes the other.

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