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For a function $f(x)$ the definition of its derivative is $$f'(x) = \lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}.$$

The derivative $f'(x)$ is supposed to be the same for $\Delta x>0$ and $\Delta x<0$. Which means that if we now set $\Delta x>0$, $$\lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim \limits_{\Delta x \to 0} \frac{f(x-\Delta x)-f(x)}{-\Delta x}$$ needs to be true.

But how do you show that the above expression is true?

I thought of the example where $f(x)=e^x$, and let $\Delta x=3$. At the point $x=0$, $$\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{e^3-1}{3}$$and$$\frac{f(x-\Delta x)-f(x)}{-\Delta x}=\frac{e^{-3}-1}{-3}.$$ They are not equal and intuitively I thought that as $\Delta x \to 0$, they will approximately equal to each other but never exactly equal, since the graph of $f(x)=e^x$ is not symmetrical about $x=0$.

Is the statement that the derivative is independent of the sign of $\Delta x$, ie the statement $$\lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim \limits_{\Delta x \to 0} \frac{f(x-\Delta x)-f(x)}{-\Delta x}$$ just an approximation then?

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  • $\begingroup$ Hint: for the "negative" direction, instead of $$ \frac{f(x-\Delta x) - f(x)}{-\Delta x} $$ I would write $$ \frac{f(x) - f(x-\Delta x)}{-\Delta x} $$ which better reflects the direction of the slope. $\endgroup$
    – Matti P.
    Jul 4, 2018 at 6:05
  • $\begingroup$ @MattiP. the $\Delta x$ below in your second expression should not have a negative sign then, if the $\Delta x $ itself is positive. $\endgroup$
    – Taenyfan
    Jul 4, 2018 at 6:07
  • $\begingroup$ The main problem here is that $\Delta x=3$ is too large. The growth rate of this function changes too quickly. Try $\Delta x= 0.1$ to see what you wanted to see. Another problem is that you seem to think $f(0)=0$. But in fact $f(0)=e^0=1$. $\endgroup$
    – Jyrki Lahtonen
    Jul 4, 2018 at 6:13
  • $\begingroup$ If the derivative of f as you approach x from from the right weren't equal to the derivative as you approach x from the left, the function would have a sharp kink in it (it wouldn't be differentiable at that point) and $f'(x)$ would be discontinous at x. For your example of $f(x) = e^x$, $f'(x) = e^x$ is definitely continuous. $\endgroup$
    – David Diaz
    Jul 4, 2018 at 6:15
  • $\begingroup$ @JyrkiLahtonen thanks, I edited my question. $\endgroup$
    – Taenyfan
    Jul 4, 2018 at 6:16

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What is equal is the limit indeed

$$\lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim \limits_{\Delta x \to 0} \frac{e^{\Delta x}-1}{\Delta x}=1$$

and

$$\lim \limits_{\Delta x \to 0} \frac{f(x-\Delta x)-f(x)}{-\Delta x}= \lim \limits_{\Delta x \to 0} \frac{e^{-\Delta x}-1}{-\Delta x}=\lim \limits_{\Delta x \to 0} \frac1{e^{\Delta x}}\frac{e^{\Delta x}-1}{\Delta x}=1$$

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  • $\begingroup$ thank you, I understand that the limits equal. Do you know how to show that the two limits will equal for any continuous function $f(x)$? $\endgroup$
    – Taenyfan
    Jul 4, 2018 at 6:26
  • $\begingroup$ Recall that continuity is not sufficient (e.f. |x| at x=0). For f differentialble it can be shown by a change of variable. As an alternative recall the geometrical meaning of the derivative as the slope of the tangent at a point, and if f is differentiable the slope of the secant line approaches the same value from both sides. $\endgroup$
    – user
    Jul 4, 2018 at 6:41

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