negative Delta y in derivatives.

Question

I'm first year in high school in Russia so please understand for noobyness and possible miss-formats.

From derivative definition we have: $$ f'(x) = \lim_{\Delta x \to 0}(\Delta y) / (\Delta x) $$ Delta y is f(x) - f(x0) which can be as a negative value depending on whether derivative is increasing, decr. or zero. but Delta usually means that we take difference between something. should we take an absolute value here? if don't, I get the derivatives. if do I ain't get it...

Thanks. I've tried to google this q I do promise.

@NapoleonTheCake "Modulus", "magnitude" and "absolute value" are the English terms — Ben Grossmann, Nov 21 '16 at 19:31
it is just a theoretical question. I mean I can definitely find this in books but I don't have time right now and I am obsessed with this question. — NapoleonTheCake, Nov 21 '16 at 19:34
I am not certain I have well understood your question, but if you take the absolute value on $\Delta x$ and $\Delta y$, you will get the absolute value of the derivative... But if you take it only on $\Delta y$ the limit doesn't exist. — Jean Marie, Nov 21 '16 at 19:37
I mean Delta y can be positive this is obvious. but can it be negative as Delta y is f(x)-f(x0) and if derivative is decreasing f(x)-f(x0) will be negative. — NapoleonTheCake, Nov 21 '16 at 19:42

score 0 · Accepted Answer · answered Nov 21 '16 at 19:40

0

As you've said, we have $\Delta y = f(x) - f(x_0)$. We then have $\Delta x = x-x_0$. So, $\Delta x < 0$ when $x < x_0$, and $\Delta x > 0$ when $x > x_0$. No absolute values to take here.

The more common presentation of this limit is its full form $$ f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} $$ It is sometimes helpful to note that this is the same thing as writing $$ f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} $$ though in practice, the first version makes computation easier.

answered Nov 21 '16 at 19:40

Ben Grossmann

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so Delta y should simply be replaced and Delta means not an absolute difference? Thank you! – NapoleonTheCake Nov 21 '16 at 19:42
That's right. It's a "signed" difference. – Ben Grossmann Nov 21 '16 at 19:43

negative Delta y in derivatives.

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