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Let's say we have $y = 10^2$

How much does $y$ increase when $x$ increases by $1$ unit?

We have the form $y=x^2$ and $\dfrac{dy}{dx}=2x \ldots$ here $x=10$, so why isn't the answer $2(10)=20$?

By simple evaluation the answer is $(11^2)-(10^2)=21$

But thinking about it from the perspective of using the general form of the derivative of $y=x^2$, we could answer it by computing $2(10.5)=21$

Finally, my question is: What is the reasoning behind the $x$ in "$2x$" being $10.5$?

FoiledIt24
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Will
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    $y=100$ is a constant function and doesn't change with time. The answer should be $0$. – CyclotomicField Feb 25 '21 at 01:49
  • "when $x$ increases by $1$ unit" What is $x$? $x$ does not appear in the problem statement. There is no $x$ in the stated equation of $y=10^2$. This might just as well be $y=10^2 + 0\cdot x$. Who is to say that $x$ should have been the base of the exponent rather than the exponent itself or something else entirely... – JMoravitz Feb 25 '21 at 02:08
  • Now... "Given that $y=x^2$, considering what happens when $x=10$ and looking at how much $y$ would increase if $x$ were increased by $1$"... it is a fluke that plugging in $10.5$ to your derivative equation worked. If this were not $x^2$ and were instead $x^3$ or otherwise it would not have worked. As for why the derivative is useless here, that's because the derivative looks at the instantaneous rate of change as opposed to how much it actually changes over a specific time interval. – JMoravitz Feb 25 '21 at 02:12

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When you say that at $x=10, \frac {dy}{dx}=20$, it means that the gradient at the point $(10,100)=20$.

But, being a parabola, the gradient increases as $x$ increases, so that by the time $x=11$, the gradient at $(11,121)$ is $22$.

The gradient of the straight line joining the two points is $21$. Your simple evaluation is actually calculating that gradient, not the gradient at $x=10$.

It just happens to be the case that for this parabola, the point at $x=10.5$ is the midway point, or average, between $x=10$ and $x=11$ and that the gradient at that point is the average gradient between the two other points.

robert timmer-arends
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Derivatives are based on infinitesimal changes. However, you can also formulate those with non-infinitesimal changes.

$$y' = \frac{f(x + h) - f(x)}{h}$$

We normally do $\lim\limits_{h\to 0}$ on this, but there's no reason you have to. You can actually set $h$ to whatever you want and it still works. So, we can find the "full-unit derivative" (where $h = 1$) as follows:

$$f(x) = x^2 \\ f'(x) = \frac{(x + h)^2 - (x)^2}{h} \\ f'(x) = \frac{x^2 + 2xh + h^2 - x^2}{h} \\ f'(x) = \frac{2xh + h^2}{h} \\ f'(x) = 2x + h $$

So, when $h = 1$, then $f'(x) = 2x +1$. So, at $x = 10$ the amount that $y$ changes for $1$ unit of $x$ is $2x +1 = 2(10) + 1 = 21$.

Technically, this is called a "difference" equation instead of a "full-unit derivative", but, you know, technicalities.

johnnyb
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