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Let $f(x) = 3x$ and if I'm not wrong $f'(x)$ is the change in $f(x)$ for a change in $x$. So the derivative of $f(x)$ is 3. So for a unit change in $x$ there is a change in $f(x)$ by 3. And this fits perfectly. When $x$ changes for 0 to 1, $f(x)$ changes from 0 to 3. And when $x$ changes from 1 to 2 $f(x)$ changes from 3 to 6 i.e, a change by 3.

I hope this idea is correct.

But for the function $f(x) = x^2$ it doesn't match properly. The derivative is 2x. When $x$ changes from 0 to 1, $f(x)$ changes from 0 to 1 and $f'(x)$ is 2. When x changes from 1 to 2, $f(x)$ changes from 1 to 4. My textbook defined the derivative of f(x) as the change in f(x) with respect to x. So did I miss something here or what is it that I'm getting wrong?

user662650
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    This question has been asked many times before: you need the change in $x$ to infinitesimal for it to be equal to $f'(x)$. As the value of the change in $x$ approaches $0$, the value approximates $f'(x)$ better and better. – Toby Mak Apr 01 '20 at 07:52
  • The thing you are missing is that the "step length" you are taking has to approach zero in the calculation of the derivative. If you just give it arbitrary values, of course the slope will change ... – Matti P. Apr 01 '20 at 07:53
  • So for let's say x changes for 1 to 1.001 so the change is by 0.001 amount. $f'(x)$ will be 1.002001 - 1/1.001-1 = 2.001. This value means that there is a change of 2.001 for a unit change in x right? Since the denominator is 1 right? – user662650 Apr 01 '20 at 08:01
  • You should use the notation "change in $x$" ($\Delta x$) and "change in $y$" ($\Delta y$), since $f'(x)$ means the derivative. – Toby Mak Apr 01 '20 at 08:20
  • I get what you mean, but you can better express it as: when the change in $x$ is $0.001$ or $\Delta x = 0.001$, $\frac{\Delta y}{\Delta x} = 2.001$. This is not equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ or $f'(x)$ though. The problem with what you are saying is that when you have a unit change in $x$, the change in $y$ is not $2.001$. When you explain it using ratios it gives a more accurate picture. – Toby Mak Apr 01 '20 at 08:23
  • But $f'(x)$ at 1 is 2 which is very close to 2.001. So does derivative of $f(x)$ at x =1 is 2 mean that a unit change in x will cause $f(x)$ to change by 2? If not, then what does it mean? And are there any resources for me to get a better idea of this? – user662650 Apr 01 '20 at 10:02

2 Answers2

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My sister lives about $450$ miles from me. When I go to visit her (not that I can right now), it takes me about $9$ hours to get there. That means my speed is $\frac {450}{9} = 50$ miles per hour. But my speedometer is crazy. Sometimes it claims I'm going about $75$ miles per hour, sometimes - particularly during lunch - it claims I'm going $0$ miles per hour. It almost never shows $50$ miles per hour.

Of course, my speedometer is not the problem. The point is, $50$ miles per hour is my average speed over the entire trip. If I made the entire trip at one speed, never slowing down or speeding up, then $50$ miles per hour would have been the speed I was travelling. But $9$ hours is a long time to drive, and my bladder isn't going to allow me to drive it straight thru. The speedometer is not measuring the average speed of the trip. It tells me how fast I am moving at that moment.

This is the distinction between a "difference quotient" and the derivative. The expression $$\dfrac {f(b) - f(a)}{b - a}$$ is a difference quotient. It tells you what the average rate of change is for $f(x)$ on the interval from $a$ to $b$. But the derivative $f'(a)$ tells you how fast $f(x)$ is changing at $x = a$. It can change at different rates elsewhere in the interval. If it so happens that the rate of change is the same everywhere, such as is the case when $f(x) = 3x$, then $f'(a)$ will be equal to the average rate of change. But for most functions, this is not the case. $f(x)$ will sometimes change faster, sometimes change slower. Then the derivative and difference quotients will not be the same.

To figure out the derivative, we approximate it with difference quotients over smaller and smaller intervals. If those approximations approach a particular value as the interval size approaches $0$, we call that value the derivative. This is written as $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ The thing to note here is the derivative is not one of the difference quotients. Rather it is the value that the difference quotients approach as the interval size decreases.

Paul Sinclair
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You write, for $f(x)=x^2$ that

When $x$ changes from $0$ to $1$, $f(x)$ changes from $0$ to $1$ and $f'(x)$ is $2$.

But that's not true! It's true that $f'(1)=2$, but you're talking about a whole interval of numbers here, you can't just evaluate $f'$ at one point and expect that it captured everything about your function. On that interval, $f'$ actually varies between $f'(0)=0$ and $f'(1)=2$.

If you want to look at a difference over an interval, you would have to take the average of the derivative over this interval which, since $f'$ happens to be linear here, is $1$ - exactly that average rate of change you observed. However, you should note that $f$ changed more quickly in the latter half of this interval than in the former - for instance $f(1/2)=1/4$, so the interval from $0$ to $1/2$ had less total change than from $1/2$ to $1$. The derivative is sensitive to what happens on small intervals and correctly recognizes that $f$ increases very slowly near $0$, but makes up for this by increasing very quickly near $1$.

Formally, saying $f'(1)=2$ means that over all small enough intervals, your velocity will be as close as you want to $2$. For instance, for quadratic functions, you can say that is $|x - 1| < \delta$ then the slope from $(1,f(1))$ to $(x,f(x))$ is within $\delta$ of $2$. This is exactly what the definition $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h$$ says, expanded out via the definition of a limit - and it shows that the derivative is really just the slope measured along sufficiently small intervals to avoid the issue of average velocities.

Milo Brandt
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