On the nature of a first derivative

Question

This is a very, very basic question. Never been very involved in math but I've been learning calculus in my free time, so here goes. I have observed some various things that happen with derivatives, but I don't know what it's telling me. I see that if I find the slope at two x-values with a derivative, then the difference in their slopes is the same as if I plugged the difference between the two x-values into the derivative. But what does the slope itself tell me? If my x value is 8, and f'(x) is 2x, then what does my slope of 16 at that point tell me? The slope at 7.99998 is going to be .00004 away from 16, but what do those two slopes actually tell me? I guess I'm having a hard time understanding the usefulness of the value given when you plug in x-values in a slope, what does the quantity (the slope) actually mean? Thanks for taking time to read this. I understand this board has lots of really juicy questions and this is hardly worth your time but I really appreciate the help!

Answer 1

Slope tells you how the function changes. Here is a simple example you may find useful. Say you are in a deserted island and no calculator and you want to calculate the square root of 17. Don't even ask me why you would want to do that.

If you start with the function $f(x)=x^2$ and you want $f(x)=17$. You take a stab at the answer and guess it should be $x=4$. You find that $f(4) = 16$ just a bit below 17. Now what?

You know the slope of $f$ at $x=4$ is $2 x = 8$. So for small change in $x$ will make $f$ change 8 times more. Now you want $f$ to increase from $16$ to $17$, i.e. an increase of $1$. So $x$ should increase by $1/8$ so you get the answer $$x=4 \frac{1}{8}$$ This is a pretty good answer of $x=4.125$. My calculator gives $4.123$

By the way, if you are stuck on a deserted island, you have bigger problems than finding the square root of 17 but you never know when calculus may come in useful!

Answer 2

You seem to be working with $f(x)=x^2$, which has $f'(x)=2x$ as its first derivative. To simplify notation, let's call the first derivative $g(x)$, instead of $f'(x)$. You should note first that your observation that $g(x_2-x_1)=g(x_2)-g(x_1)$ is not true in general. But it is true if $g(x)$ is linear, as it is in your example.

You ask: "What does the slope tell me?" You can interpret the slope in many ways depending on the specific context of your problem, but I like to interpret the slope as an approximation to the change in $f(x)$ when $x$ increases by $1$. In your example, $g(8)=16$. That value is an approximation of $\Delta y= f(9)-f(8)=81-64=17$. In other words, $f'(x) \approx \Delta y$ when $\Delta x=1$.

Answer 3

Derivatives often come up in handy in physical situations when a quantity keeps changing. In particular, rates of change come up often.

For example, we may be interested in the speed of the object. This is simple enough when it is at constant speed: we take the total distance divided by the total time travel. However, when the speed keeps changing, one way to tackle may be to let $y$ be the distance and $x$ the time. We can plot the graph and the slope, $\frac{\mathrm{d}y}{\mathrm{d}x}$ will give us the instantaneous speed. E.g., perhaps by experiment we conclude that $y=x^2$. Then $\frac{\mathrm{d}y}{\mathrm{d}x}=2x$ and when we substitute $x=8$, the value of 16 tells us the speed at time = 8 units.