# What does the output of a derivative actually say in real life?

Perhaps the wording of the question will need to be adjusted, but I do not know how else to convey what I am looking for in one sentence.

## Here is what I understand:

The derivative of a function in essence shows the "rate of change" of a function at any given point. In practical application, determining the current speed of a car at a given time along it's route, as opposed to the average speed of a car over it's entire route.

The value (output) given by the derivative function expressed mathematically is the "slope of a line",the tangent, at any given point (output) of the function as displayed on a graph, which represents the how much the output of the function has changed with the input.

But what exactly is the output of the derivative telling me about this change? If the output of the derivative of some 'x' is '2', what is this '2'? It is a slope, yes, but what is this slope telling me in real life?

Here is an example of what I am trying to understand. We will use the following function... $$f(x)=x^2$$ and its derivative... $$f'(x)=2x$$

Let us imagine I have an extremely unique and completely absurd job that lasts for a '1' hour shift each day, and that has a pay rate of f(x)=x^2 per minute, where 'x' is the number of minutes worked in the shift so far.

If we look at just the first five minutes of the shift, both the 'pay rate per minute' and the 'rate of change (slope) of pay per minute', we get... $$f(0\text{min})=0;f'(0\text{min})=0$$ $$f(1\text{min})=1;f'(1\text{min})=2$$ $$f(2\text{min})=4;f'(2\text{min})=4$$ $$f(3\text{min})=9;f'(3\text{min})=6$$ $$f(4\text{min})=16;f'(4\text{min})=8$$ $$f(5\text{min})=25;f'(5\text{min})=10$$

Now, again, the output for the derivative, if graphed, is the slope of the line at any given point along the main function.

But what does that mean in practice?

If I asked someone, "How much more money am I earning from minute '4' of my shift?", and that someone answered, "according to the derivative of the money earned, 8." , what is that? '8' what? Dollars? Percent? No unit? Just '8", or any output given by the derivative, doesn't give me any useful understanding of what is changing.

The question-answer bit I gave at the end may be worded incorrectly. (let me know what I should change it to, if so), but I hope the point of this question accross.

Also, I am not looking for proofs in the answer. Proofs are not hard to find. I am looking for answers that are not written in proofs.

• I have picked an answer for this question, but please, if you have examples you would like to share, go ahead. Jan 8, 2021 at 18:53

The derivative of $$f(x)$$ with respect to $$x$$ always has units of $$\frac{\text{units of f}}{\text{units of x}}.$$ So, for your function, if $$f(x) = x^2 \text{ dollars per minute}$$ then the derivative will be $$f'(x) = 2x \text{ dollars per minute per minute}.$$ In practice, that means that tells you how fast your pay per minute in increasing. If, for example, you say $$f'(4) = 8$$ what this really means is that a minute $$4$$ your pay per minute is increasing by $$\8$$ per minute per minute.
Let's say that we don't actually know what $$f(x)$$ is, just that $$f(4) = \16$$ per minute and $$f'(4) = \8$$ per minute per minute. That would not only tell you that your pay per minute was increasing, but that at minute $$5$$ you would be making approximately an additional $$\8$$ per minute than you were at minute $$4$$.