Eli5: why are derivatives useful?

764 viewsMathematicsOther

I don’t mean in which cases I can use them, nor how they work. I know how they work (at least at a basic level, the derivative of ax\^b is abx\^(b-1), but I mean… why is a function that does those steps useful to solve any problem? It really seems like a random choice of operations.

In: Mathematics

13 Answers

Anonymous 0 Comments

First, the derivative of x^n = n x^(n-1) thing is not what the derivative actually *is*. That’s a special case of how to calculate it in certain circumstances. The derivative is a method of calculating rates of change. The derivative of position (where you are) is velocity (how fast you’re going). The derivative of how much water is in your tank is how fast water is entering/leaving your tank. 

The equation that defines the derivative is the limit definition: limit as change in x approaches 0 of f(x+change in x) – f(x) over change in x. That is what the derivative is: how much things change over a period of time that approaches 0. All the other rules that you will learn, including the power rule you mentioned, are just tricks to make it easier to compute. 

But what is it useful for? 

Let’s say you throw a ball straight up in the air. How high will it go? Well, one way to figure it out is to say that the highest point is when it stops going up and starts going down. What does that mean in terms of math? Well it means it’s velocity switches from positive (up) to negative (down) – which means it passes through zero. Velocity is the derivative of position. So the highest point of position is when the derivative of position is zero. This type of thing matters to ballistics and similar. 

Cool. But you don’t throw balls for a living, you sell candy bars, so why do you care about throwing balls? Well, you might notice that the more you charge for your candy bars, the fewer people buy them. These candy bars cost you $1, and you’ve run a survey that says that you’ll sell 1000 per month if you sell for $2, and that for every cent you increase the price, you lose exactly one sale per month. What price will get you the most profit? 

That’s the same problem as throwing the ball: you can easily create an equation for the profit that you’ll make as a function of price. The maximum profit happens when increasing the price switches from increasing profit to decreasing profit. 

Or let’s say you have a computer, and you want it to learn to recognize pictures of cats. If something is a picture of a cat, it should output 1. If not, then 0. Well, what you can do is make a function with a whole bunch of random pieces in it, and shove a picture of a cat through it. It will give you a random number, say 0.5. You then use derivatives to figure out how to change the random pieces of your function to make that output a little bit closer to 1. Then repeat this a millions of times with different pictures of cats and non cats, and this is machine learning. 

The applications are endless. I’ve only touched on what is called optimization problems, which is just one corner of what derivatives are used for. 

You are viewing 1 out of 13 answers, click here to view all answers.