# Product Rule: Intuitive Idea and Solved Problems

The product rule is one of the essential differentiation rules. In any calculus textbook the introduction to this rule is a formal deduction using the definition of the derivative. I don't want to do that again here. You can easily find that on other websites.

Here we'll first focus on trying to get an intuitive understanding of why this rule is the way it is using simple examples. Then we'll solve some example problems applying it.

### Intuitive Idea Behind the Product Rule

Let's say you are running a business, and you are tracking your profits. A function S(t) represents your profits at a specified time t. We usually think of profits in discrete time frames. For example, your profit in the year 2015, or your profits last month.

There is nothing stopping us from considering S(t) at any time t, though. For many businesses, S(t) will be zero most of the time: they don't make a sale for a while. Then, they make a sale and S(t) makes an instant jump.

For the sake of this explanation, let's say that you business is huge, making sales every second of the day. In this case, the graph of S(t) against time would be a continuous curve.

Also, let's say that your business sells a single product: tomatoes. In this case we can express your profit S(t) as a (mathematical) product. We can write:

where u(t) is how many tomatoes you sold at time t, and v(t) is your profit for each tomato sold at time t. Both u(t) and v(t) change with time. It is obvious why u(t) changes: you sell more or less tomatoes depending of the time t.

But v(t) also changes: the market price of tomatoes and your production costs change all the time, hence your profits per sale change with time.

Now, we have the profits S(t) expressed as a profuct. A natural question you make as a business owner is: what is the trend? You want to predict how your profits will change in the future. And the tool for that is the derivative.

As we learned in the page on the intuitive idea behind the derivative, the derivative gives the rate of change of a function. The rate of change of your profit is valuable information for you. So, let's try to solve the problem of finding S'(t). This will lead us to the product rule.

Let's think a bit. S'(t) is the rate of change of your profit. For your profit to change, at least one of two things must change: your sales or your profits per sale. That is, functions u(t) or v(t).

This means that S'(t) will be a function of the rates of change of those functions: u'(t) and v'(t).

But what is the precise connection?

Let's say, for the sake of simplicity, that both u'(t) and v'(t) are positive for a specific time t. This means that sales are increasing and profits per sale also are increasing.

A concrete value of u'(t) could be one tomato per second. That is, at time t, your sales are increasing one unit per second. To know how much your profits are increasing at time t, you need to multiply that rate by the profit per sale.

For example, if at time t you are making 5 dollars per sale (v(t)=5), the increase of your profits due to u'(t) is:

That is, due to the increase of sales (u'(t)), your profits are increasing at a rate of 5 dollars per second. From this we conclude: to obtain the increase of S(t) due to u'(t) we take the product u'(t)v(t).

Now, the other factor we need to consider is v'(t). This is how your profits per sale are increasing. A concrete value of v(t) could be 1 dollar per sale per second. If at time t you make 10 sales (u(t)=10), the rate of change of your profits due to v'(t) is:

Due to the increase of profit per sale (v'(t)), your net profits are increasing at a rate of 10 dollars per second. We conclude: to obtain the increase of S(t) due to v'(t) we take the product v'(t)u(t).

The net rate of change of profits is the sum of the two factors:

We just discovered how to find the rate of change of a product, as a function of the rates of each factor. This is what is called the product rule. Given a function that is the product of two other functions:

its derivative is given by:

### A Second Way of Understanding the Product Rule

Another way of understaning why the product rule is the way it is, is using physical units. For example, if both u(t) and v(t) are in meters (m), S(t) is in meters squared (m2).

The rate of change S'(t) is in meters squared per second (m2/s). So, the product rule should result in that same unit.

u'(t) and v'(t) are in meters per second (m/s). So, the only way of obtaining (m2/s) using both u'(t) and v'(t) is multiplying them by v(t) and u(t), respectively. And then of course, we add the two factors. This results in the product rule:

### Example 1

Now that we have the powerful product rule in our toolkit, let's find the derivative of some sophisticated functions:

We can write:

So, applying the product rule:

That was easy.

### Example 2

Another easy example just to make you comfortable with the product rule:

Again, we can write:

Replacing, we have:

### Example 3

Now, this problem is a little harder:

We now have the product of three functions. What do we do?

Always remember, something that looks difficult always is the combinations of many easy things. Let's tackle this in easy chunks. We can do the following:

That is, we write:

So, we have:

Now, we know how to derive this:

We have:

So, replacing in the formula, we get:

As with the chain rule, you'll apply the product all the time, so you'll get more practice. Your next step should be to learn about the quotient rule.

If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.