Differential Calculus - The Product Rule
The product rule gives us the derivative of the product of two (or more) functions. We know that we can find the differential of a polynomial function by adding together the differentials of the individual terms of the polynomial, each of which can be considered a function in its own right. You might therefore be tempted to assume that, for a function that is the product of two functions (i.e. two functions multiplied together), we can simply multiply together the derivatives of each function. Unfortunately, it is not quite that simple. An example should serve to illustrate the point. Suppose we need to find the derivative of the function ƒ(x) = x(x^{2} + 1). This function is the product of two functions, ƒ(x) = x and ƒ(x) = x^{2} + 1. We'll start by finding the derivative of each function separately:
Obviously, if we multiply these two derivatives together, the result will be x. Multiplying out the brackets in our original function, however, we get:
x(x^{2} + 1) = x^{3} + x
Taking the derivative of x^{3} + x, we get:
This is obviously very different from the result we got by simply multiplying together the derivatives of the functions ƒ(x) = x and ƒ(x) = x^{2} + 1. In this case, multiplying out the two functions was a relatively trivial exercise, and we were able to find the derivative of the function ƒ(x) = x(x^{2} + 1) without difficulty. There will be occasions, however, when this will much more difficult or impossible to do. Fortunately, there is a simple formula (the product rule) that we can use to find the derivative of the product of two functions.
Essentially, the rule states that in order to find the derivative of the product of two functions, we take the first function multiplied by the derivative of the second function, and add it to the second function multiplied by the derivative of the first function. Let's state this more formally. Suppose we have two functions u and v. The derivative of the product of these two functions uv is given by:
Let's test the formula by applying it to the function ƒ(x) = x(x^{2} + 1), for which we have already obtained the derivatives of the individual functions and their product:
This is the result we obtained previously, as we would expect. Let's try another example. Supposing we want to find the derivative of y = (x^{2} - 4x)(3 - 2x^{3}). We'll assign these functions as follows:
u(x) = x^{2} - 4x
v(x) = 3 - 2x^{3}
Now we find the differential for each of these functions:
So we have:
Multiplying out the brackets we get:
Factorising, we get:
We can use a general form of the product rule to find the derivative of the product of more than two functions. For example, the product of the three functions u, v and w can be found using the following formula:
For the product of n functions ƒ1, . . . ƒn, we can use the following generalisation:
Don't worry if this looks a bit frightening. It's just a formal way of saying how, for a collection of n functions, we can find the derivative of the product of those functions. The first part of the formula starts off by telling us that we are looking at the derivative of something, using the notation we are familiar with. The upper case Greek letter Pi (Π) is then used to indicate that it is the derivative of a product - in this case, the product of the functions ƒ1(x) through ƒn(x). In the second part of the formula, the upper case Greek letter Sigma (Σ) is used to indicate that the derivative of the product of these functions is equal to the sum of something. We must take the derivative of each function in turn, and multiply it by the product of all the other functions. Then we sum (add together) the results.
You only really need to worry about remembering this formula if you are studying mathematics at a fairly advanced level. All you really need to know here is that, in order to find the derivative of the product of two or more functions, we need to take the derivative of each function in turn and multiply it by the product of all the other functions. Then we just add the results together. In some cases, it may be easier just to multiply out the functions and differentiate the result in the usual way. In other cases this is not an option, and we will need to use the product rule. You'll have to judge each case on its merits.