When we use the term *product,* we’re actually referring to two functions being multiplied together. The Product Rule is one of the differentiation rules in calculus that allows us to find the derivative of the product of two or more differentiable functions. This rule is really useful when we want to avoid or can’t do the multiplication before differentiation.

In other words, the Product Rule allows us to find the derivative of two differentiable functions that are being **multiplied together** by combining our knowledge of both the Power Rule and the Sum and Difference Rule for derivatives.

## Product Rule Formula and Formal Definition

The Product Rule formula for two functions is as follows:

$$\frac{\text{d}}{\text{d}x}(f(x)g(x))=f(x)\xb7\frac{\text{d}}{\text{d}x}g(x)+g(x)\xb7\frac{\text{d}}{\text{d}x}f(x)$$For two differentiable functions $f(x)$ and $g(x)$, the derivative of their product with respect to $x$ is given by:

$$\frac{\text{d}}{\text{d}x}(f(x)g(x))=f(x)\xb7\frac{\text{d}}{\text{d}x}g(x)+g(x)\xb7\frac{\text{d}}{\text{d}x}f(x)$$where $\frac{\text{d}}{\text{d}x}f(x)$ is the derivative of $f(x)$ and $\frac{\text{d}}{\text{d}x}g(x)$ is the derivative of $g(x)$.

In plain words, the rule states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function.

## Steps to Apply the Product Rule

Identify the two functions being multiplied together, $f(x)$ and $g(x)$.

Find the derivative of the first function, $\frac{\text{d}}{\text{d}x}f(x)$.

Find the derivative of the second function, $\frac{\text{d}}{\text{d}x}g(x)$.

Multiply the first function, $f(x)$, by the derivative of the second function, $\frac{\text{d}}{\text{d}x}g(x)$.

Multiply the second function, $g(x)$, by the derivative of the first function, $\frac{\text{d}}{\text{d}x}f(x)$.

Add the results from steps 4 and 5 to obtain the derivative of the product.

## Common Mistakes Using the Product Rule

It is essential to note that the derivative of a product is not (!) equal to the product of the derivatives. In other words:

$$\frac{\text{d}}{\text{d}x}(f(x)g(x))\ne \frac{\text{d}}{\text{d}x}f(x)\xb7\frac{\text{d}}{\text{d}x}g(x)$$This is something students often get mixed up with when first learning about the product rule. To make sure you don’t make this mistake, always remember to multiply each function by the derivative of the other function, then add the two results together.

## Examples

### Example 1: Binomial Function

Let’s find the derivative of the following function using the Product Rule:

$$h(x)=(3{x}^{2}+2)(4x-1)$$Using color-coding to make it easier to follow:

- $f(x)={3{x}^{2}+2}$ (blue)
- $g(x)={4x-1}$ (red)

Applying the Product Rule:

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}h(x)& =({3{x}^{2}+2)\xb7\frac{\text{d}}{\text{d}x}({4x-1)+({4x-1)\xb7\frac{\text{d}}{\text{d}x}({3{x}^{2}+2)\phantom{\rule{0ex}{0ex}}\&=({3{x}^{2}+2)\xb74+({4x-1)\xb76x\phantom{\rule{0ex}{0ex}}\&=12{x}^{2}+8+24{x}^{2}-6x\phantom{\rule{0ex}{0ex}}\&=36{x}^{2}-6x+8}}}}}}\hfill \end{array}$$### Example 2: Instantaneous Rate of Change

Suppose we want to find the derivative of $h(x)=(2{x}^{3}+1)(5x-3)$ when $x=1$.

First, we’ll use the Product Rule to find the derivative:

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}h(x)& =(2{x}^{3}+1)\xb7\frac{\text{d}}{\text{d}x}(5x-3)+(5x-3)\xb7\frac{\text{d}}{\text{d}x}(2{x}^{3}+1)\hfill \\ \hfill & =(2{x}^{3}+1)\xb75+(5x-3)\xb76{x}^{2}\hfill \\ \hfill & =10{x}^{3}+5+30{x}^{3}-18{x}^{2}\hfill \\ \hfill & =40{x}^{3}-18{x}^{2}+5\hfill \end{array}$$Now, we’ll find the value of the derivative when $x=1$:

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}h(1)& =40(1{)}^{3}-18(1{)}^{2}+5\hfill \\ \hfill & =40-18+5\hfill \\ \hfill & =27\hfill \end{array}$$So, the instantaneous rate of change of $h(x)$ at $x=1$ is 27.

### Example 3: Trigonometric Functions

Let’s find the derivative of $f(x)=\mathrm{sin}(x)\xb7\mathrm{cos}(x)$ using the Product Rule.

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}f(x)& =\mathrm{sin}(x)\xb7\frac{\text{d}}{\text{d}x}\mathrm{cos}(x)+\mathrm{cos}(x)\xb7\frac{\text{d}}{\text{d}x}\mathrm{sin}(x)\hfill \\ \hfill & =\mathrm{sin}(x)\xb7(-\mathrm{sin}(x))+\mathrm{cos}(x)\xb7\mathrm{cos}(x)\hfill \\ \hfill & =-{\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)\hfill \end{array}$$Just a quick note to say that we’ll be covering the derivatives of trigonometric functions in a separate lesson. I’m sure you’ll find this example really helpful in showing you how useful the Product Rule can be when you’re working with functions that are completely unrelated, like trigonometric functions.

### Example 4: More Than Two Functions

The Product Rule can be extended to find the derivative of the product of more than two functions. For three functions, the formula becomes:

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}(f(x)g(x)h(x))& =f(x)g(x)\xb7\frac{\text{d}}{\text{d}x}h(x)+f(x)h(x)\xb7\frac{\text{d}}{\text{d}x}g(x)+g(x)h(x)\xb7\frac{\text{d}}{\text{d}x}f(x)\hfill \end{array}$$This pattern can be extended to find the derivative of the product of any number of functions. For example, let’s find the derivative of $f(x)=({x}^{2}+1)(2x-3)(3x+4)$.

Using the extended Product Rule:

$$\begin{array}{cc}\hfill \frac{\text{d}}{\text{d}x}f(x)& =({x}^{2}+1)(2x-3)\xb7\frac{\text{d}}{\text{d}x}(3x+4)+({x}^{2}+1)(3x+4)\xb7\frac{\text{d}}{\text{d}x}(2x-3)+(2x-3)(3x+4)\xb7\frac{\text{d}}{\text{d}x}({x}^{2}+1)\hfill \\ \hfill & =({x}^{2}+1)(2x-3)\xb73+({x}^{2}+1)(3x+4)\xb72+(2x-3)(3x+4)\xb72x\hfill \\ \hfill & =3({x}^{2}+1)(2x-3)+2({x}^{2}+1)(3x+4)+2x(2x-3)(3x+4)\hfill \end{array}$$The result can be further simplified by expanding the terms and combining like terms.