# The Product Rule

Theorem 1 (The Product Rule): If $f$ and $g$ are differentiable functions, then the derivative of the product $fg$ is $\frac{d}{dx} [f(x)g(x)] = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x)$. |

We will now derive the product rule geometrically. Suppose $f$ and $g$ are positive functions that are differentiable, and denote $\Delta x$ to be the change in $f(x)$ and $g(x)$. Therefore, we can denote the change in $f(x)$ and the change in $g(x)$ to be:

(1)The diagram below illustrates the area of $f(x)g(x)$ as well as the area $(f(x) + \Delta f(x))(g(x) + \Delta g(x))$.

Thus, the change in the rectangles' area can be denoted as:

(2)If we now divide all terms by $\Delta x$ and then take $\lim_{\Delta x \to 0}$, we then get the derivative of the product $f(x)g(x)$.

(3)We note that $\lim_{\Delta x \to 0 } f(x) \frac{\Delta g(x)}{\Delta x} = f(x) \cdot \frac{d}{dx} g(x)$, $\lim_{\Delta x \to 0 } g(x) \frac{\Delta f(x)}{\Delta x} + \lim_{\Delta x \to 0 } = g(x) \cdot \frac{d}{dx} f(x)$. Furthermore, $\lim_{\Delta x \to 0 } \Delta f(x) \frac{\Delta g(x)}{\Delta x} = 0 \cdot \frac{dv}{dx}$, since as $\Delta x \to 0$, $\Delta f(x) \to 0$, and therefore:

(4)So we have derived the product rule. We will now formally prove the product rule.

**Proof:**We will do a special trick to derive the product rule. It will be necessary to perform somewhat of a "trick" in this proof, that is we will add the term $-f(x + h)g(x) + f(x + h)g(x)$ (which is really just adding 0) to the numerator. Noting that $\lim_{h \to 0} f(x+h) = f(x)$ we proceed as follows:

We will now look at some examples of apply the derivative product rule.

## Example 1

**Find the derivative of the function $f(x) = x^2 e^x$.**

Well we can clearly see that $f(x)$ consists of a product of two parts, namely $x^2$ and $e^x$, so we can apply the product rule to get that:

(6)Note: Simplification is often not necessary when applying the product rule. |

## Example 2

**Find the derivative of the function $y = 2x^3 \sin x$.**

In this example, the function $y$ is composed of two parts, namely $2x^3$ and $\sin x$, and thus applying the product rule again we get:

(7)## Example 3

**Find the derivative of the function $y = x^2 x^3$.**

You may be tempted to use the product rule, and it would work sufficiently. However, notice that both terms can be combined such that $x^2 x^3 = x^5$, to which we can applying the power rule to get that:

(8)# Derivatives for Functions of Large Products

Suppose that $f$ is not a product of two other functions, but rather of three functions (or more), such that $f(x) = g(x)h(x)m(x)$ such as the function $f(x) = x^2 e^x \sin x$. We can apply the product rule twice in the same sense and we'll eventually derive the following rule:

(9)You may be able to see a clear pattern that $f'(x)$ is equal to the derivative of the first function $g$ multiplied by the other two functions plus the derivative of the second function $h$ multiplied by the other two functions plus the derivative of the third function $m$ multiplied by the other two functions.

## Example 4

**Find the derivative of the function $f(x) = \sin x \cos x \tan x$.**

Applying the product rule as necessary we get that:

(10)