Differentiation Rules: Learn It 3

The Advanced Rules

The Product Rule

Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern.

To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[/latex], whose derivative is,

[latex]f^{\prime}(x)=2x[/latex]

which is not the same as,

[latex]\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1[/latex].

the product rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

[latex]\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))\cdot g(x)+\frac{d}{dx}(g(x))\cdot f(x)[/latex]

 

That is,

if [latex]j(x)=f(x)g(x)[/latex] then [latex]j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)[/latex]

 

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Proof

We begin by assuming that [latex]f(x)[/latex] and [latex]g(x)[/latex] are differentiable functions. At a key point in this proof we need to use the fact that, since [latex]g(x)[/latex] is differentiable, it is also continuous. In particular, we use the fact that since [latex]g(x)[/latex] is continuous, [latex]\underset{h\to 0}{\lim}g(x+h)=g(x)[/latex].

By applying the limit definition of the derivative to [latex]j(x)=f(x)g(x)[/latex], we obtain

[latex]j^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}[/latex]

 

By adding and subtracting [latex]f(x)g(x+h)[/latex] in the numerator, we have

[latex]j^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}[/latex]

 

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

[latex]j^{\prime}(x)=\underset{h\to 0}{\lim}\left(\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}\right)+\underset{h\to 0}{\lim}\left(\frac{f(x)g(x+h)-f(x)g(x)}{h}\right)[/latex]

 

Rearranging, we obtain

[latex]j^{\prime}(x)=\underset{h\to 0}{\lim}\left(\frac{f(x+h)-f(x)}{h}\cdot g(x+h)\right)+\underset{h\to 0}{\lim}\left(\frac{g(x+h)-g(x)}{h}\cdot f(x)\right)[/latex]

 

By using the continuity of [latex]g(x)[/latex], the definition of the derivatives of [latex]f(x)[/latex] and [latex]g(x)[/latex], and applying the limit laws, we arrive at the product rule,

[latex]j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)[/latex]

[latex]_\blacksquare[/latex]

As you begin using the product rule, it may be useful to remember that addition and multiplication are commutative for all real numbers.

The following properties hold for real numbers [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].

  Addition Multiplication
Commutative Property [latex]a+b=b+a[/latex] [latex]a\cdot b=b\cdot a[/latex]

This is particularly useful for our product rule because our formula consists solely of these two operations. Due to the commutative property of addition:

[latex]j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=g^{\prime}(x)f(x)+f^{\prime}(x)g(x)[/latex]

Additionally, the order in which you multiply each of these terms doesn’t matter, due to the commutative property of multiplication.

For [latex]j(x)=f(x)g(x)[/latex], use the product rule to find [latex]j^{\prime}(2)[/latex] if [latex]f(2)=3, \, f^{\prime}(2)=-4, \, g(2)=1[/latex], and [latex]g^{\prime}(2)=6[/latex].

For [latex]j(x)=(x^2+2)(3x^3-5x)[/latex], find [latex]j^{\prime}(x)[/latex] by applying the product rule. Check the result by first finding the product and then differentiating.