The Advanced Rules

Learning Outcomes

  • Use the product rule for finding the derivative of a product of functions
  • Use the quotient rule for finding the derivative of a quotient of functions
  • Extend the power rule to functions with negative exponents

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] and not [latex]\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1[/latex].

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.

Recall: commutative property of addition and multiplication for real numbers

The following properties hold for real numbers a, b, and c.

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.

Example: Applying the Product Rule to Functions at a Point

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].

Example: Applying the Product Rule to Binomials

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.

Try It

Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[/latex].

Watch the following video to see the worked solution to the above Try It.

Try It

The Quotient Rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

[latex]\frac{d}{dx}(x^2)=2x[/latex], which is not the same as [latex]\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} =\dfrac{3x^2}{1}=3x^2[/latex]

 

The Quotient Rule

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

[latex]\frac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\frac{d}{dx}(f(x))\cdot g(x)-\dfrac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}[/latex]

 

That is,

if [latex]j(x)=\dfrac{f(x)}{g(x)}[/latex], then [latex]j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}[/latex]

 

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

Example: Applying the Quotient Rule

Use the quotient rule to find the derivative of [latex]k(x)=\dfrac{5x^2}{4x+3}[/latex]

We explored the flexibility of the product rule given to the commutative property under addition and multiplication. It is worth mentioning that for the quotient rule, the order of the terms in the numerator will matter, as the commutative property does not hold under subtraction. We can see this from the example above:

[latex]{10x(4x+3)-4(5x^2)=20x^2+30x}[/latex]

however,

[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[/latex]

 

Try It

Find the derivative of [latex]h(x)=\dfrac{3x+1}{4x-3}[/latex]

Watch the following video to see the worked solution to the above Try It.

Try It

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form [latex]x^k[/latex] where [latex]k[/latex] is a negative integer.

Extended Power Rule

If [latex]k[/latex] is a negative integer, then

[latex]\dfrac{d}{dx}(x^k)=kx^{k-1}[/latex]

 

Proof

If [latex]k[/latex] is a negative integer, we may set [latex]n=−k[/latex], so that [latex]n[/latex] is a positive integer with [latex]k=−n[/latex]. Since for each positive integer [latex]n, \, x^{−n}=\frac{1}{x^n}[/latex], we may now apply the quotient rule by setting [latex]f(x)=1[/latex] and [latex]g(x)=x^n[/latex]. In this case, [latex]f^{\prime}(x)=0[/latex] and [latex]g^{\prime}(x)=nx^{n-1}[/latex]. Thus,

[latex]\frac{d}{dx}(x^{−n})=\dfrac{0(x^n)-1(nx^{n-1})}{(x^n)^2}[/latex].

 

Simplifying, we see that

[latex]\frac{d}{dx}(x^{−n})=\dfrac{−nx^{n-1}}{x^{2n}}=−nx^{(n-1)-2n}=−nx^{−n-1}[/latex].

 

Finally, observe that since [latex]k=−n[/latex], by substituting we have

[latex]\frac{d}{dx}(x^k)=kx^{k-1}[/latex]

[latex]_\blacksquare[/latex]

Example: Using the Extended Power Rule

Find [latex]\frac{d}{dx}(x^{-4})[/latex]

Example: Using the Extended Power Rule and the Constant Multiple Rule

Use the extended power rule and the constant multiple rule to find [latex]f(x)=\dfrac{6}{x^2}[/latex]

Try It

Find the derivative of [latex]g(x)=\dfrac{1}{x^7}[/latex] using the extended power rule.

Watch the following video to see the worked solution to the above Try It.

Try It