2.2: Differentiation Techniques - The Product and Quotient Rules

Section Preview

Note to the Instructor (click to expand)

Since the students will only know how to take derivatives of polynomials and expressions with rational exponents, this is a fast section. A good chunk of your time here (other than going through examples with the class) should be spent on proving the Product Rule.

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• An Overview of Functions
  • Function Arithmetic and Domain: The Product and Quotient Rules differentiate a product \(f(x)g(x)\) and a quotient \(\frac{f(x)}{g(x)}\) of functions; understanding what those combinations of functions mean is the conceptual basis for the entire section.

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Factoring Techniques
  • Factoring Expressions Involving Negative Rational Exponents: In Example 3(b), simplifying the Quotient-Rule result requires factoring/clearing the \(t^{-2/3}\) term so the denominator becomes \(3t^{2/3}(t-5)^2\).
• Simplifying Expressions
  • Simplifying Expressions Involving Negative Rational Exponents: Differentiating \(t^{1/3}\) produces \(\frac{1}{3}t^{-2/3}\), and simplifying the resulting Quotient-Rule expression in Example 3(b) requires manipulating negative rational exponents.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• 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.
• Combine the differentiation rules to find the derivative of a polynomial or rational function.

The Product Rule

Now that we have examined the basic rules, we can look at some of the more advanced ones. 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 \(f(x)=x^2\), whose derivative is \(f^{\prime}(x)=2x\) and not \(\frac{d}{dx}(x) \cdot \frac{d}{dx}(x)=1 \cdot 1=1\).

Theorem: The Product Rule

Let \(f(x)\) and \(g(x)\) be differentiable functions. Then\[\dfrac{d}{dx}(f(x)g(x))=\dfrac{d}{dx}(f(x)) \cdot g(x)+\dfrac{d}{dx}(g(x)) \cdot f(x).\nonumber\]That is,\[\text{if }p(x)=f(x)g(x),\quad \text{then }p^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x).\nonumber\]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 \(f(x)\) and \(g(x)\) are differentiable functions. At a key point in this proof, we need to use the fact that since \(g(x)\) is differentiable, it is also continuous. In particular, we use the fact that since \(g(x)\) is continuous, \(\displaystyle \lim_{h \to 0}g(x+h)=g(x)\).

By applying the limit definition of the derivative to \(p(x)=f(x)g(x)\), we obtain\[p^{\prime}(x)=\lim_{h \to 0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}.\nonumber\]By adding and subtracting \(f(x)g(x+h)\) in the numerator, we have\[p^{\prime}(x)=\lim_{h \to 0}\dfrac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}.\nonumber\]After breaking apart this quotient and applying the Sum Law for limits, the derivative becomes\[p^{\prime}(x)=\lim_{h \to 0}\dfrac{f(x+h)g(x+h)-f(x)g(x+h)}{h}+\lim_{h \to 0}\dfrac{f(x)g(x+h)-f(x)g(x)}{h}.\nonumber\]Rearranging, we obtain\[\begin{array}{rcl} p^{\prime}(x) & = & \displaystyle \lim_{h \to 0}\left(\dfrac{f(x+h)-f(x)}{h} \cdot g(x+h)\right) + \displaystyle \lim_{h \to 0}\left(\dfrac{g(x+h)-g(x)}{h} \cdot f(x)\right) \\[16pt] & = & \left(\displaystyle \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\right) \cdot \left(\displaystyle \lim_{h \to 0}\;g(x+h)\right)+\left(\displaystyle \lim_{h \to 0}\dfrac{g(x+h)-g(x)}{h}\right) \cdot f(x) \\[16pt] \end{array}\nonumber\]By using the continuity of \(g(x)\), the definition of the derivatives of \(f(x)\) and \(g(x)\), and applying the Limit Laws, we arrive at the Product Rule,\[p^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x).\nonumber\]

Q.E.D.

Example \(\PageIndex{1}\): Applying the Product Rule to Constant Functions

For \(p(x)=f(x)g(x)\), use the Product Rule to find \(p^{\prime}(2)\) if \(f(2)=3,\; f^{\prime}(2)=-4,\; g(2)=1\), and \(g^{\prime}(2)=6\).

Solution

Since \(p(x)=f(x)g(x)\), \(p^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)\), and hence \[p^{\prime}(2)=f^{\prime}(2)g(2)+g^{\prime}(2)f(2)=(-4)(1)+(6)(3)=14.\nonumber\]

Example \(\PageIndex{2}\): Applying the Product Rule to Binomials

For \(p(x)=(x^2+2)(3x^3-5x)\), find \(p^{\prime}(x)\) by applying the Product Rule. Check the result by first finding the product and then differentiating.

Solution

If we set \(f(x)=x^2+2\) and \(g(x)=3x^3-5x\), then \(f^{\prime}(x)=2x\) and \(g^{\prime}(x)=9x^2-5\). Thus, \[p^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2).\nonumber\]Simplifying, we have\[p^{\prime}(x)=15x^4+3x^2-10.\nonumber\]To check, we see that \(p(x)=3x^5+x^3-10x\) and, consequently, \(p^{\prime}(x)=15x^4+3x^2-10\).

Checkpoint \(\PageIndex{2}\)

Use the Product Rule to obtain the derivative of \(p(x)=2x^5(4x^2+x)\).

Answer

\(p^{\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5\).

Before we move on, it is important to repeat a previous warning.

Caution: The Product Rule is NOT the Product of Derivatives!

\[\dfrac{d}{dx}\left( f(x) \cdot g(x) \right) \neq \dfrac{d}{dx} \left( f(x) \right) \cdot \dfrac{d}{dx} \left( g(x) \right)\nonumber\]

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. To better grasp why we cannot simply take the quotient of the derivatives, keep in mind that\[\dfrac{d}{dx}(x^2)=2x,\text{ not }\dfrac{\dfrac{d}{dx}(x^3)}{\dfrac{d}{dx}(x)}=\dfrac{3x^2}{1}=3x^2.\nonumber\]

Theorem: The Quotient Rule

Let \(f(x)\) and \(g(x)\) be differentiable functions. Then\[\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\dfrac{d}{dx}(f(x)) \cdot g(x)-\dfrac{d}{dx}(g(x)) \cdot f(x)}{\big(g(x)\big)^2}.\nonumber\]That is, if\[q(x)=\dfrac{f(x)}{g(x)}\nonumber\]then\[q^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{\big(g(x)\big)^2}.\nonumber\]

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 to find derivatives in the next example.

Example \(\PageIndex{3}\): Applying the Quotient Rule

Find the derivative of each function.

1. \(q(x)=\frac{5x^2}{4x+3}\)
2. \(s(t) = \frac{\sqrt[3]{t}}{t - 5}\)

Solutions

1. Let \(f(x)=5x^2\) and \(g(x)=4x+3\). Thus, \(f^{\prime}(x)=10x\) and \(g^{\prime}(x)=4\). Substituting into the Quotient Rule, we have\[q^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}=\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}.\nonumber\]Simplifying, we obtain\[q^{\prime}(x)=\dfrac{20x^2+30x}{(4x+3)^2}.\nonumber\]
2. It's best to rewrite \(s\) using rational exponents, where possible.\[s(t) = \dfrac{t^{1/3}}{t - 5}\nonumber\]Now we can start to take the derivative of this quotient using the Quotient Rule. For reference, if we let \(N(t)\) and \(D(t)\) be the numerator and denominator, respectively, then\[\begin{array}{ccccccc} N(t) & = & t^{1/3} & \quad \text{ and } \quad & D(t) & = & t - 5 \\[16pt] N^{\prime}(t) & = & \dfrac{1}{3} t^{-2/3} & \quad \text{ and } \quad & D^{\prime}(t) & = & 1 \\[16pt] \end{array}\nonumber\]Therefore,\[\begin{array}{rclcl} \dfrac{d}{dt} \left( s(t) \right) & = & \dfrac{d}{dt} \left( \dfrac{N(t)}{D(t)} \right) & & \\[16pt] & = & \dfrac{N^{\prime}(t) D(t) - N(t) D^{\prime}(t)}{[D(t)]^2} & \quad & (\text{Quotient Rule}) \\[16pt] & = & \dfrac{\left( \dfrac{1}{3} t^{-2/3} \right)\left( t - 5 \right) - \left( t^{1/3} \right) \left( 1 \right)}{\left( t - 5 \right)^2} & & \\[16pt] & = & \dfrac{\left( t - 5 \right) - 3 t}{3 t^{2/3} \left( t - 5 \right)^2} & & \\[16pt] & = & \dfrac{-2t - 5}{3 t^{2/3} \left( t - 5 \right)^2} & & \\[16pt] \end{array}\nonumber\]

The previous example demonstrates the need to use our prerequisite algebra skills often when using the Quotient Rule. We will commonly have to factor, simplify, and multiply by fractions equivalent to \(1\) when using this rule.

Checkpoint \(\PageIndex{3}\)

Find the derivative of \(h(x)=\frac{3x+1}{4x-3}\).

Answer

\(h^{\prime}(x)=-\frac{13}{(4x-3)^2}\).

As with the last topic, caution is merited here.

Caution: The Quotient Rule is NOT the Quotient of Derivatives!

\[\dfrac{d}{dx} \left( \dfrac{N(x)}{D(x)} \right) \neq \dfrac{\dfrac{d}{dx}\left( N(x) \right)}{\dfrac{d}{dx} \left( D(x) \right)}\nonumber\]

Revisiting the Power Rule

Using the Quotient Rule to prove most of the General Power Rule is now possible. In the previous section, recall that we proved the Power Rule for positive integers. Still, we stated that we had to wait until later in the course to prove the Power Rule for the general case when the exponent is any real number. At this point, we can also add to our previous proof by extending the Power Rule to negative integers. This proof is provided for completeness.

Proof

Let \(f(x) = x^k\). If \(k\) is a negative integer, we may set \(n=-k\), so that \(n\) is a positive integer with \(k=-n\). Since for each positive integer \(n\),\(x^{-n}=\frac{1}{x^n}\), we may now apply the Quotient Rule by setting \(f(x)=1\) and \(g(x)=x^n\). In this case, \(f^{\prime}(x)=0\) and \(g^{\prime}(x)=nx^{n-1}\). Thus,\[\dfrac{d}{dx}(x^{-n})=\dfrac{0(x^n)-1(nx^{n-1})}{(x^n)^2}.\nonumber\]Simplifying, we see that\[\begin{array}{rcl} \dfrac{d}{dx}(x^{-n}) & = & \dfrac{-nx^{n-1}}{x^{2n}} \\[16pt] & = & -nx^{(n-1)-2n}\\[16pt] & = & -nx^{-n-1}. \\[16pt] \end{array}\nonumber\]Finally, observe that since \(k=-n\), by substituting we have\[\dfrac{d}{dx}(x^k)=kx^{k-1}.\nonumber\]

Q.E.D.

Combining Differentiation Rules

As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. By combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on, we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.

Example \(\PageIndex{4}\): Combining Differentiation Rules

For \(k(x)=3h(x)+x^2g(x)\), find \(k^{\prime}(x)\).

Solution

Finding this derivative requires the Sum Rule, the Constant Multiple Rule, and the Product Rule.\[\begin{array}{rclcl} k^{\prime}(x) & = & \dfrac{d}{dx}\big(3h(x)+x^2g(x)\big) & & \\[16pt] & = & \dfrac{d}{dx}\big(3h(x)\big)+\dfrac{d}{dx}\big(x^2g(x)\big) & \quad & \left( \text{Sum Rule} \right) \\[16pt] & = & 3\dfrac{d}{dx}\big(h(x)\big)+\left(\dfrac{d}{dx}(x^2)g(x)+\dfrac{d}{dx}(g(x))x^2\right) & \quad & \left( \text{Constant Multiple and Product Rules} \right) \\[16pt] & = & 3h^{\prime}(x)+2xg(x)+g^{\prime}(x)x^2 & & \\[16pt] \end{array}\nonumber\]

Example \(\PageIndex{5}\): Extending the Product Rule

For \(k(x)=f(x)g(x)h(x)\), express \(k^{\prime}(x)\) in terms of \(f(x),g(x),h(x)\), and their derivatives.

Solution

We can think of the function \(k(x)\) as the product of the function \(f(x)g(x)\) and the function \(h(x)\). That is, \(k(x)=(f(x)g(x)) \cdot h(x)\). Thus,\[\begin{array}{rclcl} k^{\prime}(x) & = & \dfrac{d}{dx}\big(f(x)g(x)\big) \cdot h(x)+\dfrac{d}{dx}\big(h(x)\big) \cdot \big(f(x)g(x)\big) & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & \big(f^{\prime}(x)g(x)+g^{\prime}(x)f(x)\big)h(x)+h^{\prime}(x)f(x)g(x) & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & f^{\prime}(x)g(x)h(x)+f(x)g^{\prime}(x)h(x)+f(x)g(x)h^{\prime}(x) & & \\[16pt] \end{array}\nonumber\]

Example \(\PageIndex{6}\): Combining the Quotient Rule and the Product Rule

For \(h(x)=\frac{2x^3k(x)}{3x+2}\), find \(h^{\prime}(x)\).

Solution

This procedure is typical for finding the derivative of a rational function.\[\begin{array}{rclcl} h^{\prime}(x) & = & \dfrac{\frac{d}{dx}(2x^3k(x)) \cdot (3x+2)-\frac{d}{dx}(3x+2) \cdot (2x^3k(x))}{(3x+2)^2} & \quad & \left( \text{Quotient Rule} \right) \\[16pt] & = & \dfrac{(6x^2k(x)+k^{\prime}(x) \cdot 2x^3)(3x+2)-3(2x^3k(x))}{(3x+2)^2} & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & \dfrac{-6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k^{\prime}(x)+4x^3k^{\prime}(x)}{(3x+2)^2} & & \\[16pt] \end{array}\nonumber\]