Opening the book…
- 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.
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- An Overview of Functions
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- Factoring Techniques
- Simplifying Expressions
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
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- 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 \(\frac{d}{dx}(x) \cdot \frac{d}{dx}(x)=1 \cdot 1=1\).
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
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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.
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
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Since \(p(x)=f(x)g(x)\), \(p^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)\), and hence \[p^{\prime}(2)=f^{\prime}(2)g(2)+g^{\prime}(2)f(2)=(-4)(1)+(6)(3)=14.\nonumber\]
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
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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, \(p^{\prime}(x)=15x^4+3x^2-10\).
Use the Product Rule to obtain the derivative of \(p(x)=2x^5(4x^2+x)\).
- Answer
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\(p^{\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5\).
Before we move on, it is important to repeat a previous warning.
\[\dfrac{d}{dx}\left( f(x) \cdot g(x) \right) \neq \dfrac{d}{dx} \left( f(x) \right) \cdot \dfrac{d}{dx} \left( g(x) \right)\nonumber\]
The following Interactive Element might help you visualize why the derivative of a product is \(f'g + g'f\) and not \(f'g'\).
Interact: Drag the large corner point toward the base to shrink the growth.
Observation: First, recall that \(\frac{d}{dx} (f \cdot g)\) represents how quickly the product, \(f \cdot g\), is changing at any given moment. Second, if the product, \(f \cdot g\), is changing, there is an associated change in both \(f\) and \(g\) (these are \(\Delta f\) and \(\Delta g\)). Notice from the manipulable that, as the area shrinks, the corner, \(\Delta f \cdot \Delta g\), vanishes far faster than the strips \(f \cdot \Delta g\) and \(g \cdot \Delta f\). This implies (but does not prove) that only the two strips contribute anything of note to the derivative. That is, to compute how quickly the product \(f \cdot g\) is changing, we only need to consider the blue and green areas in the manipulable above. Hence, it seems that \(\frac{d}{dx}(fg) \approx g \cdot \Delta f + f \cdot \Delta g\) and never just \(\Delta f \cdot \Delta g\).
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\[\dfrac{d}{dx}(x^2)=2x,\text{ not }\dfrac{\dfrac{d}{dx}(x^3)}{\dfrac{d}{dx}(x)}=\dfrac{3x^2}{1}=3x^2.\nonumber\]
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.
Find the derivative of each function.
- \(q(x)=\frac{5x^2}{4x+3}\)
- \(s(t) = \frac{\sqrt[3]{t}}{t - 5}\)
- Solutions
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- 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\[q^{\prime}(x)=\dfrac{20x^2+30x}{(4x+3)^2}.\nonumber\]
- 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.
Find the derivative of \(h(x)=\frac{3x+1}{4x-3}\).
- Answer
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\(h^{\prime}(x)=-\frac{13}{(4x-3)^2}\).
As with the last topic, caution is merited here.
\[\dfrac{d}{dx} \left( \dfrac{N(x)}{D(x)} \right) \neq \dfrac{\dfrac{d}{dx}\left( N(x) \right)}{\dfrac{d}{dx} \left( D(x) \right)}\nonumber\]
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.
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.
For \(k(x)=3h(x)+x^2g(x)\), find \(k^{\prime}(x)\).
- Solution
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Finding this derivative requires the Sum Rule, the Constant Multiple Rule, and the Product Rule.\[\begin{array}{rclcl} k^{\prime}(x) & = & \dfrac{d}{dx}\big(3h(x)+x^2g(x)\big) & & \\[16pt] & = & \dfrac{d}{dx}\big(3h(x)\big)+\dfrac{d}{dx}\big(x^2g(x)\big) & \quad & \left( \text{Sum Rule} \right) \\[16pt] & = & 3\dfrac{d}{dx}\big(h(x)\big)+\left(\dfrac{d}{dx}(x^2)g(x)+\dfrac{d}{dx}(g(x))x^2\right) & \quad & \left( \text{Constant Multiple and Product Rules} \right) \\[16pt] & = & 3h^{\prime}(x)+2xg(x)+g^{\prime}(x)x^2 & & \\[16pt] \end{array}\nonumber\]
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
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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,\[\begin{array}{rclcl} k^{\prime}(x) & = & \dfrac{d}{dx}\big(f(x)g(x)\big) \cdot h(x)+\dfrac{d}{dx}\big(h(x)\big) \cdot \big(f(x)g(x)\big) & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & \big(f^{\prime}(x)g(x)+g^{\prime}(x)f(x)\big)h(x)+h^{\prime}(x)f(x)g(x) & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & f^{\prime}(x)g(x)h(x)+f(x)g^{\prime}(x)h(x)+f(x)g(x)h^{\prime}(x) & & \\[16pt] \end{array}\nonumber\]
For \(h(x)=\frac{2x^3k(x)}{3x+2}\), find \(h^{\prime}(x)\).
- Solution
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This procedure is typical for finding the derivative of a rational function.\[\begin{array}{rclcl} h^{\prime}(x) & = & \dfrac{\frac{d}{dx}(2x^3k(x)) \cdot (3x+2)-\frac{d}{dx}(3x+2) \cdot (2x^3k(x))}{(3x+2)^2} & \quad & \left( \text{Quotient Rule} \right) \\[16pt] & = & \dfrac{(6x^2k(x)+k^{\prime}(x) \cdot 2x^3)(3x+2)-3(2x^3k(x))}{(3x+2)^2} & \quad & \left( \text{Product Rule} \right) \\[16pt] & = & \dfrac{-6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k^{\prime}(x)+4x^3k^{\prime}(x)}{(3x+2)^2} & & \\[16pt] \end{array}\nonumber\]
This page titled 2.2: Differentiation Techniques - The Product and Quotient Rules was last modified on Fri, 02 Oct 2026 23:00:39 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman via source content that was edited to the style and standards of the LibreTexts platform.
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