The Product and Quotient Rules

Section 2.4 The Product and Quotient Rules

We saw in Theorem 2.3.10 Combinations of Derivatives that when we want to find the derivative of a sum or difference of functions, we can just find the derivatives of each function separately, and then combine the derivatives back together (by adding or subtracting). This, hopefully, is pretty intuitive: of course a slope of a sum of things is just the slopes of each thing added together!

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In this section, we want to think about derivatives of product and quotients of functions. What happens when we differentiate a function that is really just two functions multiplied together?

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Activity 2.4.1. Thinking About Derivatives of Products.

Let’s start with two quick facts:

\begin{equation*} \ddx{x^3} =3x^2 \text{ and } \ddx{\sin(x)}=\cos(x)\text{.} \end{equation*}

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(a)

What is \(\Ddx{x^3+\sin(x)}\text{?}\) What about \(\Ddx{x^3-\sin(x)}\text{?}\)

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(b)

Based on what you just explained, what is a reasonable assumption about what \(\Ddx{x^3\sin(x)}\) might be?

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Hint.

Does it seem reasonable that we could just multiply the derivatives together, and say that \(\Ddx{x^3\sin(x)}\) was the same thing as

\begin{equation*} \ddx{x^3}\cdot \ddx{\sin(x)}\text{?} \end{equation*}

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(c)

Let’s just think about \(\Ddx{x^3}\) for a moment. What is \(x^3\text{?}\) Can you write this as a product? Call one of your functions \(f(x)\) and the other \(g(x)\text{.}\) You should have that \(x^3 = f(x)g(x)\) for whatever you used.

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(d)

Use your example to explain why, in general, \(\Ddx{f(x)g(x)}\neq \Ddx{f(x)}\cdot\Ddx{g(x)}\text{.}\)

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(e)

Let’s show another way that we know this. Think about \(\sin(x)\text{.}\) We know two things:

\begin{equation*} \sin(x) = (1)(\sin(x)) \text{ and } \ddx{\sin(x)}=\cos(x)\text{.} \end{equation*}

What, though, is \(\Ddx{1}\cdot\Ddx{\sin(x)}\text{?}\)

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(f)

Use all of this to reassure yourself that even though the derivative of a sum of functions is the sum of the derivatives of the functions, we will need to develop a better understanding of how the derivatives of products of functions work.

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I like to think about it this way: if \(\Ddx{f(x)g(x)} = f'(x)g'(x)\text{,}\) then every function’s derivative would be 0.

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In the example with the \(\sin(x)\) function, we noticed that we could write the function as \((1)(\sin(x))\text{.}\) This is true of every function!

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If \(\Ddx{f(x)g(x)} = f'(x)g'(x)\text{,}\) then we could say that for any function \(f(x)\) with a derivative \(f'(x)\text{:}\)

\begin{align*} \ddx{f(x)} \amp = \ddx{1\cdot f(x)}\\ \amp = \ddx{1}\ddx{f(x)}\\ \amp = 0 \cdot f'(x)\\ \amp =0\text{.} \end{align*}

This, obviously, can’t be true! We know of tons of functions that have non-zero slopes...most of them do!

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So, we hopefully have some motivation for building a rule to that helps us think about derivatives of products of functions.

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Subsection The Product Rule

Activity 2.4.2. Building a Product Rule.

Let’s investigate how we might differentiate the product of two functions:

\begin{equation*} \ddx{f(x)g(x)}\text{.} \end{equation*}

We’ll use an area model for multiplication here: we can consider a rectangle where the side lengths are functions \(f(x)\) and \(g(x)\) that change for different values of \(x\text{.}\) Maybe \(x\) is representative of some type of time component, and the side lengths change size based on time, but it could be anything.

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If we want to think about \(\ddx{f(x)g(x)}\text{,}\) then we’re really considering the change in area of the rectangle.

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(a)

Find the area of the two rectangles. The second rectangle is just one where the input variable for the side length has changed by some amount, leading to a different side length.

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A rectangle with the width labeled f(x) and the height labeled g(x). — Figure 2.4.1.
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A larger rectangle with the width labeled f(x+ delta x) and the height labeled g(x + delta x). The original, smaller, rectangle is marked with a dotted line inside of the new larger rectangle. — Figure 2.4.1.
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(b)

Write out a way of calculating the difference in these areas.

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(c)

Let’s try to calculate this difference in a second way: by finding the actual area of the region that is new in the second rectangle.

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A small rectangle inside a larger rectangle. The difference is broken up into three shaded in regions. A bar on the top, a bar on the right, and a corner piece. — Figure 2.4.2.
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In order to do this, we’ve broken the region up into three pieces. Calculate the areas of the three pieces. Use this to fill in the following equation:

\begin{equation*} f(x+\Delta x)g(x+\Delta x) - f(x)g(x) = \fillinmath{XXXXXXXXXXXXXXXXXXXX}\text{.} \end{equation*}

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(d)

We do not want to calculate the total change in area: a derivative is a rate of change, so in order to think about the derivative we need to divide by the change in input, \(\Delta x\text{.}\)

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We’ll also want to think about this derivative as an instantaneous rate of change, meaning we will consider a limit as \(\Delta x \to 0\text{.}\) Fill in the following:

\begin{align*} \ddx{f(x)g(x)} \amp \lim_{\Delta x \to 0} \left( \frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)}{\Delta x} \right)\\ \amp = \lim_{\Delta x\to 0} \left(\frac{\fillinmath{XXXXXXXXXXXXXXXXXXXX}}{\Delta x}\right) \end{align*}

We can split this fraction up into three fractions:

\begin{align*} \ddx{f(x)g(x)} \amp = \lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right)\\ \amp + \lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right) \\ \amp + \lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right) \end{align*}

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(e)

In any limit with \(f(x)\) or \(g(x)\) in it, notice that we can factor part out of the limit, since these functions do not rely on \(\Delta x\text{,}\) the part that changes in the limit. Factor these out.

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In the third limit, factor out either \(\displaystyle \lim_{\Delta x\to 0} \left(f(x+\Delta x) - f(x)\right)\) or \(\displaystyle \lim_{\Delta x\to 0} \left(g(x+\Delta x) - g(x)\right)\text{.}\)

\begin{align*} \ddx{f(x)g(x)} \amp = f(x)\lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right)\\ \amp + g(x)\lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right) \\ \amp +\lim_{\Delta x\to 0}\left(\fillinmath{XXXXXXXXXX}\right)\left( \lim_{\Delta x \to 0} \left(\frac{\fillinmath{XXXXXXXXXX}}{\Delta x}\right)\right) \end{align*}

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(f)

Use Definition 2.1.2 The Derivative Function to rewrite these limits. Show that when \(\Delta x\to 0\text{,}\) we get:

\begin{equation*} f(x)g'(x) + g(x)f'(x) + 0\text{.} \end{equation*}

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We can investigate this visual a bit more dynamically: see the differences in area as \(\Delta x\to 0\text{.}\) What parts are left, when \(\Delta x \to 0\text{?}\) What areas aren’t visible?

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Theorem 2.4.3. Product Rule.

If \(u(x)\) and \(v(x)\) are functions that are differentiable at \(x\) and \(f(x) = u(x)\cdot v(x)\text{,}\) then:

\begin{equation*} \ddx{f(x)} = u'(x)\cdot v(x) + u(x) \cdot v'(x)\text{.} \end{equation*}

For convenience, this often is written as:

\begin{equation*} \ddx{u\cdot v} = u'v + uv' \hspace{1cm}f \text{or}\hspace{1cm} \ddx{u\cdot v} = v \left(\dd{u}{x}\right) + u \left(\dd{v}{x}\right)\text{.} \end{equation*}

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Example 2.4.4.

Use the Product Rule to find the following derivatives.

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(a)

\(\Ddx{x^3 \sin(x)}\)

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Hint.

Use \(u=x^3\) and \(v=\sin(x)\text{.}\) Now find \(u'\) and \(v'\text{,}\) and use:

\begin{equation*} \ddx{uv} = u'v+uv'\text{.} \end{equation*}

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(b)

\(\Ddx{(x^3+4x)e^x}\)

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(c)

\(\Ddx{\sqrt{x}\cos(x)}\)

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Subsection What about Dividing?

Since we can differentiate a product of functions, then the obvious next question should be about division: if we needed to build a reasonable way of differentiating a product, shouldn’t we also need to build a new way of thinking about derivatives of quotients?

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A nice thing that we can do is think about division as really just multiplication. For instance, if we want to differentiate \(\Ddx{\dfrac{\sin(x)}{x^2}}\text{,}\) then we can just think about this quotient as really a product:

\begin{equation*} \ddx{\frac{\sin(x)}{x^2}} = \ddx{\frac{1}{x^2}\left(\sin(x)\right)}\text{.} \end{equation*}

Now we can just apply the Product Rule!

\begin{align*} \ddx{\frac{1}{x^2}\left(\sin(x)\right)} \amp =\ddx{x^{-2}\sin(x)}\\ u = \sin(x) \amp \;\;\; v = x^{-2} \\ u'=\cos(x) \amp \;\;\; v'=-2x^{-3}\\ \ddx{\sin(x)x^{-2}}\amp = x^{-2}\cos(x) + (-2x^{-3}\sin(x))\\ \amp = \frac{\cos(x)}{x^2} - \frac{2\sin(x)}{x^3} \end{align*}

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This works great! We can always think about quotients as just products of reciprocals! But the problem is that we can’t always differentiate these reciprocals (for now). We were able to differentiate \(\dfrac{1}{x^2}\) by rewriting this as just a power function (with a negative exponent).

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What about this flipped example:

\begin{equation*} \ddx{\frac{x^2}{\sin(x)}}\text{?} \end{equation*}

In order for us to do the same thing, we need to rewrite this as

\begin{equation*} \ddx{x^2\left(\sin(x)\right)^{-1}} \end{equation*}

but we don’t know how to differentiate \(\left(\sin(x)\right)^{-1}\) (yet).

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So let’s try to build a general way of differentiating quotients.

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Activity 2.4.3. Constructing a Quotient Rule.

We’re going to start with a function that is a quotient of two other functions:

\begin{equation*} f(x) = \frac{u(x)}{v(x)}\text{.} \end{equation*}

Our goal is that we want to find \(f'(x)\text{,}\) but we’re going to shuffle this function around first. We won’t calculate this derivative directly!

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(a)

Start with \(f(x)=\dfrac{u(x)}{v(x)}\text{.}\) Multiply \(v(x)\) on both sides to write a definition for \(u(x)\text{.}\)

\begin{equation*} u(x) = \fillinmath{XXXXXXXXXXXXXXXXXXXX} \end{equation*}

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(b)

Find \(u'(x)\text{.}\)

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(c)

Wait: we don’t care about \(u'(x)\text{,}\) right? We care about finding \(f'(x)\text{!}\)

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Use what you found for \(u'(x)\) and solve for \(f'(x)\text{.}\)

\begin{equation*} f'(x) = \fillinmath{XXXXXXXXXXXXXXXXXXXX} \end{equation*}

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(d)

This is a strange formula: we have a formula for \(f'(x)\) written in terms of \(f(x)\text{!}\) But we said earlier that \(f(x) = \dfrac{u(x)}{v(x)}\text{.}\)

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In your formula for \(f'(x)\text{,}\) replace \(f(x)\) with \(\dfrac{u(x)}{v(x)}\text{.}\)

\begin{equation*} f'(x) = \fillinmath{XXXXXXXXXXXXXXXXXXXX} \end{equation*}

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This formula is fine, but a little clunky. We’ll try to rewrite it in a friendlier way, but it is a bit more complex than the Product Rule.

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Theorem 2.4.5. Quotient Rule.

If \(u(x)\) and \(v(x)\) are differentiable at \(x\) and \(f(x) = \dfrac{u(x)}{v(x)}\) then:

\begin{equation*} f'(x) = \dfrac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\text{.} \end{equation*}

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For convenience, this often is written as:

\begin{equation*} \ddx{\frac{u}{v}} = \frac{vu'-uv'}{v^2}\text{.} \end{equation*}

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Example 2.4.6.

Use the Quotient Rule to find the following derivatives.

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(a)

\(\Ddx{\dfrac{\sin(x)}{x^2}}\)

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Once you have this derivative, confirm that it is the same as \(\dfrac{\cos(x)}{x^2} - \dfrac{2\sin(x)}{x^3}\text{,}\) the way that we found it using the Product Rule.

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(b)

\(\Ddx{\dfrac{x^2}{\sin(x)}}\)

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(c)

\(\Ddx{\dfrac{x+4}{x^2+1}}\)

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Subsection Derivatives of (the Rest of the) Trigonometric Functions

You might remember that of the six main trigonometric functions, we can write four of them in terms of the other two: here are the different trigonometric functions written in terms of sine and cosine functions:

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A student encounters the problem \(\ddx{e^x(x^2+1)}\) and notes that since \(\ddx{e^x} = e^x\) and \(\ddx{x^2+1}=2x\text{,}\) then \(\ddx{e^x(x^2+1)} = 2xe^x\text{.}\) Explain not only why the student is incorrect, but also what kinds of pitfalls can occur if we use this “the derivative of a product is the product of the derivatives” method. Differentiate correctly, explaining your process for the student’s benefit.

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2.

Consider the following table of values of \(f\text{,}\) \(g\text{,}\) \(f'\text{,}\) and \(g'\text{.}\)

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Table 2.4.8.

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\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
\(0\)	\(-3\)	\(\frac{1}{2}\)	\(2\)	\(-4\)
\(2\)	\(\frac{1}{7}\)	\(-1\)	\(-\frac{5}{2}\)	\(\frac{7}{3}\)

\(f(x) = \dfrac{xe^x}{x+2}\) at \((0,0)\)

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