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Lecture 8 The Product Rule

Goal: Be able to differentiate functions of the form \( f(x) g(x) \)

Question: Does \( \frac{d}{dt}(f(t) g(t)) = \frac{d}{dt}(f(t)) \cdot \frac{d}{dt}(g(t)) \)?

Answer: No! Consider \(f(t) = 1\) and \(g(t) = t\)

\[ \begin{gathered} \frac{d}{dt}(1 \cdot t) = 1 \\[0.5em] \frac{d}{dt}(1) \cdot \frac{d}{dt}(t) = 1 \cdot 0 = 0 \end{gathered} \]

To derive the product rule, imagine \(f(t)g(t)\) as the area of a rectangle with sides \(f(t)\) and \(g(t)\). Since the side lengths are changing with time, if we take a tiny step in time \((\Delta t)\), there will be change to the length \((\Delta f(t))\) and the width \((\Delta g(t))\). We want to find the change in the area \(\Delta (f(t)g(t))\).

Square of side f(t) by g(t) with shaded L-shaped strips of width delta f(t) and delta g(t)
Figure 1
A hand-drawn diagram of a rectangle representing the product \( f(t) g(t) \). The top edge of the inner rectangle is labeled \( f(t) \) and the left edge is labeled \( g(t) \). Extending to the right of the rectangle is a narrow hatched vertical strip labeled \( \Delta f(t) \), and extending below the rectangle is a narrow hatched horizontal strip labeled \( \Delta g(t) \). Together the hatched strips form an L-shaped shaded region, including the small corner square where the two strips meet, representing the increase in area \( \Delta(f(t) g(t)) \) when both factors grow by a small amount.
\begin{align*} \Delta(f(t) g(t)) =&~ \text{Area of Shaded Region}\\[1em] =&~ \Delta f(t) \cdot g(t) + f(t) \Delta g(t) + (\Delta f(t))(\Delta g(t))\\[1em] \frac{\Delta(f(t)g(t))}{\Delta t} =&~ \frac{\Delta f(t)}{\Delta t} \cdot g(t) + f(t) \cdot\frac{\Delta g(t)}{\Delta t} + \Delta f(t)\cdot\frac{\Delta g(t)}{\Delta t}\\[1em] \frac{\Delta(f(t)g(t))}{\Delta t} =&~ \frac{\Delta f(t)}{\Delta t} \cdot g(t) + f(t) \cdot\frac{\Delta g(t)}{\Delta t} + \Delta t \cdot\frac{\Delta f(t)}{\Delta t}\cdot\frac{\Delta g(t)}{\Delta t}\\[1em] \end{align*}

Take \( \Delta t \to 0 \) to get \(\displaystyle \frac{\Delta f(t)}{\Delta t} \to \dfrac{df}{dt}\) and \(\displaystyle \frac{\Delta g(t)}{\Delta t} \to \dfrac{dg}{dt}\). This gets us the equations in the theorem below.

Theorem (Product Rule)

Let \( f(x) \) and \( g(x) \) be differentiable functions. Then,

\begin{align*} \frac{d}{dx}(f(x) g(x)) =&~ \frac{df}{dx} \cdot g(x) + f(x) \cdot \frac{dg}{dx} \quad \text{(In Leibniz notation)}\\[0.5em] [f(x) g(x)]' =&~ f'(x) g(x) + f(x) g'(x)\quad\text{(In Newton/Prime notation)} \end{align*} There is a mnemonic to help memorize this formula. If we consider \(f(x)\) as the "left" function and \(g(x)\) as the "right" function, then to take the derivative it is "right d-left plus left d-right" (where "d" refers to taking the derivative).

Example 1

As a sanity check, we already know \(\frac{d}{dx}(x^3) = 3x^2 \). Let's use the Product Rule to differentiate \(x^3\) and see if we get the same answer. \begin{align*} \frac{d}{dx}(x^3) =&~ \frac{d}{dx}(x \cdot x^2)\\[0.5em] =&~ [x]' \cdot x^2 + x \cdot [x^2]' \\[0.5em] =&~ 1 \cdot x^2 + x \cdot 2x \\[0.5em] =&~ x^2 + 2x^2\\[0.5em] =&~ 3x^2 \checkmark \end{align*}

NOTE:

By continuing Example 1, this gives us a way of computing \( \frac{d}{dx}(x^n) \) for an integer \( n \ge 0 \).

\[ \begin{gathered} x^4 = x \cdot x^3 \\[0.5em] x^5 = x \cdot x^4 \\[0.5em] \vdots \end{gathered} \]

Ie, this shows the Power Rule is true for non negative integer powers.

Example 2

\begin{align*} \frac{d}{dx}(2 \sin x \cos x) =&~ [2 \sin x]' \cos x + 2 \sin x [\cos x]' \\[0.5em] =&~ 2 \cos x \cos x + 2 \sin x (-\sin x) \\[0.5em] =&~ 2 [\cos^2 x - \sin^2 x] \end{align*}

Example 3

Let \( h(x) = \underbrace{(x^{10} + x + 1)}_{f} \underbrace{(x^2 + 9)}_{g} \)

  1. Compute \( h'(x) \)

    Another way to tackle these kinds of problems (especially if your function is very complicated) is to find \(f(x), f'(x), g(x)\), and \(g'(x)\) on their own:

    \begin{align*} f(x) =&~ x^{10} + x + 1 \longrightarrow f'(x) = 10x^9 + 1 \\[0.5em] g(x) =&~ x^2 + 9 \longrightarrow g'(x) = 2x \end{align*} Then, combine them at the end using the Product Rule: \begin{align*} h'(x) =&~ f'(x) g(x) + f(x) g'(x) \\[0.5em] =&~ (10x^9 + 1)(x^2 + 9) + (x^{10} + x + 1)(2x) \end{align*}

  2. Compute \( h'(1) \)

    \[ h'(1) = 11 \cdot 10 + 3 \cdot 2 = 116 \]

Example 4

Compute \( h'(x) \) if \( h(x) = (x+1)(2x+1)(3x+1) \)

\begin{align*} h'(x) =&~ [x+1]'(2x+1)(3x+1) + (x+1)\left[ (2x+1)(3x+1) \right]' \\[0.5em] =&~ (2x+1)(3x+1) + (x+1)\left[ [2x+1]'(3x+1) + (2x+1)[3x+1]' \right] \\[0.5em] =&~ (2x+1)(3x+1) + (x+1)\left[ 2(3x+1) + (2x+1)\cdot 3 \right] \\[0.5em] =&~ 1 \cdot (2x+1)(3x+1) + (x+1)\cdot 2 \cdot (3x+1) + (x+1)(2x+1)\cdot 3 \end{align*} In other words, we can take the derivative if there are more than 2 functions being multiplied together. Though, it requires using the Product Rule more than once.

Example 5

Find the equation of the line tangent to \(h(x) = (\sqrt{x} + 4)\left( x^{\frac{3}{4}} - 2x \right)\) at \(x = 16\).

We need to find \(h'(16)\). One way to do it is to compute \(h'(x)\), then plug in 16 (like what we did in Example 3(b)). Another way is to write \(h'(16) = f'(16)g(16) + f(16)g'(16)\); here, we find \(f(16), f'(16), g(16),\) and \(g'(16)\) seprately: \begin{align*} f(x) =&~ \sqrt{x} + 4\\[0.5em] f(16) =&~ 4 + 4 \\[0.5em] =&~ 8 \end{align*} \begin{align*} f'(x) =&~ \frac{1}{2\sqrt{x}}\\[0.5em] f'(16) =&~ \frac{1}{2\cdot 4} \\[0.5em] =&~ \frac{1}{8} \end{align*} \begin{align*} g(x) =&~ x^{\frac{3}{4}} - 2x \\[0.5em] g(16) =&~ 8 - 32 \\[0.5em] =&~ -24 \end{align*} \begin{align*} g'(x) =&~ \frac{3}{4}x^{-\frac{1}{4}} - 2\\[0.5em] =&~\frac{3}{4 \cdot \sqrt[4]{x}} - 2\\[0.5em] g'(16) =&~ \frac{3}{8} - 2 \\[0.5em] =&~ -\frac{13}{8} \end{align*} Then, combine everything using the Product Rule: \begin{align*} \text{Slope}: \; h'(16) =&~ f'(16) g(16) + f(16) g'(16) \\[0.5em] =&~ \frac{1}{8}(-24) + 8\left( \frac{-13}{8} \right) \\[0.5em] =&~ -3 - 13 \\[0.5em] =&~-16 \end{align*} We now finish the problem: \begin{align*} \text{Point}: \; x\text{-coord}: \; x =&~ 16\\[0.5em] y\text{-coord}: \; h(16) =&~ f(16) g(16) \\[0.5em] =&~ 8(-24)\\[0.5em] =&~-192 \\[0.5em] \text{Equation}: \; y + 192 =&~ -16(x - 16)\\[0.5em] y =&~ -16x + (256 - 192)\\[0.5em] y =&~ -16x + 64 \end{align*}

Example 6

When does \( y = 10x^5 e^x \) have a horizontal tangent line?

\begin{align*} y' =&~ 50x^4 e^x + 10x^5 e^x \\[0.5em] =&~ 10x^4 e^x (5+x)\\[0.5em] \overset{\text{set}}{=}&~ 0 \end{align*} If a product equals 0, then at least one of the factors is 0. I.e., either \[\begin{gathered}10x^4 = 0 \quad \text{OR} \quad e^x = 0 \quad \text{OR} \quad 5 + x = 0 \\[0.5em] \boxed{x = 0} \qquad \text{Never} \qquad \boxed{x = -5} \end{gathered} \]

Example 7

Compute \( h'(x) \) if \(h(x) = 6e^x \sin x - 13 e^x \cos x\)

The main idea here is we want to factor out that \(e^x\) to get \(h(x)=e^x(6\sin x - 13 \cos x)\). Now, as opposed to needing to do the Product Rule twice, we only need to do it once. \begin{align*} h'(x) =&~ e^x (6 \sin x - 13 \cos x) + e^x (6 \cos x + 13 \sin x) \\[0.5em] =&~ e^x (6 \sin x - 13 \cos x + 6 \cos x + 13 \sin x) \\[0.5em] =&~ e^x (19 \sin x - 7 \cos x) \end{align*}