Compute each of the derivatives below. Assume \(y\) is a function of \(x\text{.}\)
\begin{equation*}
1. \quad \left[x^6 e^{3x}\right]' \hspace{5em} 2. \quad \frac{d}{dx}\left[e^{x^3} y\right]
\end{equation*}
Section 6.1 Product Rule, A Review
Since the integrating factor method relies heavily on the product rule, let’s make sure we remember how it works.
Subsection The Product Rule
In calculus, the product rule tells us how to take the derivative of two multiplied functions. If \(f(x)\) and \(g(x)\) are differentiable, then
\begin{equation*}
\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + f'(x) \cdot g(x)\text{.}
\end{equation*}
To refresh your memory of how it works, try the following practice problems.
Checkpoint 62. 📖❓ Remembering the Product Rule.
Subsection The Product Rule in Reverse
Now that we’ve recalled how the product rule works, let’s try running it in reverse.
Suppose you’re given the expression:
\begin{equation}
\frac{1}{x} \sin x + \cos x \ln x\text{.}\tag{6.1}
\end{equation}
Can this be written as the derivative of a product? That is, can you find functions \(f(x)\) and \(g(x)\) so that
\begin{equation*}
\frac{d}{dx} \left[ f(x) g(x) \right] = \frac{1}{x} \sin x + \cos x \ln x?
\end{equation*}
Here’s a tip: the product rule always combines four pieces, two functions and their derivatives. If you can spot \(f\) and \(g\) in the terms, the rest should fall into place. In this case, trying \(f(x) = \sin x\) and \(g(x) = \ln x\) works out perfectly.
Let’s practice this reverse thinking.
🌌 Example 63.
Rewrite each expression as a derivative of a product:
\begin{equation*}
1. \quad e^{x}\cos x + e^{x}\sin x \hspace{5em} 2. \quad \frac{e^{x^2}}{x} + 2 x \ln x e^{x^2}
\end{equation*}
Solution.
First, select an \(f\) and \(g\) in different terms, like so
\begin{equation*}
e^x \
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}\cos x}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}e^x}}} \
\sin x\text{.}
\end{equation*}
For this to work, \(f'\) should be \(\sin x\text{,}\) next to our chosen \(g\text{.}\) However,
\begin{equation*}
f' = [\cos x]' = -\sin x \ne \sin x \quad
\end{equation*}
So this labeling doesn’t work, and we instead try the labels
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}e^x}}} \
\cos x
+
e^x \
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}\sin x}}}\text{.}
\end{equation*}
Noting that \(f' = [e^x]' = e^x\) is next to \(g\) and \(g' = [\sin x]' = \cos x\) is next to \(f\) confirms this is a valid product rule. Therefore, we can reverse the product rule as
\begin{equation*}
e^x \cos x + e^x \sin x = [e^x \sin x]'\text{.}
\end{equation*}
To make this easier, let’s separate the terms as
\begin{equation*}
\frac{1}{x} e^{x^2} + \ln x (2 x e^{x^2})\text{.}
\end{equation*}
So, as before, we first set \(f\) and \(g\)
\begin{equation*}
\frac{1}{x} \
\underset{\Large f}{ \underset{\uparrow}{\ul{\vphantom{|}e^{x^2}}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}\ln x}}} \
2 x e^{x^2}\text{,}
\end{equation*}
then verify the derivatives of \(f\) and \(g\) are in the other terms,
\begin{equation*}
f' = \left[e^{x^2}\right]' = 2 x e^{x^2} \quad \qquad g' = \left[\ln x\right]' = \frac{1}{x} \quad \text{.}
\end{equation*}
Therefore,
\begin{equation*}
\frac{e^{x^2}}{x} + 2 x \ln x e^{x^2} = \left[e^{x^2} \ln x\right]'\text{.}
\end{equation*}
Let’s now try expressions involving a variable function, like \(y(x)\text{.}\) These are the ones we’ll actually encounter when solving differential equations.
🌌 Example 64.
Rewrite each expression in the form \([f(x) y(x)]'\text{.}\)
\begin{equation*}
1. \quad e^{9x}\frac{dy}{dx} + 9y e^{9x} \hspace{5em} 2. \quad e^{1/x}y' - \frac{e^{1/x}}{x^2}y
\end{equation*}
Solution.
Since \(dy/dx\) and \(y\) are in separate terms, setting \(g\) to \(y\) gives you \(g'\) and \(f\) for free since
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{e^{9x}}}} \
\underset{\Large g'}{ \underset{\uparrow}{\ul{\vphantom{|}\frac{dy}{dx}}}}
+
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}} \
(9 e^{9x})\text{.}
\end{equation*}
The only thing you need to check is
\begin{equation*}
f' = [e^{9x}]' = 9e^{9x} \quad
\end{equation*}
which is sitting right next to \(g\text{.}\) Therefore, we have
\begin{equation*}
e^{9x}\frac{dy}{dx} + 9 e^{9x} y = \left[e^{9x} y\right]'\text{.}
\end{equation*}
Again, set \(g\) to \(y\) and the rest fall into place as
\begin{equation*}
\underset{\Large f}{ \underset{\uparrow}{\ul{e^{1/x}}}} \
\underset{\Large g'}{ \underset{\uparrow}{\ul{\vphantom{|}y'}}}
+
\left(-\frac{e^{1/x}}{x^2}\right) \
\underset{\Large g}{ \underset{\uparrow}{\ul{\vphantom{|}y}}}\text{.}
\end{equation*}
The only thing left to do is verify the derivative of \(f\text{:}\)
\begin{equation*}
f' = \left[e^{1/x}\right]' = e^{1/x}\left[\frac{1}{x}\right]^\prime = e^{1/x}\left(-\frac{1}{x^2}\right) = -\frac{e^{1/x}}{x^2} \quad \text{.}
\end{equation*}
So, we can reverse the product rule as
\begin{equation*}
e^{1/x}y' - \frac{e^{1/x}}{x^2}y = \left[e^{1/x} y\right]'\text{.}
\end{equation*}
Checkpoint 65. 📖❓ What’s Missing in the Product Rule.
Fill in the missing parts of each product rule below.
\(\ds\frac{d}{dx}\big[e^{3x} \cdot \) \(\big] = \) \(\ds\frac{1}{x} \cdot e^{3x} + \) \(e^{3x} \cdot \ln x \)
While reversing the product rule isn’t a standard technique, practicing it a few times helps build the intuition you’ll need when working with integrating factors.
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