Section B.1 Product Rule
The next technique for solving DEs we will study relies heavily on you knowing the Product Rule for differentiation inside and out, forward and backward (literally!). These problems are intended to help you review the Product Rule. (Don’t forget about the chain rule, too!) Let’s warm up by practicing using the product rule.
- Write down the product rule.\begin{equation*} \frac{d}{dx}\Big( f(x)\cdot g(x) \Big) = f(x) \cdot g'(x) + g(x) \cdot f'(x) \end{equation*}
- Evaluate the following derivatives.
- \(\ds \frac{d}{dx}\Big( e^{4x}\cos x \Big) \qquad\)
Solution.
\begin{align*} \frac{d}{dx}\Big( e^{4x}\cos x \Big) =\amp\ e^{4x} \cdot\frac{d}{dx}\Big(\cos x \Big) + \cos x\cdot \frac{d}{dx}\Big(e^{4x}\Big) \\ =\amp\ e^{4x} \cdot (-\sin x) + \cos x\cdot e^{4x} \cdot \frac{d}{dx}(4x)\\ =\amp\ -e^{4x} \sin x + \cos x\cdot e^{4x} \cdot (4)\\ =\amp\ -e^{4x} \sin x + 4 e^{4x} \cos x\\ =\amp\ e^{4x} (4\cos x - \sin x) \end{align*}Answer.
\begin{equation*} e^{4x} (4\cos x - \sin x) \end{equation*} - \(\ds \frac{d}{dx}\Big( x^5\ln (7x^2) \Big) \qquad\)
Solution.
\begin{align*} \frac{d}{dx}\Big( x^5\ln (7x^2) \Big) =\amp\ x^5 \cdot\frac{d}{dx}\Big(\ln(7x^2) \Big) + \ln(7x^2)\cdot \frac{d}{dx}\Big( x^5 \Big)\\ =\amp\ x^5 \cdot \frac{1}{7x^2}\cdot\frac{d}{dx}(7x^2) + \ln(7x^2)\cdot (5x^4)\\ =\amp\ \frac{14x^6}{7x^2} + 5x^4\ln(7x^2)\\ =\amp\ 2x^4 + 5x^4\ln(7x^2) \end{align*}Answer.
\begin{equation*} 2x^4 + 5x^4\ln(7x^2) \end{equation*} - \(\ds \frac{d}{dx}\Big( x\arctan x \Big) \qquad\)
Solution.
\begin{align*} \frac{d}{dx}\Big( x\arctan x \Big) =\amp\ x \cdot\frac{d}{dx}\Big(\arctan x \Big) + \arctan x\cdot \frac{d}{dx}\Big( x \Big)\\ =\amp\ x \cdot\frac{1}{1+x^2} + \arctan x\cdot 1\\ =\amp\ \frac{x}{1+x^2} + \arctan x \end{align*}Answer.
\begin{equation*} \frac{x}{1+x^2} + \arctan x \end{equation*} -
Suppose \(y\) is a function of \(x\) and use the product rule to rewrite the following:\(\ds \frac{d}{dx}\Big( \cos x \cdot y(x) \Big) = \)
Answer.
\begin{equation*} \frac{d}{dx}\Big( \cos x \cdot y(x) \Big) = \cos x \cdot \frac{dy}{dx} + y(x) \cdot (-\sin x) \end{equation*} -
Suppose \(z\) is a function of \(t\) and use the product rule to rewrite the following:\(\ds \frac{d}{dt}\Big( e^{5t} \cdot z \Big) = \)
Answer.
\begin{align*} \frac{d}{dt}\Big( e^{5t} \cdot z \Big) =\amp\ e^{5t} \cdot \frac{dz}{dt} + z \cdot (e^{5t} \cdot 5) \\ =\amp\ e^{5t} \cdot \frac{dz}{dt} + z \cdot 5e^{5t} \end{align*}
Now let’s look at the product rule "in the other direction"... Or we can think about this as "un-doing" the product rule.
For example, if we have the expression
\begin{equation*}
e^{(t^3)} \cdot \frac{dy}{dt} + 3t^2e^{(t^3)}\cdot y,
\end{equation*}
then we can see that if we think identify \(e^{(t^3)}\) as
\begin{equation*}
f(t) = e^{(t^3)}
\end{equation*}
then
\begin{equation*}
f'(t) = 3t^2e^{(t^3)}
\end{equation*}
Then we might label the expression as follows:
\begin{equation*}
\ub{e^{(t^3)}}_{f(t)} \cdot \ub{\frac{dy}{dt}}_{g'(t)} + \ub{3t^2e^{(t^3)}}_{f'(t)}\cdot \ub{y}_{g(t)}
\end{equation*}
This now looks like the result of having taken the derivative of a product. That is:
\begin{align*}
\ub{e^{(t^3)}}_{f(t)} \cdot \ub{\frac{dy}{dt}}_{g'(t)} + \ub{3t^2e^{(t^3)}}_{f'(t)}\cdot \ub{y}_{g(t)}
=\amp\ f(t) \cdot g'(t) + f'(t) \cdot g(t)\\
=\amp\ \frac{d}{dt}\Big( f(t) \cdot g(t) \Big)\\
=\amp\ \frac{d}{dt}\Big( e^{(t^3)} \cdot y \Big)
\end{align*}
Now you try some.
Rewrite each of the following as the derivative of a product:
- \(\ds x^{-4} \frac{dy}{dx} - 4x^{-5}y = \frac{d}{dx}\Big(\qquad ? \qquad\Big)\qquad\)
Answer.
\begin{equation*} \ub{x^{-4}}_{f} \ub{\frac{dy}{dx}}_{g'} + \ub{- 4x^{-5}}_{f'}\ub{y}_{g} = \frac{d}{dx}\Big(\,\ob{\ub{x^{-4}}_{f}\ub{y}_{g}}^{?}\,\Big) \end{equation*} - \(\ds e^{-3x} y' - 3e^{-3x}y = \frac{d}{dx}\Big(\qquad ? \qquad\Big)\qquad\)
Answer.
\begin{equation*} \ub{e^{-3x}}_{f} \ub{y'}_{g'} + \ub{- 3e^{-3x}}_{f'}\ub{y}_{g} = \frac{d}{dx}\Big(\,\ob{\ub{e^{-3x}}_{f}\ub{y}_{g}}^{?}\,\Big) \end{equation*} - \(\ds \sec(\theta) \frac{dr}{d\theta} + \sec(\theta) \tan(\theta) r = \frac{d}{dx}\Big(\qquad ? \qquad\Big)\qquad\)
Answer.
\begin{equation*} \ub{\sec(\theta)}_{f} \ub{\frac{dr}{d\theta}}_{g'} + \ub{\sec(\theta) \tan(\theta)}_{f'}\ub{r}_{g} = \frac{d}{d\theta}\Big(\,\ob{\ub{\sec(\theta)}_{f}\ub{r}_{g}}^{?}\,\Big) \end{equation*}
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