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Product Rule for Derivatives
In Calculus and its applications we often encounter functions that are
expressed as the product of two other functions, like the following examples:
- $h(x) = x e^x = (x)(e^x),$
- $h(x) = x^2 \sin x = (x^2)(\sin x),$
- $h(x) = e^{-x^2} \cos 2x = (e^{-x^2})(\cos 2x).$
In each of these examples, the values of the function $h$ can be written in the
form
$$
h(x) = f(x) g(x)
$$
for functions $f(x)$ and $g(x)$. If we know the derivative of $f(x)$ and
$g(x)$, the Product Rule provides a formula for the derivative of
$h(x) = f(x) g(x)$:
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$h'(x) = \left[f(x)g(x)\right]' = f'(x) g(x) + f(x) g'(x).$
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Proof
We illustrate this rule with the following examples.
- If $h(x) = x e^x $ then
\begin{eqnarray*}
h'(x) &=& (x)' e^x + x (e^x)'\\
&=& e^x + xe^x.
\end{eqnarray*}
- If $h(x) = x^2 \sin x $ then
\begin{eqnarray*}
h'(x) &=& (x^2)' \sin x + (x^2)(\sin x)'\\
&=& 2x \sin x + x^2 \cos x.
\end{eqnarray*}
- If $h(x) = e^{-x^2} \cos 2x $ then
\begin{eqnarray*}
h'(x) &=& (e^{-x^2})' \cos 2x + e^{-x^2} (\cos 2x)' \\
&=& -2xe^{-x^2} \cos 2x -2e^{-x^2} \sin 2x.
\end{eqnarray*}
Key Concepts
Product Rule
Let $f(x)$ and $g(x)$ be differentiable at $x$. Then $h(x) = f(x)g(x)$ is
differentiable at $x$ and
$h'(x) = f'(x)g(x) + f(x)g'(x)$.
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