# Product Rule

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Product Rule for Derivatives – HMC Calculus Tutorial

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 $h(x) = f(x) g(x)$:

$h'(x) = \left[f(x)g(x)\right]’ = f'(x) g(x) + f(x) g'(x).$

The derivative $h'(x)$ is given by the limit formula: $$h'(x) = \lim_{\Delta x \to 0} \frac{h(x + \Delta x) – h(x)}{\Delta x},$$ provided the limit exists. We now express $h$ using the product of $f$ and $g$, \begin{eqnarray*} h'(x) & = & \lim_{\Delta x \to 0} \frac{h(x + \Delta x) – h(x)}{\Delta x}\\ & = & \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) – f(x)g(x)}{\Delta x}. \end{eqnarray*} To make further progress, we need to relate our limit formula to the limit formulas for the derivatives $f'(x)$ and $g'(x)$, namely $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) – f(x)}{\Delta x} \quad \mbox{and} \quad g'(x) = \lim_{\Delta x \to 0} \frac{g(x + \Delta x) – g(x)}{\Delta x}.$$ To relate these formulas to the limit for $h'(x)$, we use the “trick” of adding and subtracting the term $f(x)g(x + \Delta x)$ in the numerator, and then simplifying: \begin{eqnarray*} h'(x) & = & \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) – f(x)g(x)}{\Delta x}\\ & = & \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) + (f(x)g(x + \Delta x) – f(x)g(x + \Delta x)) – f(x)g(x)}{\Delta x}\\ &=& \lim_{\Delta x \to 0} \left[\left(\frac{f(x + \Delta x) – f(x)}{\Delta x}\right) g(x + \Delta x) + \left( \frac{g(x + \Delta x) – g(x)}{\Delta x} \right) f(x) \right]\\ & = & \lim_{\Delta x \to 0} \frac{f(x + \Delta x) – f(x)}{\Delta x} \lim_{\Delta x \to 0} g(x + \Delta x) + f(x) \lim_{\Delta x \to 0} \frac{g(x + \Delta x) – g(x)}{\Delta x}\\ & = & f'(x) g(x) + f(x) g'(x). \end{eqnarray*} The last two steps are justified by assuming that $f'(x)$ and $g'(x)$ exist.

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 $h'(x) = f'(x)g(x) + f(x)g'(x)$.