
# 3.3: The Product Rule

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Consider the product of two simple functions, say $$f(x)=(x^2+1)(x^3-3x)$$. An obvious guess for the derivative of $$f$$ is the product of the derivatives of the constituent functions: $$(2x)(3x^2-3)=6x^3-6x$$.

Is this correct? We can easily check, by rewriting $$f$$ and doing the calculation in a way that is known to work. First, $$f(x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x$$, and then $$f'(x)=5x^4-6x^2-3$$. Not even close! What went "wrong''? Well, nothing really, except the guess was wrong.

So the derivative of $$f(x)g(x)$$ is NOT as simple as $$f'(x)g'(x)$$. Surely there is some rule for such a situation? There is, and it is instructive to "discover'' it by trying to do the general calculation even without knowing the answer in advance.

\eqalign{ {d\over dx}(&f(x)g(x)) = \lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x) - f(x)g(x)\over \Delta x}\cr& =\lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x) + f(x+\Delta x)g(x)- f(x)g(x)\over \Delta x}\cr & =\lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)\over \Delta x} + \lim_{\Delta x \to0} {f(x+\Delta x)g(x)- f(x)g(x)\over \Delta x}\cr & =\lim_{\Delta x \to0} f(x+\Delta x){ g(x+\Delta x)-g(x)\over \Delta x} + \lim_{\Delta x \to0} {f(x+\Delta x)- f(x)\over \Delta x}g(x)\cr & =f(x)g'(x) + f'(x)g(x)\cr }

A couple of items here need discussion. First, we used a standard trick, "add and subtract the same thing'', to transform what we had into a more useful form. After some rewriting, we realize that we have two limits that produce $$f'(x)$$ and $$g'(x)$$. Of course, $$f'(x)$$ and $$g'(x)$$ must actually exist for this to make sense. We also replaced $$\lim_{\Delta x\to0}f(x+\Delta x)$$ with $$f(x)$$---why is this justified?

What we really need to know here is that $$\lim_{\Delta x\to 0}f(x+\Delta x)=f(x)$$, or in the language of section 2.5, that $$f$$ is continuous at $$x$$. We already know that $$f'(x)$$ exists (or the whole approach, writing the derivative of $$fg$$ in terms of $$f'$$ and $$g'$$, doesn't make sense). This turns out to imply that $$f$$ is continuous as well. Here's why:

\eqalign{ \lim_{\Delta x\to 0} f(x+\Delta x) &= \lim_{\Delta x\to 0} (f(x+\Delta x) -f(x) + f(x))\cr& = \lim_{\Delta x\to 0} {f(x+\Delta x) -f(x)\over \Delta x}\Delta x + \lim_{\Delta x\to 0} f(x)\cr& =f'(x)\cdot 0 + f(x) = f(x)\cr }

To summarize: the product rule says that

${d\over dx}(f(x)g(x)) = f(x)g'(x) + f'(x)g(x).$

Returning to the example we started with, let

$f(x)=(x^2+1)(x^3-3x).$

Then

$f'(x)=(x^2+1)(3x^2-3)+(2x)(x^3-3x)=3x^4-3x^2+3x^2-3+2x^4-6x^2= 5x^4-6x^2-3,$

as before. In this case it is probably simpler to multiply $$f(x)$$ out first, then compute the derivative; here's an example for which we really need the product rule.

Example $$\PageIndex{1}$$

Compute the derivative of $$f(x)=x^2\sqrt{625-x^2}$$.

Solution

${d\over dx}\sqrt{625-x^2}={-x\over\sqrt{625-x^2}}.\nonumber$
\begin{align*} f'(x)&=x^2{-x\over\sqrt{625-x^2}}+2x\sqrt{625-x^2} \\[5pt]&= {-x^3+2x(625-x^2)\over \sqrt{625-x^2}} \\[5pt] &= {-3x^3+1250x\over \sqrt{625-x^2}}. \end{align*}