# 3.3: The Product Rule

- Page ID
- 467

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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**

\[ {d\over dx}\sqrt{625-x^2}={-x\over\sqrt{625-x^2}}.\nonumber\]

Now

\[\begin{align*} f'(x)&=x^2{-x\over\sqrt{625-x^2}}+2x\sqrt{625-x^2} \\[4pt]&= {-x^3+2x(625-x^2)\over \sqrt{625-x^2}} \\[4pt] &= {-3x^3+1250x\over \sqrt{625-x^2}}. \end{align*}\]

## Contributors

Integrated by Justin Marshall.