Consider the product of two simple functions, say $\ds f(x)=(x^2+1)(x^3-3x)$. An obvious guess for the derivative of $f$ is the product of the derivatives of the constituent functions: $\ds (2x)(3x^2-3)=6x^3-6x$. Is this correct? We can easily check, by rewriting $f$ and doing the calculation in a way that is known to work. First, $\ds f(x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x$, and then $\ds f'(x)=5x^4-6x^2-3$. 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 $\ds \lim_{\Delta x\to0}f(x+\Delta x)$ with $f(x)$—why is this justified?

What we really need to know here is that $\ds \lim_{\Delta x\to 0}f(x+\Delta x)=f(x)$, 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 $\ds f(x)=(x^2+1)(x^3-3x)$. Then $\ds f'(x)=(x^2+1)(3x^2-3)+(2x)(x^3-3x)=3x^4-3x^2+3x^2-3+2x^4-6x^2= 5x^4-6x^2-3$, 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 3.3.1 Compute the derivative of $\ds f(x)=x^2\sqrt{625-x^2}$. We have already computed $\ds {d\over dx}\sqrt{625-x^2}={-x\over\sqrt{625-x^2}}$. Now $$f'(x)=x^2{-x\over\sqrt{625-x^2}}+2x\sqrt{625-x^2}= {-x^3+2x(625-x^2)\over \sqrt{625-x^2}}= {-3x^3+1250x\over \sqrt{625-x^2}}. $$

## Exercises 3.3

In 1–4, find the derivatives of the functions using the product rule.

**Ex 3.3.1**
$\ds x^3(x^3-5x+10)$
(answer)

**Ex 3.3.2**
$\ds (x^2+5x-3)(x^5-6x^3+3x^2-7x+1)$
(answer)

**Ex 3.3.3**
$\ds \sqrt{x}\sqrt{625-x^2}$
(answer)

**Ex 3.3.4**
$\displaystyle {\sqrt{625-x^2}\over x^{20}}$
(answer)

**Ex 3.3.5**
Use the product rule to compute the derivative of $\ds f(x)=(2x-3)^2$.
Sketch the function. Find an equation of the tangent line to the curve at
$x=2$. Sketch the tangent line at $x=2$.
(answer)

**Ex 3.3.6**
Suppose that $f$, $g$, and $h$ are differentiable functions.
Show that $(fgh)'(x) = f'(x) g(x)h(x) + f(x)g'(x) h(x) + f(x) g(x)
h'(x)$.

**Ex 3.3.7**
State and prove a rule to compute $(fghi)'(x)$,
similar to the rule in the previous problem.

Remark 3.3.2 {Product notation} Suppose $\ds f_1 , f_2 , \ldots f_n$ are functions. The product of all these functions can be written $$ \prod _{k=1 } ^n f_k.$$ This is similar to the use of $\ds \sum$ to denote a sum. For example, $$\prod _{k=1 } ^5 f_k =f_1 f_2 f_3 f_4 f_5$$ and $$ \prod _ {k=1 } ^n k = 1\cdot 2 \cdot \ldots \cdot n = n!.$$ We sometimes use somewhat more complicated conditions; for example $$\prod _{k=1 , k\neq j } ^n f_k$$ denotes the product of $\ds f_1$ through $\ds f_n$ except for $\ds f_j$. For example, $$\prod _{k=1 , k\neq 4} ^5 x^k = x\cdot x^2 \cdot x^3 \cdot x^5 = x^{11}.$$

**Ex 3.3.8**
The **generalized product rule**
says that if $\ds f_1 , f_2 ,\ldots ,f_n$ are differentiable functions at
$x$ then
$${d\over dx}\prod _{k=1 } ^n f_k(x) =
\sum _{j=1 } ^n \left(f'_j (x) \prod _{k=1 , k\neq j} ^n
f_k (x)\right).$$
Verify that this is the same as your answer to the previous problem
when $n=4$,
and write out what this says when $n=5$.