# Untitled Page 8

- Page ID
- 148346

## 2.3 The product rule

When I first learned calculus, it seemed to me that if the derivative of 3*t* was 3, and the derivative of
7*t* was 7, then the derivative of *t* multiplied by *t* ought to be just plain old *t*, not 2*t*. The reason there's
a factor of 2 in the correct answer is that *t*^{2} has two reasons to grow as *t* gets bigger: it grows because
the first factor of *t* is increasing, but also because the second one is. In general, it's possible to find
the derivative of the product of two functions any time we know the derivatives of the individual functions.

### The product rule

If *x* and *y* are both functions of *t*, then the derivative of their product is

The proof is easy. Changing *t* by an infinitesimal amount *dt* changes the product *xy* by
an amount

whose standard part is the result to be proved.

### Example 7

◊ Find the derivative of the function *t*sin *t*.

◊

Figure h gives the geometrical interpretation of the product rule. Imagine that the king, in his castle
at the southwest corner of his rectangular kingdom, sends out a line of infantry to expand his territory
to the north, and a line of cavalry to take over more land to the east. In a time interval *dt*,
the cavalry, which moves faster, covers a distance *dx* greater than that covered by the infantry,
*dy*. However, the strip of territory conquered by the cavalry, *y* *dx*, isn't as great as it could have been,
because in our example *y* isn't as big as *x*.

A helpful feature of the Leibniz notation is that one can easily use it to check whether the units of an
answer make sense. If we measure distances in meters and time in seconds, then *xy* has units of
square meters (area), and so does the change in the area, d(*xy*). Dividing by *dt* gives the
number of square meters per second being conquered. On the right-hand side of the product rule,
*dx*/*dt* has units of meters per second (velocity), and multiplying it by
*y* makes the units square meters per second, which is consistent with the left-hand side. The
units of the second term on the right likewise check out. Some beginners might be tempted to
guess that the product rule would be d(*xy*)/*dt*=(*dx*/*dt*)(*dy*/*dt*), but the Leibniz
notation instantly reveals that this can't be the case, because then the units on the left,
m^{2}/s, wouldn't match the ones on the right, m^{2}/s^{2}.

Because this unit-checking feature is so helpful, there is a special way of writing a second derivative in the Leibniz notation. What Newton called x¨, Leibniz wrote as

Although the different placement of the 2's on top and bottom seems strange and inconsistent to many
beginners, it actually works out nicely.
If *x* is a distance, measured in meters, and *t* is a time, in units of seconds, then the
second derivative is supposed to have units of acceleration, in units of meters per second per
second, also written (m/s)/s, or m/s^{2}. (The acceleration of falling
objects on Earth is 9.8 m/s^{2} in these units.) The Leibniz notation is meant to suggest
exactly this: the top of the fraction looks like it has units of meters, because we're not squaring *x*,
while the bottom of the fraction looks like it has units of seconds squared, because it looks like we're
squaring *dt*. Therefore the units come out right. It's important to realize, however, that
the symbol d isn't a number (not a real one, and not a hyperreal one, either), so we can't really
square it; the notation is not to be taken as a literal statement about infinitesimals.

### Example 8

A tricky use of the product rule is to find the derivative of √t. Since √t can be written as*t*

^{1/2}, we might suspect that the rule d(

*t*

^{k})/

*dt*=

*kt*

^{k-1}would work, giving a derivative

^{1}/

_{2}t^{-1/2}=1/(2√t). However, the method from ch. 1 used to prove that rule proved on p.140 only work if

*k*is an integer, so the best we could do would be to confirm our conjecture approximately by graphing or numerical estimation.

Using the product rule, we can write f(t)=d√t/dt for our unknown derivative, and back into the result using the product rule:

But *dt*/*dt*=1, so f(t)=1/(2√t) as claimed.

The trick used in example 16 can also be used to prove that the power
rule d(*x*^{n})/*dx*=*nx*^{n-1} applies to cases where *n* is an integer less than 0, but
I'll instead prove this on page 41 by a technique that doesn't depend on
a trick, and also applies to values of *n* that aren't integers.