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2.3 The Product Rule

When I first learned calculus, it seemed to me that if the derivative of 3t was 3, and the derivative of 7t was 7, then the derivative of t multiplied by t ought to be just plain old t, not 2t. The reason there's a factor of 2 in the correct answer is that t2 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

\[ \frac{d(xy)}{dt} = \frac{dx}{dt}\cdot y + x \cdot \frac{dy}{dt} . \]

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

\begin{align} & (x+dx)(y+dy)-xy \\ &= ydx + xdy + dxdy \end{align}

and dividing by \(dt\) makes this into

\[ \frac{dx}{dt}\cdot y + x \cdot \frac{dy}{dt} + \frac{dx dy}{dt} \]

whose standard part is the result to be proved.

Example 15

◊ Find the derivative of the function tsin t.

\begin{align} \frac{d(t\sin t)}{dt} &= t\cdot \frac{d(\sin t)}{dt}+\frac{dt}{dt}\cdot\sin t \\ &= t\cos t+\sin t \end{align}

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.

Figure H. A geometrical interpretation of the product rule.

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, m2/s, wouldn't match the ones on the right, m2/s2.

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 \(\ddot x\), Leibniz wrote as

\[\frac{d^2 x}{dt^2} .\]

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/s2. (The acceleration of falling objects on Earth is 9.8 m/s2 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 16

A tricky use of the product rule is to find the derivative of \(\sqrt{t}\). Since \(\sqrt{t}\) can be written as t1/2, we might suspect that the rule d(tk)/dt=ktk-1 would work, giving a derivative \(\frac{1}{2}t^{-1/2}=1/(2\sqrt{t})\) . 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 \(f(t)=d\sqrt{t}/dt\) for our unknown derivative, and back into the result using the product rule:

\begin{align} \frac{dt}{dt} &= \frac{d(\sqrt{t}\sqrt{t})}{dt} \\ &= f(t)\sqrt{t}+\sqrt{t}f(t) \\ &= 2f(t)\sqrt{t} \end{align}

But dt/dt=1, so \(f(t)=1/(2\sqrt{t})\) as claimed.

The trick used in example 16 can also be used to prove that the power rule d(xn)/dx=nxn-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.