Differentiating Products

Differentiating Products of Functions

Many functions are built by multiplying two simpler functions. For example, $ y(x) = x(x + 1)
$ is the product of the linear function $ x
$ and the function $
x + 1.
$

On this page, we will verify the

Product Rule

$(yz)^\prime = y^\prime z + y z^\prime$

As always, we'll compute the average rate of change of the function over the interval $
x_0
$ to $
x_0 + h.
$ This average rate of change is

$\frac{y(x_0 + h)z(x_0+h) - y(x_0)z(x_0)}{h}$

Our argument will use the tangent line approximation to compute the approximate average rate of change. Remember that

$\begin{eqnarray*} y(x_0 + h) & \cong & y(x_0) + y^\prime(x_0) h \\ z(x_0 + h) & \cong & z(x_0) + z^\prime(x_0) h \end{eqnarray*}$

and that these approximations are very good when $ h $ is small.

If we multiply these two approximations together, we see that

$y(x_0+h)z(x_0 + h) \cong y(x_0)z(x_0) + (y^\prime(x_0) z(x_0) + y(x_0)z^\prime(x_0))h + y^\prime(x_0)z^\prime(x_0)h^2.$

This means that the average rate of change is

$\frac{y(x_0 + h)z(x_0+h) - y(x_0)z(x_0)}{h} \cong y^\prime(x_0) z(x_0) + y(x_0)z^\prime(x_0) + y^\prime(x_0) z^\prime(x_0) h.$ Now as $
h
$

becomes very small, the approximation improves and the last term above becomes increasingly small. This says that

$(yz)^\prime = y^\prime z + y z^\prime.$

Example:

Consider the function $
y(x) = x(x + 1).
$ We may compute the derivative
$\frac{dy}{dx} = \frac{dx}{dx}(x + 1) + x\frac{d(x+1)}{dx} = 1(x+1) + x(1) = 2x + 1.$
Notice that in this example, the derivative depends on the value of $
x
$ in which we are interested.
Below, the graph of the function $ y(x) = x(x+1)
$ is shown on the left while its derivative $ 2x + 1
$ is shown on the right.

Questions:

Can you explain the relationship between these two graphs?
What does it mean when the derivative is negative and what does it mean when it is positive?
What happens at the point where the derivative equals zero?