Section 4-4-3

Chapter 4
Differential Calculus and Its Uses

4.4 The Product Rule

4.4.3 More on the Growth Rate of
Energy Consumption

We can use our energy consumption example to interpret the Product Rule. Recall that U. S. population in millions is approximated by

P = P (t) = P_{0} e^{k t},

where time `t` is measured in years from 1982, `k=0.00917`, and `P_0=232. Per capita energy consumption (in millions of BTU's per person) is approximated by

E = E (t) = m t + b,

where `m=3.4` and `b=302`. So total energy use, `T=Ttext[(]t text[)]` is approximated by the product

T = P E = P_{0} e^{k t} (m t + b) .

We wanted to calculate how fast total energy use in the U. S. is growing.

We know the growth rate of population,

\frac{d P}{d t} = k P_{0} e^{k t},

and of per capita energy consumption,

\frac{d E}{d t} = m .

The Product Rule then tells us that

$\frac{d T}{d t}$	$= P \frac{d E}{d t} + E \frac{d P}{d t}$
	$= (P_{0} e^{k t}) m + (m t + b) (k P_{0} e^{k t})$
	$= P_{0} (k m t + k b + m) e^{k t} .$

When we substitute numeric values for `k`, `m`, and `P_0`, we find an explicit formula for the derivative:

\frac{d T}{d t} = (7.23 t + 1430) e^{0.00917 t} .

If we want to know, for example, the growth rate of U.S. energy consumption in 1985, we substitute `t=3` to get about `1.5` billion BTU's per year. (Check the details.)

More important than specific numbers is the formula

\frac{d T}{d t} = P \frac{d E}{d t} + E \frac{d P}{d t},

which tells us that the energy consumption growth rate has two components. The first, `P dEtext[/]dt` represents population times growth rate of per capita consumption. The second, `E dPtext[/]dt` is per capita consumption times growth rate of the population.

Example 1

Find the derivative of `f text[(]x text[)]=x^3e^(4x)`.

Solution We solve this problem in both functional and variable notation to make this point: You don't have to do this. Choose whichever notation you find convenient and comfortable for the task at hand.

If we set `gtext[(]x text[)]=x^3,` and `htext[(]x text[)]=e^(4x),` then `g' text[(]x text[)]=3x^2` and `h' text[(]x text[)]=4e^(4x)`. Thus,

f ′ ( x )	= g ( x ) h ′ ( x ) + h ( x ) g ′ ( x )
	$= x^{3} (4 e^{4 x}) + e^{4 x} (3 x^{2})$
	$= (4 x^{3} + 3 x^{2}) e^{4 x} .$

If we set `u=x^3,` `v=e^(4x),`and `w=u v`, then `(du)/(dx)=3x^2` and `(dv)/(dx)=4e^(4x)`. Thus,

$\frac{d w}{d x}$	$= u \frac{d v}{d x} + v \frac{d u}{d x}$
	$= x^{3} (4 e^{4 x}) + e^{4 x} (3 x^{2})$
	$= (4 x^{3} + 3 x^{2}) e^{4 x} .$

In subsequent examples, we will use whichever notation we prefer — or mix them if the occasion calls for it.

Checkpoint 1

Contents for Chapter 4

Chapter 4 Differential Calculus and Its Uses

4.4 The Product Rule

Chapter 4
Differential Calculus and Its Uses