5128_Ch03_pp098-184
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118 Chapter 3 Derivatives<br />
EXAMPLE 1 Differentiating a Polynomial<br />
Find d p<br />
if p t<br />
dt<br />
3 6t 2 5 3 t 16.<br />
SOLUTION<br />
By Rule 4 we can differentiate the polynomial term-by-term, applying Rules 1 through 3<br />
as we go.<br />
d p d<br />
t dt<br />
d t<br />
3 d<br />
6t d t<br />
2 d<br />
d<br />
( t<br />
5 3 ) t d<br />
16 Sum and Difference Rule<br />
dt<br />
3t 2 6 • 2t 5 3 0<br />
Constant and Power Rules<br />
3t 2 12t 5 3 Now try Exercise 5.<br />
EXAMPLE 2<br />
Finding Horizontal Tangents<br />
Does the curve y x 4 2x 2 2 have any horizontal tangents? If so, where?<br />
SOLUTION<br />
The horizontal tangents, if any, occur where the slope dydx is zero. To find these<br />
points, we<br />
(a) calculate dydx:<br />
d y d<br />
x dx<br />
d x<br />
4 2x 2 2 4x 3 4x.<br />
(b) solve the equation dydx 0 for x:<br />
4x 3 4x 0<br />
4xx 2 1 0<br />
x 0, 1, 1.<br />
The curve has horizontal tangents at x 0, 1, and 1. The corresponding points on the<br />
curve (found from the equation y x 4 2x 2 2) are 0, 2, 1, 1, and 1, 1. You<br />
might wish to graph the curve to see where the horizontal tangents go.<br />
Now try Exercise 7.<br />
The derivative in Example 2 was easily factored, making an algebraic solution of the<br />
equation dydx 0 correspondingly simple. When a simple algebraic solution is not possible,<br />
the solutions to dydx 0 can still be found to a high degree of accuracy by using<br />
the SOLVE capability of your calculator.<br />
[–10, 10] by [–10, 10]<br />
Figure 3.18 The graph of<br />
y 0.2x 4 0.7x 3 2x 2 5x 4<br />
has three horizontal tangents. (Example 3)<br />
EXAMPLE 3<br />
Using Calculus and Calculator<br />
As can be seen in the viewing window 10, 10 by 10, 10, the graph of<br />
y 0.2x 4 0.7x 3 2x 2 5x 4 has three horizontal tangents (Figure 3.18).<br />
At what points do these horizontal tangents occur?<br />
continued