5128_Ch03_pp098-184
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
Section 3.6 Chain Rule 151<br />
Finding dy/dx Parametrically<br />
If all three derivatives exist and dxdt 0,<br />
d y dydt<br />
. (3)<br />
dx<br />
d xdt<br />
2<br />
1<br />
0<br />
y<br />
<br />
t –4<br />
1<br />
(1 – 2, 1)<br />
x sec t, y tan t,<br />
– –2<br />
<br />
t – 2<br />
Figure 3.43 The hyperbola branch in<br />
Example 6. Equation 3 applies for every<br />
point on the graph except 1, 0). Can you<br />
state why Equation 3 fails at 1, 0)?<br />
2<br />
x<br />
EXAMPLE 6<br />
Differentiating with a Parameter<br />
Find the line tangent to the right-hand hyperbola branch defined parametrically by<br />
x sec t, y tan t, p 2 t p 2 <br />
at the point 2, 1, where t p4 (Figure 3.43).<br />
SOLUTION<br />
All three of the derivatives in Equation 3 exist and dxdt sec t tan t 0 at the indicated<br />
point. Therefore, Equation 3 applies and<br />
d y dydt<br />
<br />
dx<br />
d xdt<br />
sec2<br />
t<br />
<br />
se c t tan t<br />
s ec<br />
t<br />
<br />
tan<br />
t<br />
Setting t p4 gives<br />
csc t.<br />
d y<br />
<br />
dx<br />
|tp4<br />
csc p4 2.<br />
The equation of the tangent line is<br />
y 1 2x 2<br />
y 2x 2 1<br />
y 2x 1. Now try Exercise 41.<br />
Power Chain Rule<br />
If f is a differentiable function of u, and u is a differentiable function of x, then substituting<br />
y f u into the Chain Rule formula<br />
d y d<br />
<br />
dx<br />
y<br />
d u<br />
• du<br />
d <br />
x<br />
leads to the formula<br />
d<br />
f u f u d u<br />
.<br />
d x<br />
dx<br />
Here’s an example of how it works: If n is an integer and f u u n , the Power Rules<br />
(Rules 2 and 7) tell us that f u nu n1 . If u is a differentiable function of x, then we can<br />
use the Chain Rule to extend this to the Power Chain Rule:<br />
d<br />
u d x<br />
n nu n1 d u d<br />
. u dx<br />
d u<br />
n nu n1