5128_Ch03_pp098-184

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