Multivariate Calculus - Bruce E. Shapiro

LECTURE 12. THE CHAIN RULE 83
The problem is not finished because the answer we have derived depends on x, y,
and t, and should only depend on t. To finish the problem, we need to substitute
the expressions we are given for x and y as a function of t, namely x = sin t and
y = t 2 . This leads to
dw
dt
= 4x 3 y 2 cos t + 4x 4 yt
= 4(sin t) 3 (t 2 ) 2 cos t + 4(sin t) 4 (t 2 )t
= 4t 4 sin 3 t cos t + 4t 2 sin 4 t
= 4t 2 sin 3 t(t 2 cos t + sin t)
Example 12.2 Find du/dr, as a function of r, using the chain rule for u =
z √ x + y, x = e 3r , y = 12r and z = ln r.
Solution.
du
dr
= ∂u dx
∂x
= ∂
∂x
=
=
=
=
dr + ∂u dy
∂y dr + ∂u dy
∂z dr
( √ ) d
z x + y
dr (e3r ) + ∂ ∂y
+ ∂ ( √ ) d
z x + y (ln r)
∂z
dr
[
z(1/2)(x + y) −1/2] [
e 3r (3) ] +
+ (√ x + y ) (1/r)
3ze 3r
2 √ x + y + 12z √ x + y
2 √ x + y + r
3ze 3r
2 √ x + y + √ 6z √ x + y
+ x + y r
(
z
√ x + y
) dy
dr (12r)
√
3 ln(r)e 3r
2 √ e 3r + 12r + 6 ln r e
√
e 3r + 12r + 3r + 12r
r
[
z(1/2)(x + y) −1/2] (12)
Example 12.3 Find dw/dt, as a function of t, using the chain rule, for w = xy +
yz + xz, x = t 2 , y = 1 − t 2 , and z = 1 − t.
Math 250, Fall 2006 Revised December 6, 2006.