8—Multivariable Calculus 181In the first factor of the first term,∆f/∆x, the variablexis changed butyis not. In the first factor
of the second term, the reverse holds true. The limit of this expression is then
lim
∆t→ 0∆f
∆t
=
df
dt
=
(
∂f
∂x
)
ydx
dt
+
(
∂f
∂y
)
xdy
dt
(8.4)
If these manipulations look familiar, it’s probably because they mimic the procedures of section1.5.
That case is like this one, with the special valuesx≡y≡t.
Example: (When you want to check out an equation, you should construct an example so that
it reveals a lot of structure without requiring a lot of calculation.)
f(x,y) =Axy^2 , and x(t) =Ct^3 , y(t) =Dt^2
First do it using the chain rule.
df
dt
=
(
∂f
∂x
)
ydx
dt
+
(
∂f
∂y
)
xdy
dt
=
(
Ay^2
)(
3 Ct^2
)
+
(
2 Axy
)(
2 Dt
)
=
(
A(Dt^2 )^2
)(
3 Ct^2
)
+
(
2 A(Ct^3 )(Dt^2 )
)(
2 Dt
)
= 7ACD^2 t^6
Now repeat the calculation by first substituting the values ofxandyand then differentiating.
df
dt
=
d
dt
[
A(Ct^3 )(Dt^2 )^2
]
=
d
dt
[
ACD^2 t^7
]
= 7ACD^2 t^6
What iffalso has an explicittin it:f
(
t,x(t),y(t)
)
? That simply adds another term. Remem-ber,dt/dt= 1.
df
dt
=
(
∂f
∂t
)
x,y+
(
∂f
∂x
)
y,tdx
dt
+
(
∂f
∂y
)
x,tdy
dt
(8.5)
Sometimes you see the chain rule written in a slightly different form. You can change coordinatesfrom(x,y)to(r,φ), switching from rectangular to polar. You can switch from(x,y)to a system such
as(x′,y′) = (x+y,x−y). The function can be expressed in the new coordinates explicitly. Solve for
x,yin terms ofr,φorx′,y′and then differentiate with respect to the new coordinate. OR you can
use the chain rule to differentiate with respect to the new variable.
(∂f
∂x′
)
y′=
(
∂f
∂x
)
y(
∂x
∂x′
)
y′+
(
∂f
∂y
)
x(
∂y
∂x′
)
y′ (8.6)
This is actually not a different equation from Eq. (8.4). It only looks different because in addition tot
there’s another variable that you have to keep constant:t→x′, andy′is constant.
Example: When you switch from rectangular to plane polar coordinates what is∂f/∂φin terms
of thexandyderivatives?
x=rcosφ, y=rsinφ, so
(
∂f
∂φ
)
r=
(
∂f
∂x
)
y(
∂x
∂φ
)
r+
(
∂f
∂y
)
x(
∂y
∂φ
)
r=(
∂f
∂x
)
y(−rsinφ) +
(
∂f
∂y
)
x