8—Multivariable Calculus 213Sometimes you see the chain rule written in a slightly different form. You can change coordinates from
(x,y)to(r,θ), 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 forx,yin terms ofr,θor
x′,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.
Suppose that I knowf(x,y)and I want to find the derivative offwith respect tox′holdingy′fixed. The
equation ( 5 ) tells you the differentialdf in terms of the coordinatesdxanddy. All that I have to do now is to
compute these in terms ofdx′anddy′. The coordinatexis a function ofx′andy′, so apply the same Eq. ( 5 )
to them.
dx=(
∂x
∂x′)
y′dx′+(
∂x
∂y′)
x′dy′ and dy=(
∂y
∂x′)
y′dx′+(
∂y
∂y′)
x′dy′I want the derivative off with respect tox′holdingy′constant, so that means that I’m going to computedx
anddygiven thatdy′is zero, then substitute the values ofdxanddyinto the Eq. ( 5 ) fordf. The second term
in each of the preceding equations is zero becausedy′= 0.
df=df(x,y,dx,dy) =(
∂f
∂x)
ydx+(
∂f
∂y)
xdy ( 5 )df=(
∂f
∂x)
y(
∂x
∂x′)
y′dx′+(
∂f
∂y)
x(
∂y
∂x′)
y′dx′ (8)and the derivative that I want,(∂f/∂x′)y′is the coefficient ofdx′in this equation.
Example: When you switch from rectangular to plane polar coordinates what is∂f/∂θin terms of thex
andyderivatives?
x=rcosθ, y=rsinθ, so
(
∂f
∂θ
)
r=
(
∂f
∂x)
y(
∂x
∂θ)
r+
(
∂f
∂y)
x(
∂y
∂θ)
r=(
∂f
∂x)
y(−rsinθ) +(
∂f
∂y)
x(rcosθ)Iff(x,y) =x^2 +y^2 this better be zero, because I’m finding howfchanges whenris held fixed. Check it out;
it is.