PARTIAL DIFFERENTIATION
Thus, from (5.17), we may write
∂
∂x=cosφ∂
∂ρ−
sinφ
ρ∂
∂φ,
∂
∂y=sinφ∂
∂ρ+
cosφ
ρ∂
∂φ.
Now it is only a matter of writing
∂^2 f
∂x^2=
∂
∂x(
∂f
∂x)
=
∂
∂x(
∂
∂x)
f=
(
cosφ∂
∂ρ−
sinφ
ρ∂
∂φ)(
cosφ∂
∂ρ−
sinφ
ρ∂
∂φ)
g=
(
cosφ∂
∂ρ−
sinφ
ρ∂
∂φ)(
cosφ∂g
∂ρ−
sinφ
ρ∂g
∂φ)
=cos^2 φ∂^2 g
∂ρ^2+
2cosφsinφ
ρ^2∂g
∂φ−
2cosφsinφ
ρ∂^2 g
∂φ∂ρ+sin^2 φ
ρ∂g
∂ρ+
sin^2 φ
ρ^2∂^2 g
∂φ^2and a similar expression for∂^2 f/∂y^2 ,
∂^2 f
∂y^2=
(
sinφ∂
∂ρ+
cosφ
ρ∂
∂φ)(
sinφ∂
∂ρ+
cosφ
ρ∂
∂φ)
g=sin^2 φ∂^2 g
∂ρ^2−
2cosφsinφ
ρ^2∂g
∂φ+
2cosφsinφ
ρ∂^2 g
∂φ∂ρ+cos^2 φ
ρ∂g
∂ρ+
cos^2 φ
ρ^2∂^2 g
∂φ^2.
When these two expressions are added together the change of variables is complete and
we obtain
∂^2 f
∂x^2+
∂^2 f
∂y^2=
∂^2 g
∂ρ^2+
1
ρ∂g
∂ρ+
1
ρ^2∂^2 g
∂φ^2.
5.7 Taylor’s theorem for many-variable functionsWe have already introduced Taylor’s theorem for a functionf(x) of one variable,
in section 4.6. In an analogous way, the Taylor expansion of a functionf(x, y)of
two variables is given by
f(x, y)=f(x 0 ,y 0 )+∂f
∂x∆x+∂f
∂y∆y+1
2![
∂^2 f
∂x^2(∆x)^2 +2∂^2 f
∂x∂y∆x∆y+∂^2 f
∂y^2(∆y)^2]
+···, (5.18)where ∆x=x−x 0 and ∆y=y−y 0 , and all the derivatives are to be evaluated
at (x 0 ,y 0 ).