168 3 Real Analysis
means that its graph (which is a surface inR^3 ) admits a tangent plane at each point. For a
three-variable function, the graph is a three-dimensional manifold in a four-dimensional
space, and differentiability means that at each point the graph admits a three-dimensional
tangent hyperplane.
The tilting of the tangent (hyper)plane is determined by the slopes in the directions
of the coordinate axes, and these slopes are the partial derivatives of the function. We
denote the partial derivatives offby∂f∂x,∂f∂y,∂f∂z. They are computed by differentiating
with respect to the one variable while keeping the others fixed. This being said, let us
start with the examples.
Euler’s theorem.A functionz(x, y)is calledn-homogeneous ifz(tx, ty)=tnz(x, y)
for allx, y∈Randt> 0. Assume thatz(x, y)isn-homogeneous withnan integer.
Then for allk≤n+ 1 ,
∑kj= 1(
k
j)
xjyk−j∂kz
∂xj∂yk−j=n(n− 1 )···(n−k+ 1 )z.Proof.We first prove the casek=1. Differentiating the relationz(tx, ty)=tnz(x, y)
with respect toy, we obtain
t∂z
∂y(tx, ty)=tn∂z
∂y(x, y),which shows that∂z∂yis(n− 1 )-homogeneous.
Replacexby 1,ybyyx, andtbyxin the homogeneity condition, to obtainz(x, y)=
xnz( 1 ,yx). Differentiating this with respect toxyields
∂z
∂x
(x, y)=nxn−^1 z(
1 ,
y
x)
+xn∂z
∂y(
1 ,
y
x)
·
(
−
y
x^2)
.
Because∂y∂zis(n− 1 )-homogeneous, the last term is just−yx∂z∂y(x, y). Moving it to the
right and multiplying through byxgives the desired
x∂z
∂x
+y∂z
∂y
=nz.Now we prove the general case by induction onk, withk=1 the base case. To simplify
the notation, set
(k
j)
=0ifj<0orj>k. The induction hypothesis is∑j(
k
j)
xjyk−j
∂kz
∂xj∂yk−j=n(n− 1 )···(n−k+ 1 )z,