6.9. THE DERIVATIVE AND THE CARTESIAN PRODUCT 155

Definition 6.9.3 Let g : U ⊆∏ni=1 Xi→ Y , where U is an open set. Then the map

z→ g (x1, · · · ,xi−1,z,xi+1, · · · ,xn)

is a function from the open set in Xi,

{z : x= (x1, · · · ,xi−1,z,xi+1, · · · ,xn) ∈U}

to Y . When this map is differentiable, its derivative is denoted by Dig (x). To aid in thenotation, for v ∈ Xi, let θ iv ∈∏

ni=1 Xi be the vector (0, · · · ,v, · · · ,0) where the v is in the

ith slot and for v ∈∏ni=1 Xi, let vi denote the entry in the ith slot of v. Thus, by saying

z→ g (x1, · · · ,xi−1,z,xi+1, · · · ,xn)

is differentiable is meant that for v ∈ Xi sufficiently small,

g (x+θ iv)−g (x) = Dig (x)v+o(v) .

Note Dig (x) ∈L (Xi,Y ) .

As discussed above, we have the following definition of C1 (U) .

Definition 6.9.4 Let U ⊆ X be an open set. Then f : U →Y is C1 (U) if f is differ-entiable and the mapping x→ Df (x) , is continuous as a function from U to L (X ,Y ).

With this definition of partial derivatives, here is the major theorem. Note the resem-blance with the matrix of the derivative of a function having values in Rm in terms of thepartial derivatives.

Theorem 6.9.5 Let g,U,∏ni=1 Xi, be given as in Definition 6.9.3. Then g is C1 (U)

if and only if Dig exists and is continuous on U for each i. In this case, g is differentiableand

Dg (x)(v) = ∑k

Dkg (x)vk (6.14)

where v = (v1, · · · ,vn) .

Proof: Suppose then that Dig exists and is continuous for each i. Note ∑kj=1 θ jv j =

(v1, · · · ,vk,0, · · · ,0) . Thus ∑nj=1 θ jv j = v and define ∑

0j=1 θ jv j ≡ 0. Therefore,

g (x+v)−g (x) =n

∑k=1

[g

(x+

k

∑j=1

θ jv j

)−g

(x+

k−1

∑j=1

θ jv j

)](6.15)

=n

∑k=1

[(g

(x+

k

∑j=1

θ jv j

)−g (x+θ kvk)

)−

(g

(x+

k−1

∑j=1

θ jv j

)−g (x)

)]

+n

∑k=1

(g (x+θ kvk)−g (x))

If hk (x) ≡ g(x+∑

k−1j=1 θ jv j

)− g (x) then the top sum is ∑

nk=1hk (x+θ kvk)−hk (x)

and from the definition of hk, ∥Dhk (x)∥ < ε a sufficiently small ball containing x. Thus