724 CHAPTER 22. THE DERIVATIVE

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 ) .

Definition 22.8.4 Let U ⊆ X be an open set. Then f : U → Y is C1 (U) if f is differentiableand 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 22.8.5 Let g,U,∏ni=1 Xi, be given as in Definition 22.8.3. Then g is C1 (U) if and

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

Dg(x)(v) = ∑k

Dkg(x)vk (22.8.15)

where v = (v1, · · · ,vn) .

Proof: Suppose then that Dig exists and is continuous for each i. Note that

k

∑j=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

)](22.8.16)

Consider the terms in this sum.

g

(x+

k

∑j=1

θ jv j

)−g

(x+

k−1

∑j=1

θ jv j

)= g(x+θ kvk)−g(x)+ (22.8.17)

(g

(x+

k

∑j=1

θ jv j

)−g(x+θ kvk)

)−

(g

(x+

k−1

∑j=1

θ jv j

)−g(x)

)(22.8.18)