4.4. THE USUAL FORM OF THE CHAIN RULE 95
Then from the definition of partial derivatives,
limh→0
f (h,0)− f (0,0)h
= limh→0
0−0h
= 0
and
limh→0
f (0,h)− f (0,0)h
= limh→0
0−0h
= 0
However f is not even continuous at (0,0) which may be seen by considering the behaviorof the function along the line y = x and along the line x = 0. By Lemma 4.1.4 this impliesf is not differentiable. Therefore, it is necessary to consider the correct definition of thederivative given above if you want to get a notion which generalizes the concept of thederivative of a function of one variable in such a way as to preserve continuity wheneverthe function is differentiable.
4.4 The Usual Form of the Chain RuleLet z≡ g(y) and let y = f(x). Assuming Dg(f(x)) exists and Df(x) both exist and then wehave x ∈U ⊆Rn and y ∈V ⊆Rm where U,V are open sets and f(V )⊆V with g : V →Rp.What is the matrix of g◦ f(x) = g(y)? Say g has values in Rp. From the chain rule above,and the description of the matrix of the derivative in Theorem 4.3.1,(
∂z∂x1
∂z∂x2
· · · ∂z∂xn
)= (4.5)
(∂z∂y1
∂z∂y2
· · · ∂z∂ym
)(∂y∂x1
∂y∂x2
· · · ∂y∂xn
)Now from the way we multiply matrices, to find the i jth entry of the matrix on the right,one multiplies the ith row of the left matrix with the jth column of the matrix on the right.Thus the i jth entry of the matrix on the right is
∑k
∂ zi
∂yk
∂yk
∂x j
and by the chain rule, Theorem 4.2.1, this equals the i jth entry of the matrix of 4.5. That is,
∂ zi
∂x j= ∑
k
∂ zi
∂yk
∂yk
∂x j
This is stated as the following proposition.
Proposition 4.4.1 Let z≡ g◦ f(x) and let y≡ f(x) , then assuming Df(x) exists andDg(f(x)) exists, then ∂ zi
∂x j= ∑k
∂ zi(y)∂yk
∂yk∂x j
assuming that all functions make sense. That isf : U→ f(U)⊆V and g : V →Rp for U,V open sets in Rn and Rm respectively. Also, sincethis holds for each i, ∂z
∂x j= ∑k
∂z∂yk
∂yk∂x j
.
Some people like to dispense with the summation sign and write instead ∂z∂x j
= ∂z∂yk
∂yk∂x j
where it is understood that summation takes place on the repeated index.