6.16. IMPLICIT FUNCTION THEOREM 127

Using the standard rules of differentiation, for (x,y) ̸= (0,0) ,

fx = yx4− y4 +4x2y2

(x2 + y2)2 , fy = xx4− y4−4x2y2

(x2 + y2)2

Now

fxy (0,0) ≡ limy→0

fx (0,y)− fx (0,0)y

= limy→0

−y4

(y2)2 =−1

while

fyx (0,0) ≡ limx→0

fy (x,0)− fy (0,0)x

= limx→0

x4

(x2)2 = 1

showing that although the mixed partial derivatives do exist at (0,0) , they are not equalthere.

6.16 Implicit Function TheoremThe implicit function theorem is one of the greatest theorems in mathematics. There aremany versions of this theorem. However, I will give a very simple proof valid in finitedimensional spaces.

Theorem 6.16.1 (implicit function theorem) Suppose U is an open set in Rn×Rm. Letf : U → Rn be in C1 (U) and suppose

f(x0,y0) = 0, D1f(x0,y0)−1 ∈L (Rn,Rn) . (6.16.26)

Then there exist positive constants, δ ,η , such that for every y ∈ B(y0,η) there exists aunique x(y) ∈ B(x0,δ ) such that

f(x(y) ,y) = 0. (6.16.27)

Furthermore, the mapping, y→ x(y) is in C1 (B(y0,η)).

Proof: Let

f(x,y) =

f1 (x,y)f2 (x,y)

...fn (x,y)

 .