Implict Function Theorem
Eugene M. Cliff
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Motivation
Implicit Function Theorems are among the most useful tools in advanced
calculus. In the simplest setting suppose we have a smooth real-valued function of two variables, say f (x, y). We have a point in the plane such that
f (xo , yo ) = 0 and we want to ’solve’ this for y as a function of x. Can this
be done ? That is, is there a real-valued function of a single real-variable,
say h(x), so that yo = h(xo ) and so that f (x, h(x)) ≡ 0, at least around the
value xo ? The implicit function theorem gives us this and more. In fact it
tells us how to compute the derivative of h. Indeed, we can do this in terms
of various derivatives of f without ever actually getting our hands on h.
We can anticipate what is needed by proceeding formally with a threeterm Taylor expansion for f
f (xo + δx, yo + δy ) = f (xo , yo ) +
∂f
∂f
δx +
δy
∂x
∂y
The hypothesis is that the first term on the right is zero and for a given δx
we want to find the appropriate δy to maintain f ≡ 0. The formula suggests
that
∂f
∂f
δy = −(
δx)/
,
∂x
∂y
and we expect to encounter trouble when the denominator term vanishes
( ∂∂yf = 0). Moreover, if we hypothesize that this denominator term is nonzero then further formal manipulations suggest that
(δy/δx) = −(
∂f ∂f
)/
.
∂x ∂y
This ought to be the derivative of the implicitly defined function h. It is.
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A few examples are in order:
Example 1
f (x, y ) = x2 + y 2 ,
xo = y o = 0
In this case we compute ∂∂yf = 0 and we expect trouble. In fact, the origin is
an isolated zero for the function f and there are no neighboring points with
f (x, y ) = 0. No explicit function h exists.
Example 2
f (x, y ) = −x + sin y ;
xo = 1, yo = π/2
In this case we compute ∂∂yf = 0 and we expect trouble. In fact, for x values
slightly larger than xo there is no (real) y value to make f = 0, while for x
values slightly smaller than xo there are two (real) y value. Again, no explicit
function h exists.
Example 3
f (x, y ) = −x + tan y ;
xo = 1, yo = π/4
In this case we compute ∂∂yf = sec2 (π/4) = 2 and we expect no trouble. In
addition, we have ∂∂xf = −1. so that we expect that the implicit function h
exists and that its derivative (at the point x0 is
h0(xo ) = −(
∂f ∂f
)/
= 1/ sec2 (π/4) = cos2(π/4) = 1/2.
∂x ∂y
In this case the required function h is the familiar arc-tangent ( y = tan −1( x)).
This trick (implicit differentiation) can be used to figure out the derivative
of the arc-tan function (if you remember what the derivative of tan is).
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An Implicit Function Theorem
We are given a mapping f : D ⊂ Rn × Rm 7→ ×Rm that is C 1 on the region
D , and a point (x0 , y0) ∈ D such that f (xo , yo) = 0 ∈ Rm. Suppose that the
(m × m) Jacobian matrix
∂f1
∂y
∂y1
∂f1
∂y2
∂f2
∂y2
...
...
..
.
∂f1
∂ym
∂f2
ym
∂fm
∂y1
∂fm
∂y2
...
∂fm
∂ym
" ∂f21
..
.
..
.
..
.
#
|(x0 , y0 )
is non-singular. Then there are neighborhoods Sx of x0 and Sy of y0, such
that for any x
ˆ ∈ Sx the equation f (ˆ
x, y) = 0 has a unique solution y =
h(ˆ)
x ∈ Sy . Moreover, the mapping h : Sx 7→ Rm is differentiable with
∂h
∂f
=−
∂x
∂y
!−1
!
∂f
·
.
∂x
References
[1] Iterative Solution of Nonlinear Equations in Several Variables, Ortega,
J.M. and Rheinboldt, W.C., Academic Press, 1970.
[2] Advanced Calculus, Taylor, A.E. and Mann, W.R., Wilely, 1972
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