# Problem - Implicit function theorem

1. Jun 7, 2015

### ORF

1. The problem statement, all variables and given/known data

I have 3 functions which define 3 constraints:
A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

I'm asked to calculate the partial of a function h, with respect to the variable w; with h=h(x,y,z).

2. Relevant equations
A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

3. The attempt at a solution

a. With B, make y=z in all of the equations:
A(x,y)=0 (no lineal)
C(w,x,y)=w-(x+2y)=0
h=h(x,y,y)=h(x,y)

b. The subindex point out the variable of the partial; apply the chain rule
h_w = h_x * x_w + h_y * y_w,

c. For the partial derivatives x_w and y_w we can use the implicit function theorem:

x_w = - C_w / C_x = 1,
y_w = - C_w / C_y = 1/2
with C_w = \frac{\partial C}{\partial w}, etc

Doubt: if I make y=z at first the result if different from the case I make y=z at the end. What case is the correct?

Simplied example:
h(x,y)=f(x)+g(y); the constraint is x=y; and w=x+y.

a. Making x=y at the end
h_w = f_x * x_w + g_y * y_w = f(x)_x + g(y)_y
x=y,
h_w = f(x)_x + g(x)_x

b. Making x=y at first. The variable w=2x, h(x,x) = f(x) + g(x)
h_w = f_x * x_w + g_x * x_w
x_w = 1/2
h_w = 0.5 f(x)_x + 0.5 g(x)_x

Which is the correct?

Greetings!

2. Jun 7, 2015

### Ray Vickson

You can write three equations:
$$A_x dx + A_y dy + A_z dz = 0\\ dy - dz = 0\\ dw - dx - dy - dz = 0$$
Solving for $dx, dy, dz$ in terms of $dw$ allows you to express $dh = h_x dx + h_y dy + h_z dz$ solely in terms of $dw$.

If you don't like using $dv$ on a variable $v$, think instead of a finite, but small, increment $\Delta v$ and first-order Taylor expansions in terms of $\Delta x, \, \Delta y, \, \Delta z, \, \Delta w$. You want the first-order Taylor expansion of $h(x,y,z)$ in terms of $\Delta w$.

Solvability of the linear system above requires a condition on $A(x,y,z)$; that is the same condition needed for applicability of the implicit function theorem to the multivariate linear system $F_i(x,y,z,w) = 0, i = 1,2,3$.

Last edited: Jun 7, 2015