How to Show Partial Derivative ∂z/∂x for Implicit Functions?

[brendan]

Homework Statement

Assume the F(x,y,z) = 0 defines z implicitly as a function of x anf y. Show that

Homework Equations

∂z/∂x = -(∂F/∂x)/(∂F/∂z)

The Attempt at a Solution

I know this question is asking about the Implicit function theorem

So I start with F(x,y,z) =0

define it for z = F(x,y)

gives F(x,y,f(x,y))=0.

My problem is where to start to show that F(x,y,f(x,y)=0 shows ∂z/∂x = -(∂F/∂x)/(∂F/∂z)

How do I show the partial derivative ∂z/∂x of F(x,y,f(x,y)) ?
regards
Brendan

 

[Mark44]

Homework Statement

Assume the F(x,y,z) = 0 defines z implicitly as a function of x anf y. Show that

Homework Equations

∂z/∂x = -(∂F/∂x)/(∂F/∂z)

The Attempt at a Solution

I know this question is asking about the Implicit function theorem

So I start with F(x,y,z) =0

define it for z = F(x,y)

gives F(x,y,f(x,y))=0.

My problem is where to start to show that F(x,y,f(x,y)=0 shows ∂z/∂x = -(∂F/∂x)/(∂F/∂z)

How do I show the partial derivative ∂z/∂x of F(x,y,f(x,y)) ?
regards
Brendan

F(x, y, f(x, y)) = 0, so, taking the partial with respect to x of both sides, and using the chain rule, you get:
$$\frac{\partial F}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0$$

After you simplify the left side above ($\partial x/\partial x$ is 1, and since x and y are independent variables in this problem, $\partial y/\partial x$ is 0), it's easy to show what you're asked to show.