If $z=g(x, y)$ is defined by $f(x, y, z)=0$ near the point $(a, b, c)$, find $\frac{\partial g}{\partial x} (a, b)$.

Question

Problem: Suppose $f=f(x, y, z): \mathbb{R}^3\to\mathbb{R}$ is continuously differentiable, $f(a, b, c)=0$ and $\frac{\partial f}{\partial z}(a, b, c)\neq0$. If $z=g(x, y)$ is defined by $f(x, y, z)=0$ near the point $(a, b, c)$, Show that $\frac{\partial g}{\partial x}(a, b)=-\frac{\frac{\partial f}{\partial x}(a, b, c)}{\frac{\partial f}{\partial z}(a, b, c)}$.

My Question: I don't believe the Chain Rule can be of any use here. By the Implicit Function Theorem, the $g(x, y)$ is indeed differentiable; but then how exactly do I establish the connection between $\frac{\partial g}{\partial x}$ and $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial z}$?

Any HINT would be greatly appreciated.

Answer 1

Define $\hat g: N_r(a, b)\to\mathbb{R}^3$ by $\hat g(x, y)=(x, y, g(x, y))$, $h:N_r(a, b)\to\mathbb{R}$ by $h=f\circ \hat g$, where $N_r(a, b)$ is the neighborhood of $(a, b)$ such that $g$ exists and is differentiable.

Note $\hat g$ and $f$ are both differentiable on $N_r(a, b)$, thus $h$ is differentiable on $N_r(a, b)$. Now apply the chain rule.

Answer 2

Let us define $$ f(x, y, z) = z - g(x, y) $$

Then $$ f(x, y, z) = 0 \iff z = g(x, y) $$

We find that $$ {\partial f \over \partial x}(x,y,z) = - {\partial g \over \partial x}(x,y) $$ $$ {\partial f \over \partial y}(x,y,z) = - {\partial g \over \partial y}(x,y) $$ $$ {\partial f \over \partial z}(x,y,z) = 1 $$

Hence, it follows that $$ {\partial f \over \partial x}(a, b, c) = - {\partial g \over \partial x}(a, b) $$ $$ {\partial f \over \partial y}(a, b, c) = - {\partial g \over \partial y}(a, b) $$ $$ {\partial f \over \partial z}(a, b, c) = 1 $$

We note that $$ -{ {\partial f \over \partial x}(a, b, c) \over {\partial f \over \partial z}(a, b, c)} = -{- {\partial g \over \partial x}(a, b) \over 1} = {\partial g \over \partial x}(a, b) $$

Hence we have established the assertion.