partial differentation of implicit function I have a problem with this:
$f(x,y,z)=\exp(xyz)$ with $g(1,1)=\ln2$ and $f(x,y,g(x,y))=2$.
The task is to calculate the partial derivatives $\frac{\partial g}{\partial x}(1,1)$ and $\frac{\partial g}{\partial y}(1,1)$.
I've already found an equation $\frac{\partial g}{\partial x} = -\frac{\partial f / \partial x}{\partial f/ \partial g}$ which should work, but how do I calculate the denominator of the right side? I don't have any formula for $g$ given. Is it a directional derivate?
 A: The denominator is the derivative of $f$ with respect to $z$, as you can see in http://en.wikipedia.org/wiki/Implicit_function_theorem#Regularity
In your case, if $y \neq 0$,
$$\frac{\partial g}{\partial x} = -\frac{yz \exp(xyz)}{xy \exp(xyz)}.$$
Rewriting so this is not implicit:
$$\frac{\partial g}{\partial x} = -\frac{g(x,y)}{x}.$$
A: Let $\phi(x,y) =f(x,y,g(x,y))$. If $\phi(x,y) = 2$, then
${\partial \phi(x,y) \over \partial x} = {\partial f(x,y,g(x,y)) \over \partial x}+ {\partial f(x,y,g(x,y)) \over \partial z} {\partial g(x,y) \over \partial x} =0$.
This gives $2 \ln 2 + 2 {\partial g(x,y) \over \partial x} = 0$ which gives ${\partial \phi(x,y) \over \partial x} = -\ln 2$.
A: Let $h(x,y,z)=f(x,y,z)-2=e^{xyz}-2$, then $h(1,1,\ln 2)=0, h_z=\dfrac{\partial }{\partial z}h(1,1,\ln 2)=2\ne 0.$ By implicit function theorem, we may write $z=g(x,y)$ for $(x,y)$ near $(1,1)$, and $g_x=-\dfrac{h_x}{h_z}=-\dfrac{z}{x}, g_y=-\dfrac{h_y}{h_z}=-\dfrac{z}{y}$. Substitute in the values $x=1,y=1,z=\ln 2$, we have $g_x(1,1)=g_y(1,1)=-\ln 2$.
