How do I take the implicit partial derivative when the variable is not equal to the equation?

Question

So my homework problem is as follows: $$ \text{Consider the equation }xz^2-6yz+4log(z)=-1 \text{ as defining }z\text{ implicitly as a function of } x \text{ and } y \text{.}\\ \text{The values of } \frac{\partial z}{\partial x} \text{ and } \frac{\partial z}{\partial y} \text{ at } (5,1,1) \text{ are _____ and _____. }\\ \text{Remember, in webwork "log" means the natural logarithm.} $$

All the examples I've seen where we take the partial derivative and assume a variable is a function of the other variables, we have the whole equation equal to that variable. (eg $z = f(x, y) = x^2 + 7xy + 1$ or something else, this is an arbitrary example).

I thought that maybe I need to solve for $z$ first, and then I could take the implicit partial at that point. I couldn't see any obvious way to solve for $z$ algebraically, so I figured the quadratic equation was the route to go and started writing out:

$$ z = \frac{-6y \pm \sqrt((-6y)^2-4(x)(\text{(how do I get }c\text{ from }4log(z)\text{ here?}))}{(2x)} $$

So I'm thinking that overall, there's probably a simple rule or solution that I'm overlooking which is leading me on a wild goosechase.

What is the correct first step I need to take so I can perform the partial derivative?

EDIT: The comment from GFauxPas made me think that perhaps all I needed to do was use prime notation instead, but I'm still stuck.

$$ \frac{\partial z}{\partial x} = z^2+2xzz'-6yz'+\frac{4z'}{z} = 0 \\ \text{Solving for }z'\text{ gave me}\\ z'=\frac{z^3}{2xz^2-6yz+4} = \frac{1}{8} $$ When we plugin 5,1,1 that is, but that isn't showing as the correct answer. Am I at least on the right track?

EDIT: I missed a negative. This is the correct answer: $$ \frac{\partial{z}}{\partial{x}} = - \frac{1}{8} $$

Answer 1

The answer is to use $prime\text{ }notation$.

First we find $\frac{\partial z}{\partial x}$

$$ \begin{align} \frac{\partial z}{\partial x} = z^2+2xzz'-6yz' + \frac{4z'}{z} = 0 \\ 2xzz'-6yz' + \frac{4z'}{z} = -z^2 \\ \text{(multiply each term by }z\text{)}\\ 2xz^2z' -6yzz' + 4z'= -z^3\\ \frac{\partial z}{\partial x} = z' = \frac{-z^3}{2xz^2-6yz+4} \end{align} $$

Then we find $\frac{\partial z}{\partial y}$

$$ \begin{align} \frac{\partial z}{\partial y} = 2xzz'-6z-6yz'+\frac{4z'}{z} = 0 \\ 2xzz'-6yz'+\frac{4z'}{z} = 6z \\ \text{(multiply each term by }z\text{, again)}\\ 2xz^2z'-6yzz' +4z' = 6z^2 \\ \frac{\partial z}{\partial y} = z' = \frac{6z^2}{2xz^2-6yz+4} \end{align} $$

When we plug in our point $(5, 1, 1)$ we get the answer $$ \begin{align} \frac{\partial z}{\partial x} = \frac{-1}{8} \text{ and } \frac{\partial z}{\partial y} = \frac{3}{4} \end{align} $$

Answer 2

What you have to use is simply the total derivative and partial differentitation.

You have $$F=f(x,y,z)=0$$ The total derivative is $$dF=\frac{\partial F}{\partial x}\,dx+\frac{\partial F}{\partial y}\,dy+\frac{\partial F}{\partial z}\,dz=F'_x\,dx+F'_y\,dy+F'_z\,dz=0$$ from which, as usual $$\frac{\partial z}{\partial x}=-\frac {F'_x}{F'_z}$$ $$\frac{\partial z}{\partial y}=-\frac {F'_y}{F'_z}$$

In your case $$F=xz^2-6yz+4\log(z)+1=0$$ then $F'_x=z^2$, $F'_y=-6z$, $F'_z=2xz-6y+\frac 4z$. So, after simplifications, $$\frac{\partial z}{\partial x}=\frac{z^3}{-2 x z^2+6 y z-4} $$ $$\frac{\partial z}{\partial y}=\frac{3 z^2}{x z^2-3 y z+2}$$

So, for the point $(5, 1, 1)$ we obtain $$ \begin{align} \frac{\partial z}{\partial x} = -\frac{1}{8} ~~~~~~~~~~~~ \frac{\partial z}{\partial y} = \frac{3}{4} \end{align} $$