# How do you find dz/dx and dz/dy of ln(xy+yz+xz) = 5 if x, y, and z are all positive?

The result is: $\frac{\partial z}{\partial x} = - y , \frac{\partial z}{\partial y} = - x$.

First, it's important to note that $z$ is an implicit function of the variables $x$ and $y$. So the notation used to describe derivatives of $z$ with respect to $x$ and $y$ should be the notation of partial derivatives: $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. The operational rules for partial derivatives are identical to the rules concerning total derivatives.

Also, since we have the information that all variables are always positive, the expression is well defined (recall that logarithms are only defined for positive numbers).

Now, taking the partial derivative with respect to $x$ of the expression given (we apply the operator to both sides of the equation), we get:

$\frac{\partial}{\partial x} \left[\ln \left(x y + y z + x z\right)\right] = \frac{\partial}{\partial x} 5$

Applying the chain rule to the left side of the equation, we have:

$\frac{1}{x y + y z + x z} \left(y + \frac{\partial z}{\partial x}\right) = 0 \iff y + \frac{\partial z}{\partial x} = 0 \iff$
$\iff \frac{\partial z}{\partial x} = - y$

Doing the same, now with respect to y:

$\frac{\partial}{\partial y} \left[\ln \left(x y + y z + x z\right)\right] = \frac{\partial}{\partial y} 5 \iff$
$\iff \frac{1}{x y + y z + x z} \left(x + \frac{\partial z}{\partial y}\right) = 0 \iff x + \frac{\partial z}{\partial y} = 0 \iff$
$\iff \frac{\partial z}{\partial y} = - x$