# [resolved] Partial Derivative Relationships

1. Oct 26, 2014

### hotvette

I'm trying to come up with an expression for $\partial y / \partial x$ where $z = f(x,y)$. By observation (i.e. evaluating several sample functions), the following appears to be true:
$$\begin{equation*} \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} = 0 \end{equation*}$$
but I can't find anything to prove it. I can sort of get there using the differential formula
$$dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy$$
then dividing by $dx$ and reasoning that $dz/dx = 0$ because $z$ is constant when evaluating $\partial y / \partial x$, but I suspect this isn't a valid approach.

My questions are:

(1) Is the above relationship between $\partial z / \partial x$, $\partial z / \partial y$, and $\partial y / \partial x$ true?

(2) How to prove it?

P.S. Here is an example:
\begin{align} z = x^3y && y = zx^{-3} && \frac{\partial z}{\partial x} = 3x^2y && \frac{\partial z}{\partial y} = x^3 && \frac{\partial y}{\partial x} = -3zx^{-4} = -3x^3yx^{-4} = -3y/x \end{align}
$$\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} = 3x^2y + x^3 \cdot \left(-3y/x\right) = 3x^2y - 3x^2y = 0$$

Last edited: Oct 26, 2014
2. Oct 26, 2014

### Fredrik

Staff Emeritus
Let x,y,z be variables that satisfy a constraint of the form $z=f(x,y)$. For all x,y,z such that x and z have values such that this constraint can be solved for y, the variables also satisfy a constraint of the form $y=g(x,z)$, and we have $z=f(x,y)=f(x,g(x,z))$. What you have discovered is essentially just that the value of the right-hand side is independent of the value of x, as it must be since it's equal to z.

The only caveat is that the map $(x,z)\mapsto f(x,g(x,z))$ must be differentiable for your calculation to make sense. For this to be the case, f must be a nice enough function. In particular, it must be such that the domain of the map $(x,z)\mapsto f(x,g(x,z))$ isn't something ridiculous like a singleton set or a curve $z=C(x)$.

Let h be the function such that $h(x,z)=f(x,g(x,z))$ for all $x,z$ such that the original constraint can be solved for y. For all $x,z$ such that h is differentiable, we have
$$0=D_1h(x,z)=D_1 f(x,g(x,z))+D_2 f(x,g(x,z)) D_1g(x,z) =\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}.$$

3. Oct 26, 2014

### hotvette

Makes perfect sense when viewed as a composite function. I'm surprised this relation isn't mentioned in my advanced calculus book. Doesn't seem to me to be such an obscure item.

It came up because I'm trying to find out where $\frac{\partial y}{\partial x} =0$ which in my case is very difficult to evaluate whereas $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ are straightforward. I discovered that I could just find where $\frac{\partial z}{\partial y} = 0$ but I didn't know why it worked. Now I do. Thanks.