# Partial Derivatives Using Chain Rule

1. Mar 24, 2016

### Amrator

1. The problem statement, all variables and given/known data
Suppose ω = g(u,v) is a differentiable function of u = x/y and v = z/y.
Using the chain rule evaluate $$x \frac{\partial ω}{\partial x} + y \frac {\partial ω}{\partial y} + z \frac {\partial ω}{\partial z}$$

2. Relevant equations

3. The attempt at a solution
u = f(x,y)
v = h(y,z)
∂ω/∂x = (∂ω/∂u)(∂u/∂x)
∂ω/∂y = (∂ω/∂u)(∂u/∂y)
∂ω/∂z = (∂ω/∂v)(∂v/∂z)

Is this correct so far? Because I don't know how I would take the partials w.r.t. u and v.

2. Mar 24, 2016

### BvU

Yes. And $\ {\partial \omega \over \partial u} = \ {\partial g \over \partial u}\$ etc. Which you don't have to work out any further (because you don't know what g is). But you do know what u and v are, so those have to be evaluated.

3. Mar 24, 2016

### andrewkirk

A way to write it to make it easier is to define function $\omega:\mathbb{R}^3\to \mathbb{R}$ to be the composition of $g$ with $f$ by $\omega(x,y,z)=g(f(x,y),f(y,z))$. Note that we don't need $h$ because $f$ and $h$ are the same function, which just returns the quotient of its first argument to its second.

Then the chain rule for $x$ is

$$\frac{\partial \omega}{\partial x}=\frac{\partial g}{\partial u}(u,v)\bigg|_{(u,v)=(f(x,y),f(y,z))}\frac{\partial f}{\partial x}(x,y) +\frac{\partial g}{\partial v}(u,v)\bigg|_{(u,v)=(f(x,y),f(y,z))}\frac{\partial f}{\partial x}(y,z)$$

and similarly for the other two. In some cases, one of the two terms will be zero, for instance $\frac{\partial f}{\partial x}(y,z)=0$ because $x$ is not amongst the arguments to $f$ in that expression, but in some cases both terms may be nonzero.

4. Mar 25, 2016

### Amrator

How can f(x,y) and f(y,z) be the same function if they are functions of different variables?

5. Mar 25, 2016

### Staff: Mentor

The variables aren't important. It's what the function does with the variables that is important. For example g(t) = t2 + 1 is the same as g(u) = u2 + 1. In this example, g maps a number to the square of that number plus 1.

6. Mar 25, 2016

### Amrator

$x(∂ω/∂x)+y(∂ω/∂y)+z(∂ω/∂z)=x(∂g/∂u)(∂f/∂x)+y(∂g/∂u)(∂f/∂y)+y(∂g/∂v)(∂f/∂y)+z(∂g/∂v)(∂f/∂z) =x(1/y)(∂g/∂u)−y(x/y^2)(∂g/∂u)−y(z/y^2)(∂g/∂v)+z(1/y)(∂g/∂v)=u(∂g/∂u)-u(∂g/∂u)-v(∂g/∂v)+v(∂g/∂v)=0$

7. Mar 25, 2016

### andrewkirk

Looks right to me

8. Mar 25, 2016

### Amrator

Thanks, everyone.