Partial Derivatives Using Chain Rule

[Amrator]

Homework Statement

Suppose ω = g(u,v) is a differentiable function of u = x/y and v = z/y.
Using the chain rule evaluate $$x \frac{\partial ω}{\partial x} + y \frac {\partial ω}{\partial y} + z \frac {\partial ω}{\partial z}$$

Homework Equations

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.

 

[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.

 

[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.

 

[Amrator]

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

 

[Mark44]

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

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.

 

[Amrator]

Is this the correct answer?

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

 

[andrewkirk]

Looks right to me

 

[Amrator]

Thanks, everyone.