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Partial Derivatives Using Chain Rule

  1. Mar 24, 2016 #1
    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. jcsd
  3. Mar 24, 2016 #2

    BvU

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    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.
     
  4. Mar 24, 2016 #3

    andrewkirk

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    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.
     
  5. Mar 25, 2016 #4
    How can f(x,y) and f(y,z) be the same function if they are functions of different variables?
     
  6. Mar 25, 2016 #5

    Mark44

    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.
     
  7. Mar 25, 2016 #6
    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##
     
  8. Mar 25, 2016 #7

    andrewkirk

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    Looks right to me :smile:
     
  9. Mar 25, 2016 #8
    Thanks, everyone.
     
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