Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chain rule for implicit differentiation

  1. Jan 21, 2012 #1
    I have derivative dx/dt = y(u(t)) * z(u(t)) + u(t)
    Now, what is dx/du ? I know the chain rule should help, but I am stuck :-(
     
  2. jcsd
  3. Jan 22, 2012 #2

    fluidistic

    User Avatar
    Gold Member

    Maybe [itex]\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}[/itex].
     
  4. Jan 23, 2012 #3
    Thanks fluidistics.

    Ok, so, what I get is:

    [itex]\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}=\left[y(u(t)) \cdot z(u(t)) + u(t)\right] \cdot \frac{dt}{du(t)}[/itex]

    Now, what the heck is this [itex]\frac{dt}{du(t)}[/itex] ? Any idea how should I treat it?
    Is it just a reciprocal of [itex]\frac{du(t)}{dt}[/itex]? In other words, can I just divide the original function with the time derivative of u(t)?

    [itex]\frac{dx}{du}=\frac{dx}{dt} \cdot \frac{dt}{du}=\left[y(u(t)) \cdot z(u(t)) + u(t)\right] \cdot \left[\frac{du(t)}{dt}\right]^{-1}[/itex]

    Thanks for help and comments! Its is almost... uhhhh... 15 years since I studied this stuff in university.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Chain rule for implicit differentiation
  1. The chain rule (PDE) (Replies: 1)

  2. Chain rule (Replies: 1)

Loading...