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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook