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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


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    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.
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