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Multivariable Chain rule for higher order derivatives

  1. May 9, 2013 #1
    Hello,

    Given is the function

    f = f(a,b,t), where a=a(b) and b = b(t). Need to express first and second order derivatives.

    [itex]\frac{\partial f}{\partial a}[/itex] and [itex]\frac{\partial f}{\partial b}[/itex] should be just that, nothing more to it here, correct?

    But

    [itex]\frac{df}{dt}[/itex] = [itex]\frac{\partial f}{\partial a}[/itex] [itex]\frac{da}{db}[/itex] [itex]\frac{db}{dt}[/itex] + [itex]\frac{\partial f}{\partial b}[/itex] [itex]\frac{db}{dt}[/itex] + [itex]\frac{\partial f}{\partial t}[/itex], by the chain rule, correct?

    I need to express [itex]\frac{\partial f}{\partial t}[/itex], but the above chain rule puts the total derivative [itex]\frac{df}{dt}[/itex] in the expression and it gets messy. I mean, how do I express

    [itex]\frac{\partial f}{\partial t}[/itex]?

    Then I need also [itex]\frac{\partial^2 f}{\partial t^2}[/itex], [itex]\frac{\partial^2 f}{\partial a^2}[/itex] and [itex]\frac{\partial^2 f}{\partial b^2}[/itex].

    Anyone well versed in partial derivatives?
     
  2. jcsd
  3. May 12, 2013 #2
    Let me try to clear it up a bit

    Let f = f(a,b,t) where a=a(b(t)), b=b(t)

    Then

    [itex]\frac{df}{dt}[/itex]=[itex]\frac{∂f}{∂a}[/itex][itex]\frac{da}{db}[/itex][itex]\frac{db}{dt}[/itex] + [itex]\frac{∂f}{∂b}[/itex][itex]\frac{db}{dt}[/itex] + [itex]\frac{∂f}{∂t}[/itex], correct?

    How does one express

    [itex]\frac{d^2f}{dt^2}[/itex]=?
     
  4. May 13, 2013 #3
    Okay, Let me simplify this to

    Let f = f(a,b) where a=a(b)

    Then

    [itex]\frac{df}{db}=\frac{∂f}{∂a}\frac{da}{db} + \frac{∂f}{∂b}[/itex], correct?

    How does one express

    [itex]\frac{d^2f}{db^2}[/itex]=?
     
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