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

I Chain rule found in MIT video

  1. Jul 14, 2016 #1
    While solving an equation, the lecturer was using substitution in this video:


    x=au was subbed in for Psi at timestamp 39:27
    d/dx = (1/a)(d/du). I get that.
    But then the second derivative is stated as being
    d2/dx2 = (1/a2)(d2/du2)

    How is it (1/a2) if we do not know if there is an "a" in the function "u"?
     
  2. jcsd
  3. Jul 15, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What do you mean? If you are fine with (d/dx) = (1/a)(d/du), how can you doubt (d/dx)^2 = (1/a^2)(d/du)^2?
     
  4. Jul 15, 2016 #3
    It is not the result squared. It is the second derivative.
     
  5. Jul 15, 2016 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, which is exactly what I wrote. Taking the second derivative is equivalent to applying the derivative operator twice. This is standard operator notation.
     
  6. Jul 15, 2016 #5
    I guess I'm missing it or just thinking of it wrong:

    x = au
    dx = a du
    dx2 = adu2 (since "a" is constant)

    So, why is it a2?
     
  7. Jul 15, 2016 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This makes no sense whatsoever, it is not clear what you mean by ##dx^2##. In standard notation, this would be either the differential of ##x^2## or the differential ##dx## squared. Neither behaves as you imply.
     
  8. Jul 15, 2016 #7
    My notation is off. What I was trying to ask is why is the second derivative of "x" not "a" times the second derivative of "u"? If "a" is not a variable of "u", where is the "a^2" coming from?
     
  9. Jul 16, 2016 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The second derivative of ##x## or with respect to ##x##? Your original question was the derivative with respect to ##x##.

    The following might help: You agree that the first derivative of a function wrt ##x## follows the rule
    $$\frac{df}{dx} = \frac{1}{a} \frac{df}{da}$$
    if ##x = au## and ##a## is a constant. Now, the second derivative is just applying the first derivative twice and so
    $$
    \frac{d^2f}{dx^2} = \frac{d}{dx}\frac{df}{dx} = \frac{d}{dx} \frac{1}{a}\frac{df}{du} = \frac{1}{a} \frac{d}{dx}\frac{df}{du}.
    $$
    Now, see ##df/du## just as any function ##g##. Regardless of what the function ##g## is, we have agreed that the first derivative is given by ##dg/dx = (1/a) dg/du## and so
    $$
    \frac{1}{a} \frac{d}{dx}\frac{df}{du} = \frac{1}{a} \frac{dg}{dx} = \frac{1}{a^2}\frac{dg}{du} = \frac{1}{a^2} \frac{d}{du}\frac{df}{du} = \frac{1}{a^2} \frac{d^2f}{du^2}.
    $$

    Also consider the following example: Let ##f(x) = x^2##, then if ##x = au## we find that ##df/dx = 2x## and ##d^2f/dx^2 = 2##. We also find that ##d^2f/du^2 = d(x^2)/du^2 = d(a^2u^2)/du^2 = a^2d(u^2)/du^2 = 2a^2##. Clearly, the relation ##d^2f/dx^2 = (1/u^2) d^2f/du^2## is satisfied in this case.
     
  10. Jul 16, 2016 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The way I would look at it is simply:

    ##\frac{d}{du} = a \frac{d}{dx}##

    And:

    ##\frac{d^2}{du^2} = \frac{d}{du} \frac{d}{du} = (a \frac{d}{dx})(a \frac{d}{dx}) = a^2 \frac{d^2}{dx^2} ##

    In fact, you could replace the derivatives by any linear operators:

    If ##U = aX## then ##U^2 = a^2 X^2##

    For any linear operators ##U## and ##X##
     
  11. Jul 16, 2016 #10
    Thanks for all the help. My issue I guess is that I was looking at it as the derivative of a constant function instead of it being a linear operator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Chain rule found in MIT video
  1. The chain rule (PDE) (Replies: 1)

  2. Chain rule (Replies: 1)

  3. Chain rule (Replies: 8)

Loading...