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Derivative latex help

  1. Aug 8, 2004 #1
    I posted the same thread twice. Oops. :tongue2:
    Last edited: Aug 8, 2004
  2. jcsd
  3. Aug 8, 2004 #2

    [tex]x = e^u[/tex], where u is a function of x.

    Using the chain rule:
    [tex]\frac{dy}{du} = e^u\frac{dy}{dx}[/tex]

    Using the product rule:
    [tex]\frac{d^2y}{du^2} = \frac{d}{du}(e^u\frac{dy}{dx}) = e^u\frac{dy}{dx}+e^u\frac{d^2y}{dx^2}\cdot\frac{dx}{du}[/tex]

    Why is it [tex]\frac{dx}{du}[/tex]?
    Last edited: Aug 8, 2004
  4. Aug 8, 2004 #3


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    What's this "y"-thingy?
    It doesn't appear in your first line
  5. Aug 8, 2004 #4


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    (latex hint: you can use [ itex ] tags for formulas that go in a paragraph)

    (Did you mean [itex]y = e^u[/itex]?)

    Anyways, the chain rule says that:

    \frac{dp}{dq} = \frac{dp}{dr} \frac{dr}{dq}

    In your calculation, you had to compute:

    \frac{d}{du} \left( \frac{dy}{dx} \right)

    So, throw it into the chain rule and see what you get.
  6. Aug 8, 2004 #5
    Bleh, I'm new to latex so I accidentally pressed the post thread button instead of the preview post one.

    Anyway, let me elaborate.

    [tex]x = e^u[/tex], where u is a function of x.

    Using the chain rule:
    [tex]\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du} = e^u \frac{dy}{dx}[/tex]

    Now, using the product rule:
    [tex]\frac{d^2y}{du^2} = \frac{d}{du}(\frac{dy}{du}) = \frac{d}{du}(e^u \frac{dy}{dx}) = e^u \frac{dy}{dx} + e^u \frac{d^2}{dx^2} \cdot \frac{dx}{du}[/tex]

    My question is:
    Shouldn't [itex]\frac{dx}{du}[/itex] be [itex]\frac{dy}{du}[/itex]?
    Last edited: Aug 8, 2004
  7. Aug 8, 2004 #6
    Nevermind. I see where I went wrong.
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