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Derivative of a derivative with respect to something else

  1. Jan 27, 2008 #1
    1. The problem statement, all variables and given/known data

    If [tex]x=e^t[/tex], find [tex]\frac{dy}{dx}[/tex] in terms of [tex]\frac{dy}{dt}[/tex] and hence prove [tex]\frac{d^2y}{dx^2}=e\stackrel{-2t}{}(\frac{d^2y}{dt^2}-\frac{dy}{dt})[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Well i have done the first bit:




    Thats fine so now to get the second derivative using product rule:


    now my problem is what is [tex]\frac{d(\frac{dy}{dt})}{dx}[/tex]?

    would greatly appreciate a step by step for that part so i can finally understand this one!
  2. jcsd
  3. Jan 27, 2008 #2


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    consider finding [tex]\frac{d}{dx}(y)[/tex]

    [tex]\frac{d}{dx}=\frac{d}{dy} \frac{dy}{dx}[/tex]

    so now you will have
    [tex]\frac{d}{dx}(y)=\frac{d}{dy}(y) \frac{dy}{dx}=1\frac{dy}{dx}[/tex]
  4. Jan 27, 2008 #3


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    You again invoke chain rule:

    [tex]\frac{dx}{dt}\cdot \frac{d(\frac{dy}{dt})}{dx}= \frac{d(\frac{dy}{dt})}{dt}=\frac{d^2 y}{dt^2}[/tex]

    That and a little algebra should get you home.
  5. Jan 27, 2008 #4
    im sorry but i still dont understand how to do it...
    rockfreak: what happens now when i change y in your expression to [tex]\frac{dy}{dt}[/tex]?

    jam: if that was true then i would get [tex]\frac{d^2y}{dx^2}=-x\stackrel{-2}{}\frac{dy}{dt}+x\stackrel{-1}{}\frac{d^2y}{dt^2}[/tex]

    which would produce the wrong answer...
  6. Jan 27, 2008 #5
    ok sorry i think i got it...correct me if im wrong:

    i could write [tex]\frac{d(\frac{dy}{dt})}{dx}=\frac{dt}{dx}\frac{d(\frac{dy}{dt})}{dt}=x\stackrel{-1}{}\frac{d^2y}{dt^2}[/tex]

    thanks guys!
  7. Jan 27, 2008 #6


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    You still follow the same rules...in fact, you could use the convention from Newtonian mechanics that a dot over a quantity represents its time derivative:

    [tex] \dot y = \frac{dy}{dt} [/tex]

    To make things less cumbersome. However, rockfreak essentially wrote down that:

    [tex] \frac{d \dot y}{dx} = \frac{d \dot y}{dx} [/tex]

    which isn't that useful. Jambaugh had the correct advice

    Nope, you would not get that. Check it again. You confused x with t at some point.

    Let's go over jambaugh's method for clarity. Using the chain rule, and keeping in mind that the dot denotes differentiation w.r.t. time, we can write:

    [tex] \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d \dot y}{dx} [/tex]

    [tex] = \frac{d \dot y} {dt} \frac{dt}{dx} = \frac{d^2y}{dt^2} \frac{dt}{dx} [/tex]
  8. Jan 27, 2008 #7
    yeh cheers i think i got that...pretty simple actually once you know what to look for
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