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Homework Help: Please help with chain rule

  1. Sep 19, 2006 #1
    I need help solving this problem. It is in my text book but no answer is provided in the appendix.

    If [tex]y=2t+3[/tex] and [tex]x=t^{2}-t[/tex], find [tex]\frac{dy}{dx}[/tex]

    In theory this should be fairly straight forward! Simply find [tex]\frac{dy}{dt}[/tex] and [tex]\frac{dt}{dx}[/tex] and multiply both derivatives together to find [tex]\frac{dy}{dx}[/tex] , but I am having some problems.

    I tried solving [tex]x=t^{2}-t[/tex] for t, but that gets ugly pretty fast. I am out of ideas, can someone point me in the right direction? Thanks again in advance.
     
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  3. Sep 19, 2006 #2

    arildno

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    Remember the rule for the derivative of the inverse function..
     
  4. Sep 19, 2006 #3
    Sorry

    I am sorry but I have never heard of that rule. I just looked through my textbook and I cannot find any mention of it :( This is for grade 12 calculus. could you just elaborate a little bit? sorry and thanks!
     
  5. Sep 19, 2006 #4
    dy/dt =2

    dx/dt=2t-1

    dt/dx= 1/2t-1

    dy/dx=2/(2t-1)
     
  6. Sep 19, 2006 #5
    ahhhh that all makes perfect sense now!!! But wouldn't dt/dx = -1/2t-1... :)
     
  7. Sep 19, 2006 #6

    HallsofIvy

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    Parentheses!! Since x= t2- t dx/dt= 2t- 1 and so dt/dx= 1/(2t-1). I see no reason for a negative sign there.
     
  8. Sep 20, 2006 #7
    Yea I realized that a mintute after I said it.. No need for a perpendicular derivative! lol, don't know what I was thinking, but anyways I understand 100% now. Thanks! Just to make this clear, does this apply to alll instances where you have [tex]\frac{dy}{dx}[/tex] and want to find [tex]\frac{dx}{dy}[/tex] or vice versa? Thanks. I would imagine it would but I just want to make sure.
     
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