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Homework Help: Question Need help

  1. Oct 7, 2007 #1
    Derivative trigonometric functions

    1. The problem statement, all variables and given/known data
    Find d[tex]^{2}[/tex]x/dt[tex]^{2}[/tex] as a function of x if dx/dt=xsinx

    2. Relevant equations

    3. The attempt at a solution

    I tried to solve the problem by taking a second derivative of dx/dt=xsinx
    but I was not sure how to start. Also, it is confusing because
    doesn't have to be dx/dt=tsint in order to make sense (since i should
    take the derivatice w/ respect to t, not x)?

    I'm totally lost here, please help me with the start,
    and i'll try to figure out the rest!!
    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2


    User Avatar
    Science Advisor

    Fortunately your question "doesn't have to be dx/dt=tsint in order to make sense?" made me look at the question again! I had thought you were just asking for the second derivative of x sin(x) and obviously you aren't.

    Yes, you need to differentiate with respect to t, not x. But dx/dt does NOT have to be an explicit function of x. Since x itself is a function of t, x sin(x) is an implicit function of t. If you are told that dx/dt= f(x), use the chain rule:
    [tex]\frac{d^2 x}{dt^2}= \frac{df}{dt}= \frac{df}{dx}\frac{dx}{dt}[/tex]
    Just like "implicit differentiation" except this time you already know what dx/dt is.
  4. Oct 7, 2007 #3

    This is what i did.
    since x=f(t)
    so, x'=dx/dt

    if i take the second derivative of dy/dt,
    it'll be like
    (dy/dt)'= 1*(dx/dt)sinx + x*cosx* (dx/dt)

    the red part is where i used the chain rule.
    since I don't know the actual fucntion of x, I just set the derivative of it as
    dx/dt (because i'm taking derivative w/ respect to t)

    and since this is basically the derivative of of product, i used
    a product rule to do the rest.

    I simplfied it and ended up with
    am i right..? my answer looks quite a bit complicated,,
  5. Oct 7, 2007 #4


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    Science Advisor
    Homework Helper

    That looks right.
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