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Calculus Chain Rule

  1. Aug 23, 2011 #1
    Solve the following:

    d/dt cos(theta)
    d/dt t sin(theta)
    d/dt r cos (theta)
    d/dt r^2 (theta)
    d/dt e^ (-3x)
    d/dt (x^2 + y^2)

    I would assume all by the second one are 0 since your solving for terms dt and not theta, x, y, or r... I don't think its right at all. I know it goes something like this:
    d/dt f(x) = dy/dx * dx/dt
    I just am not sure how to grasp what I'm doing wrong.
  2. jcsd
  3. Aug 23, 2011 #2


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    Is theta a function of t?
  4. Aug 23, 2011 #3
    what do you mean?
  5. Aug 23, 2011 #4


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    is [itex]\theta =\theta (t)[/itex], otherwise the derivative will be non-zero.
  6. Aug 23, 2011 #5
    all it says its differential calculus and gives the problem as I stated above
  7. Aug 23, 2011 #6


    Staff: Mentor

    From the title of the thread ("Calculus Chain Rule"), I think it's reasonable to assume that [itex]\theta[/itex] is a differentiable function of t, and that you are meant to use the chain rule.
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