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Math Help - Parametric Derivative Question

  1. #1
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    Parametric Derivative Question

    Let x=t2 + t, and let y = sin t.

    a. Find d/dt (dy/dx) as a function of t.

    b. Find d/dx (dy/dx) as a function of t.

    I understand how to do (a) part. Using the rule dy/dx = (dy/dt)/(dx/dt), I get dy/dx = (cos t) /(2t + 1). Using the Quotient Rule,
    I get d/dt (dy/dx) = [-(2t +1)sin t - 2 cos t] / (2t +1)2.

    How do I find d/dx (dy/dx)? I think it would involve the Chain Rule but not sure where to start.
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  2. #2
    MHF Contributor
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    Re: Parametric Derivative Question

    We know that

    $\dfrac d {dx} y = \dfrac {dy}{dx}=\dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}}$

    so it stands to reason that

    $\dfrac d {dx} \dfrac {dy}{dx} = \dfrac {\dfrac d {dt} \dfrac {dy}{dx}}{\dfrac {dx}{dt}}=$

    $-\dfrac {\sin(2t+1)+2\cos(t)}{(2t+1)^3}$
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