Derivative of Parametric Functions

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M8-4: tangent to parametric curves (part II)

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Key Questions

  • For the parametric equations

    #{(x=x(t)),(y=y(t)):}#,

    we can find the derivative

    #{dy}/{dx}={{dy}/{dt}}/{{dx}/{dt}}={y'(t)}/{x'(t)}#.

  • Let #{(x=x(t)),(y=y(t)):}#.

    First Derivative

    #{dy}/{dx}={{dy}/{dt}}/{{dx}/{dt}}={y'(t)}/{x'(t)}#

    Second Derivative

    #{d^2y}/{dx^2}=d/{dx}[{y'(t)}/{x'(t)}]=1/{{dx}/{dt}}{d}/{dt}[{y'(t)}/{x'(t)}]#

    by Quotient Rule,

    #=1/{x'(t)}cdot{y''(t)x'(t)-y'(t)x''(t)}/{[x'(t)]^2}#

    #={y''(t)x'(t)-y'(t)x''(t)}/{[x'(t)]^3}#

    I hope that this was helpful.

  • To find the derivative of a parametric function, you use the formula:

    #dy/dx = (dy/dt)/(dx/dt)#, which is a rearranged form of the chain rule.

    To use this, we must first derive #y# and #x# separately, then place the result of #dy/dt #over #dx/dt#.


    #y=t^2 + 2#

    #dy/dt = 2t# (Power Rule)


    #x=tsin(t)#

    #dx/dt = sin(t) + tcos(t)# (Product Rule)


    Placing these into our formula for the derivative of parametric equations, we have:

    #dy/dx = (dy/dt)/(dx/dt) = (2t)/(sin(t)+tcos(t))#

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