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If x=t^3 and y=3t^5, where t is any real number, what is the derivative of y with respect to x at x=125 ?

  • calculus -

    t=∛x =x^(1/3) =>
    y=3x^5 = 3x^(5/3).

    dy/dx =3•(5/3)•x^(2/3) =
    =5 •125^(2/3)=525=125

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