Calculus-derivatives

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What is the slope of the tangent of each curve at given point

y=√(16x^3), (4,32)



If I find the derivative of y=√(16x^3)

Will it be 16/-x^3

  • Calculus-derivatives -

    nope. √x^3 = x^(3/2)

    so,

    y = √16x^3 = 4x^(3/2)
    y' = 4(3/2)x^(1/2) = 6√x

    y'(4) = 6√4 = 12

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