Given that y = 60x + 3x^2 – 4x^3, calculate: (a) the gradient of the tangent to the curve of y at the point where x = 1​

Question

Given that y = 60x + 3x^2 – 4x^3, calculate:
(a) the gradient of the tangent to the curve of y at the
point where x = 1​

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Claire 6 days 2021-10-13T01:40:28+00:00 1 Answer 0

Answers ( )

    0
    2021-10-13T01:42:15+00:00

    The gradient of the tangent to the curve of y = 60x + 3x² – 4x³ at the  point where x = 1​ is 54

    Step-by-step explanation:

    To find the gradient of a tangent of a curve at point (a , b)

    • Differentiate the equation of the curve to find y’
    • Substitute the value of x in y’ by a to find the gradient of the tangent (m)
    • Remember that the differentiation of ax^{n} is anx^{n-1}

    ∵ The equation of the curve is y = 60x + 3x² – 4x³

    – Find y’

    ∵ The differentiation of 60x is 60

    ∵ The differentiation of 3x² is 6x

    ∵ The differentiation of -4x³ is -12x²

    ∵ y’ = 60 + 6x – 12x²

    – To find the gradient of the curve at x = 1 substitute x by 1 in y’

    ∵ m = 60 + 6(1) – 12(1)²

    ∴ m = 60 + 6 – 12

    ∴ m = 54

    The gradient of the tangent to the curve of y = 60x + 3x² – 4x³ at the  point where x = 1​ is 54

    Learn more:

    Learn more about differentiation in brainly.com/question/4279146

    #LearnwithBrainly

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