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

1. 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 is ∵ 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