Work out the gradient of the curve y=x^3(x-3) at the point (3,17)

First simplify the equation of the curve y= x^4 - 3x^3 .The gradient is the differential.To differentiate, bring down the power and take one from it.x^4 becomes 4x^3-3x^3 becomes (-3x3)= -9x^2dy/dx= 4x^3 - 9x^2Coordinates are written in (x,y) form. Hence x=3.Gradient at x=3 = 4x^3 - 9x^2 = 4(3^3) - 9(3^2) = 108 - 81 = 27

Related Further Mathematics GCSE answers

All answers ▸

If the equation of a curve is x^2 + 9x + 8 = y, then differentiate it.


Solve these simultaneous equations: 3xy = 1, and y = 12x + 3


Express (7+ √5)/(3+√5) in the form a + b √5, where a and b are integers.


f(x) = 2x^3+6x^2-18x+1. For which values of x is f(x) an increasing function?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences