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

SM
Answered by Sophie M. Further Mathematics tutor

3817 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

l1 and l2 are tangents of a circle. l1 intersects the circle at (3-√3,5) with a gradient of √3, and l2 intersects the circle at (3+√2,4+√2) with a gradient of -1. Find the centre of the circle, and hence find the radius of the circle.


Factorise 6x^2 + 7x + 2


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


What is differentiation used for?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning