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

4139 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Use differentiation to show the function f(x)=2x^3–12x^2+25x–11 is an increasing function for all values of x


Find the coordinates of the minimum point of the function y=(x-5)(2x-2)


Find the coordinates of any stationary points of the curve y(x)=x^3-3x^2+3x+2


A circle has equation x^{2}-8x+y^{2}-6y=d. A line is tangent to this circle and passes through points A and B, (0,17) and (17,0) respectively. Find the radius of the circle.


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