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

4095 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

A=(1,a;0,1/2) B=(1,-1;0,2) AB=I, calculate the value of a.


A curve is defined by the equation y = (x + 3)(x – 4). Find the coordinates of the turning point of the curve.


A curve is mapped by the equation y = 3x^3 + ax^2 + bx, where a is a constant. The value of dy/dx at x = 2 is double that of dy/dx at x = 1. A turning point occurs when x = -1. Find the values of a and b.


In a chess club there are x boys and y girls. If ten more boys join and one more girl joins, there is an equal amount of boys and girls. Knowing that y = 2x+2, Calculate x and y. [4 marks]


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