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 ▸

Find any stationary points in the function f(x) = 3x^2 + 2x


Find the stationary point of 3x^2+7x


Find the x and y coordinates of the minimum of the following equation: y = x^2 - 14x + 55.


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.


We're here to help

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

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy