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

3187 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Why is it that when 'transformation A' is followed by 'transformation B', that the combined transformation is BA and not AB?


Let y = (4x^2 + 3)^4. Find dy/dx.


Lengths of two sides of the triangle and the angle between them are known. Find the length of the third side and the area of the triangle.


In the expansion of (x-7)(3x**2+kx-3) the coefficient of x**2 is 0. i) Find the value of k ii) Find the coefficient of x. iii) write the fully expanded equation in terms of x


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences