A curve has the equation y = 2x cos(3x) + (3x^2-4) sin(3x). Find the derivative in the form (mx^2 + n) cos(3x)

y = 2x cos(3x) + (3x2-4) sin(3x)
dy/dx = (2x x -sin(3x) x 3) + (2 x cos(3x)) + (6x sin(3x)) + ((3x2-4) cos(3x) x 3)
dy/dx = -6x sin(3x) + 2 cos (3x) + 6x sin(3x) + (9x2-12) cos(3x)
dy/dx = (9x2-12 + 2) cos (3x) = (9x2-10) cos (3x)
m = 9n = -10

TL
Answered by Thomas L. Maths tutor

8972 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

I'm trying to integrate f(x)=sin(x) between 0 and 2 pi to find the area between the graph and the axis but I keep getting 0, why?


The cubic polynomial f(x) is defined by f(x) = 2x^3 -7x^2 + 2x + 3. Given that (x-3) is a factor of f(x), express f(x) in factorised form.


How do you use the chain rule?


Find the integral of [ 2x^4 - (4/sqrt(x) ) + 3 ], giving each term in its simplest form


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