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

8082 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The line l1 has equation y = −2x + 3. The line l2 is perpendicular to l1 and passes through the point (5, 6). (a) Find an equation for l2 in the form ax + by + c = 0, where a, b and c are integers.


Find the stationary point of y=3x^2-12x+29 and classify it as a maximum/minimum


Find the equation of the line through the following points: (-2, -3) and (1, 5)


Find the tangent and normal to the curve y=(4-x)(x+2) at the point (2, 8)


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences