Differentiate: f(x)=2(sin(2x))^2 with respect to x, and evaluate as a single trigonometric function.

f(x) = 2sin2(2x)Therefore, using the chain rule: f'(x)=2 x 2cos(2x) x 2sin(2x)(The 2 at the front arises from the constant 2, at the start of f(x), the 2cos(2x) comes from differentiating sin2(2x), then the 2sin(2x) comes from decreasing the original power of the sine function by 1 and multiplying by the constant in the function, 2)Therefore, f'(x)=6cos(2x)sin(2x)As we know 2sin(x)cos(x)=sin(2x) (double-angle formula), we can simplify f'(x) into f'(x)=3sin(4x)

SH
Answered by Sam H. Maths tutor

4679 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Show that 2sin(2x)-3cos(2x)-3sin(x)+3=sin(x)(4cos(x)+6sin(x)-3)


sin(x)/(cos(x)+1) + cos(x)/(sin(x)+1) = 1


I don't fully understand the purpose of integration. Could you please explain it to me?


What is the gradient of y = xcos(x) at x=0?


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