uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone
hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
This is product rule. The 2 is just a coefficient of the function sinxcosx. When you take the derivative of sinxcosx, the 2 stays out in front.
(2sinxcosx)'
= 2(sinxcosx)'
= 2[(sinx)'cosx + sinx(cosx)']
= 2[cos^2(x) - sin^2(x)]
= 2[cos(2x)] = 2cos(2x)
[Thanks for pointing out my mistake, captainlewis ]