How do you differentiate f(x)= 2x*sinx*cosx?

2 Answers
Aug 6, 2015

f'(x)=2sinxcosx+2xcos^2x-2xsin^2x

Explanation:

Use the product rule:

f=ghk => f'=g'hk+gh'k+ghk'

With:
g=2x => g'=2x
h=sinx => h'=cosx
k=cosx => k'=-sinx

We then have:
f'(x)=2sinxcosx+2xcos^2x-2xsin^2x

Aug 6, 2015

f'(x)=2sin(x)cos(x)+2x(cos^2(x)-sin^2(x))

Explanation:

f'(x)=(2x)' cdot (sin(x) cdot cos(x))+2x cdot (sin(x) cdot cos(x))'

(2x)'=2

(sin(x) cdot cos(x))'=sin(x)' cdot cos(x)+sin(x) cdot cos(x)'
=cos(x) cdot cos(x)+sin(x) cdot (-sin(x))
=cos^2(x)-sin^2(x)

f'(x)=2sin(x)cos(x)+2x(cos^2(x)-sin^2(x))