How do you differentiate #f(x)= 2xsinxcosx#?

Question

4314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-06T22:12:09

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

#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#

Answer 2 · 2015-08-06T22:17:06

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

#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))#

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