differentiate: xsin(x)cos(x)
double product rule?
Hmm yes you could do it that way, product rule with 3 terms. Might be easier to use the double angle formula for sine. So you can get rid of one of the trig terms.
is the answer -x(cosxsinx)?
Hmm I don't think so.. not sure where you're getting that from D: Unless you did some simplification that I'm not seeing quite yet. Start with this: \[\sin x \cos x=\frac{ 1 }{ 2 } \sin 2x\] Double Angle formula for Sine.
how did you get the sin2x?
Double Angle formula for Sine.
is it (1/2) +xcos(2x)+(1/2)cos(2x)-(sinx)^2+(cosx)^2
<:o
Kat just look up the double angle formulas if you're not sure where it came from :3 it'll simplify your problem very nicely. \[\large x( \sin x \cos x)=\frac{ 1 }{ 2 }x \sin 2x\] This is much easier to differentiate.
ok i understand it now. thanks!
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