Why is the derivative of y=sinxcos2x: y'= cosxcos2x-2sinxsin2x

Please show all steps and don't skip too much. I am new to finding the derivatives of trig functions and it feels odd to me still.

First let's word it like this:
y=ab
y'=a(b') + b(a')

the derivative of sinx is cosx and the derivative of cosx is -sinx
the derivative of cos2x is -sin2x(2(1)) because you find the derivative of cos and then what's inside the cos

so
y=sinxcos2x
y'= sinx(-sin2x)(2(1)) + cos(2x)cosx(1)

Sadly calculus isn't the easiest thing to explain by typing it out, hope I helped.

OK the multiplication rule for derivative is

derivative of xy is derivative x times y + derivative y times x

derivative of sinx is cosx, and derivative of cosx is -sinx, and using the chain rule, derivative of cos2x is -sin2x times the derivative of 2x which is 2, so -2cos2x, is the derivative of the second term (the cos2x),
Now put it all togather to get the derivetive of the first term times the second plus the derivative of the second term times the original first term.

y=sinxcos2x:
y'= cosx (cos2x) + (sinx)(-2sin2x)

y'= cosx cos2x-2sinx sin2x

step by step solution:
http://symbolab.com/math/solver/step-by-…

hope this helps

y = (sinx) *(cos2x)
y' = (sinx)' * (cos2x) + (cos2x)' * sinx
y' = (cosx) * (cos2x) + (-sin2x *(2x)') * (sinx)
y' = (cosx) * (cos2x) - 2sinx * sin(2x) (answer)