Calculus: trig derivative: product rule or identity substitution?

Tutoring calculus, trig identities can give choices when performing derivatives. The tutor mentions an example.

Example: find derivative of y=sinxcosx

Solution method 1: product rule (see my post here.)

y’ = (sinx)’cosx + sinx(cosx)’ = cosxcosx – sinxsinx = (cosx)^2 – (sinx)^2

Solution method 2: identity substitution (sinxcosx= (sin2x)/2) with chain rule

y’ = [(sin2x)/2]’ = (1/2)(sin2x)’ = (1/2)(cos2x)(2) = cos2x

Note that (cosx)^2 – (sinx)^2 = cos2x

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, Part One, Third Edition. Toronto: D. C. Heath and Company, 1989.