How do you find the first derivative for #f(x)= (sin x)(cos x)#?

1 Answer
Apr 11, 2015

Using the multiplication rule for derivatives, which states that the derivative of a product #(f*g)'# equals the following:

#(f*g)'=f'*g+f*g'#

Since the derivative of #sin(x)# is #cos(x)#, and the derivative of #cos(x)# is #-sin(x)#, we have that

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

This would of course be a perfect answer, but I think that noticing that #cos^2(x) - sin^2(x)=cos(2x)# would be more elegant.