Question

Why does the derivative of f(x)=sinxcosx equal f'(x)=cos2x and not f'(x)=cos^2x-sin^2x ?

Answer 1

By identity rule
cos^2x-sin^2x = cos2x

Answer 2

yea, that is similar

Answer 3

Wait, but isn't sin^2x+cos^x=1, sin^2x=1-cos^2x, and cos^2x=1-sin^2x? I don't see how cos^2x-sin^2x= cos2x

Answer 4

use this formula :
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
so, cos(2x)=cos(x+x)=cos(x)cos(x)-sin(x)sin(x)=(cos(x))^2-((sin(x))^2

Answer 5

Whoa, awesome! Thanks! :D

Answer 6

welcome :D