Question

How do you find the derivative of:

y= 2sinx cosx

Answer 1

You probably did it correctly but did not match the answer in the book.

Can be done in two ways:

1.
dy/dx = 2sinx(-cosx) + 2cosxcox
= 2cos^2 x - 2 sin^2 x

2. recognize that 2sinxcosx = sin 2x

so y = 2sinxcox
= sin 2x
dy/dx = 2cosx


but remember cos 2x = cos^2 x - sin^2 x
so the two answers are the same

Answer 2

You need to use the multiplication rule:
y = 2 sinx cosx
y' = 2 d/dx (sinx * cosx)
y' = 2 [ (sinx * -sinx) + (cosx * cosx) ]
y' = 2 [ -sin^2 x + cos^2 x ]
y' = 2cos^2 x - 2sin^2 x