Derivative of sin^2(x)

How do you get sin(2x) as the derivative? Thanks

Question 1. Using the Chain Rule
f(x) = sin²(x)
f'(x) = 2sin(x)cos(x)dx (answer)
f'(x) = sin(2x)dx (answer as an identity)

Using (u) substitution
u = sin(x)
du = cos(x)dx

f(x) = u²
f'(x) = 2udu
f'(x) = 2sin(x)cos(x)dx (answer)
f'(x) = sin(2x)dx (answer as an identity)

Note: If you look at a trigonometric identity sheet, you will see that 2sin(x)cos(x) is = sin(2x)

y = sin^2(x)
dy/dx = 2sin(x) d/dx cos(x)
= 2sin(x)cos(x)
= sin(2x) answer//

You have a function f(g(x)) with f(y)=y^2 und g(x)=sin(x)
--> product rule: f(g(x))'= f '(g(x))*g'(x) =>2sin(x)cos(x)

f(x) = sin²(x)

f'(x) = 2sinxcosx = sin2x

thats 2*sinx+cosx
because derivative of sin=cos how i got 2sinx from sin^2=2sin^2-1=2sin^1=2sin