Need a correct differential value. EMERGENCY

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx((-1+cos(x)) (-1+sin(x)))
The derivative of f(x) is f'(x):
f'(x) = d/dx((-1+cos(x)) (-1+sin(x)))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x)-1 and v = sin(x)-1:
f'(x) = (-1+cos(x)) (d/dx(-1+sin(x)))+(d/dx(-1+cos(x))) (-1+sin(x))
Differentiate the sum term by term:
f'(x) = (d/dx(-1+cos(x))) (-1+sin(x))+(-1+cos(x)) d/dx(-1)+d/dx(sin(x))
The derivative of -1 is zero:
f'(x) = (d/dx(-1+cos(x))) (-1+sin(x))+(-1+cos(x)) (d/dx(sin(x))+0)
Simplify the expression:
f'(x) = (-1+cos(x)) (d/dx(sin(x)))+(d/dx(-1+cos(x))) (-1+sin(x))
The derivative of sin(x) is cos(x):
f'(x) = (d/dx(-1+cos(x))) (-1+sin(x))+(-1+cos(x)) cos(x)
Differentiate the sum term by term:
f'(x) = (-1+cos(x)) cos(x)+(-1+sin(x)) d/dx(-1)+d/dx(cos(x))
The derivative of -1 is zero:
f'(x) = (-1+cos(x)) cos(x)+(-1+sin(x)) (d/dx(cos(x))+0)
Simplify the expression:
f'(x) = (-1+cos(x)) cos(x)+(d/dx(cos(x))) (-1+sin(x))
The derivative of cos(x) is -sin(x):
f'(x) = (-1+cos(x)) cos(x)+(-1+sin(x)) -sin(x)
Expand the left hand side:
Answer: |  f'(x) = (-1+cos(x)) cos(x)-(-1+sin(x)) sin(x)

Here's another way:

f(x)  = (cos x  - 1)(sinx - 1)     expand this

f(x)  = cosxsinx  - cos x - sinx - 1

f(x)  = (1/2)sin 2x  - cos x - sin x  - 1      take the derivative

f '(x)  = (1/2) *2 cos 2x  + sin x - cos x

f '(x)  =    cos 2x + sin x - cos x