How to prove that derivative of cosx is -sinx ?

We know that the function f(x) = cos x could be
differentiated over the real numbers set.

We'll write the
definition of the derivative at a given point x0.

f'(x0) =
lim (cos x - cos x0/(x - x0), if x approaches to x0

We'll
transform the difference of cosines into a product:

cos x -
cos x0 = 2sin[(x+x0)/2]*sin[(x0-x)/2]

We'll re-write the
limit:

f'(x0) = lim
2sin[(x+x0)/2]*sin[(x0-x)/2]/(x-x0)

f'(x0) = lim
2sin[(x+x0)/2]* lim sin[(x0-x)/2]/2*[(x-x0)/2]

f'(x0) =
-(2/2)*lim sin[(x+x0)/2]*lim sin[(x -
x0)/2]/[(x-x0)/2]

We'll recognize the remarkable limit lim
sin[(x - x0)/2]/[(x-x0)/2] = 1

f'(x0) = -lim sin[(x+x0)/2]
= -sin (x0 + x0)/2 = -sin (2*x0/2)

f'(x0) = - sin
x0

Therefore, the derivative of the cosine
function is: (cos x)' = - sin x.