Why is the differential of cos x equal to -sin x?

We'll take the function f(x) = cos
x.

Now, we'll consider the
ratio:

R(x) =
[f(x)-f(x0)]/(x-x0)

We'll substitute f(x) and f(x0) by the
expression of the function:

f(x) = cos
x

f(x0) = cos
x0.

 [f(x)-f(x0)] = cos x - cos
x0

We'll transform the difference into a
product:

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

If we'll calculate the limit of the ratio R(x),
we'll get the value of the first derivative of the function, in the point x =
x0.

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

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

But lim sin a/a =
1

lim sin [(x-x0)/2]/(x-x0)/2 =
1

lim R(x) =  -lim  sin
[(x+x0)/2]*1

lim R(x) =  -lim  sin
[(x+x0)/2]

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

But lim R(x) =
f'(x0)

f'(x0) = -sin x0, when f(x) = cos
x.