# How to prove that derivative of cosx is -sinx ?

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### 1 Answer

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.**