# Using the definition of derivative, how do you prove that (cos x)' = -sin x?

Remembering the cosine difference-to-product formula, that says:

and the fundamental limit:

than:

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together with the fundamental tigonometric limits:

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To prove that (cos x)' = -sin x using the definition of derivative, we start with the definition of the derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Let f(x) = cos x. Substituting this into the definition:

(cos x)' = lim(h->0) [cos(x + h) - cos x] / h

Using the trigonometric identity for the cosine of a sum:

(cos x)' = lim(h->0) [(cos x * cos h - sin x * sin h) - cos x] / h

(cos x)' = lim(h->0) [(cos x * cos h - cos x) - (sin x * sin h)] / h

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x * sin h)] / h

Using the limit definition of cosine and sine functions:

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x * (1 - (h^2)/2) - sin x * h)] / h

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x + (sin x * (h^2)/2) - sin x * h)] / h

(cos x)' = lim(h->0) [(-2 * sin^2(x/2) * sin^2(h/2) - sin x * h + (sin x * (h^2)/2))] / h

As h approaches 0, sin h / h approaches 1, and (h^2)/2 approaches 0:

(cos x)' = -sin x

Therefore, (cos x)' = -sin x.

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