Prove, using the definition of derivative, that if f(x) = cos(x), then f'(x) = -sin(x)

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Chapter 3, 3.3 - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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By using the definition of the derivative,

`lim_(h -> 0) [ cos(x + h) - cos(x) ] / h`

`=lim_(h -> 0)[ cos(x)cos(h) - sin(x)sin(h) - cos(x) ] / h`

`=lim_(h -> 0)[ cos(x) [ cos(h) - 1 ] - sin(x)sin(h) ] / h`

=`lim_(h -> 0)[ cos(x) [cos(h) - 1]/h ] - lim_(h -> 0) (sin(x)sin(h)/h)`

=`cos(x)lim_(h -> 0)[ [cos(h) - 1]/h ] - sin(x)lim_(h -> 0) (sin(h)/h)`

as

`lim_(h -> 0)[ [cos(h) - 1]/h ] =0 ` and

`lim_(h -> 0) (sin(h)/h) = 1`

= `cos(x) (0) -sin(x) (1)`

= `-sin(x)`

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