Question

How do you find f'(x) using the definition of a derivative for `f(x)=cosx`?

Answer 1

`(cosx)'=-sinx`

Explanation:

According to the definition:

`f'(x_0)=lim_{h->0}(f(x_0+h)-f(x_0))/h`

If we use this definition for `f(x)=cosx` we get:

`f'(x_0)=lim_{h->0}(cos(x_0+h)-cosx_0)/h`

`f'(x_0)=lim_{h->0}(-2sin((x_0+h+x)/2)*sin((x_0+h-x_0)/2))/h`

`f'(x_0)=lim_{h->0}(-2sin(x_0+h/2)sin(h/2))/h`

`f'(x_0)=lim_{h->0}(-sin(x_0+h/2)sin(h/2))/(h/2)`

`f'(x_0)=lim_{h->0}(-sin(x_0+h/2))*lim_{h->0}sin(h/2)/(h/2)`

For further calculation we will use the identity:

`lim_{x->0}sinx/x=1`

So the left limit is `-sinx_0`, and the right is `1`. Finally we get, that

`f'(x_0)=-sinx_0`

Answer 2

Here is an alternative using `cos(A+B) = cosAcosB-sinAsinB`

Explanation:

`f(x) = cosx`

`f'(x) = lim_(hrarr0)(cos(x+h)-cosx)/h`

`= lim_(hrarr0)(cosxcos h-sinxsin h-cosx)/h`

`= lim_(hrarr0)(cosxcos h-cosx-sinxsin h)/h`

`= lim_(hrarr0)(cosxcos h-cosx)/h-(sinx sin h)/h)`

`= lim_(hrarr0)(cosx(cos h-1)/h-sinx (sin h)/h)`

`= [lim_(hrarr0)cosx] [lim_(hrarr0)(cos h-1)/h]-[lim_(hrarr0)sinx][ lim_(hrarr0)(sin h)/h]`
(Provided that these 4 limits exist, which they do.)

`= [0][1]-[sinx][1] = -sinx`

That is: `f'(x) = -sinx`