Question

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

Answer 1

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

`cosalpha-cosbeta=-2sin((alpha+beta)/2)sin((alpha-beta)/2)`

and the fundamental limit:

`lim_(xrarr0)sinx/x=1`, or in the most general writing:

`lim_(f(x)rarr0)sinf(x)/f(x)=1`

than:

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

`=lim_(hrarr0)(-2sin((x+h+x)/2)sin((x+h-x)/2))/h=`

`=lim_(hrarr0)(-2sin(x+h/2)sin(h/2))/h=`

`=lim_(hrarr0)(-sin(x+h/2)sin(h/2))/(h/2)=`

`=lim_(hrarr0)-sin(x+h/2)*lim_(hrarr0)sin(h/2)/(h/2)=-sinx*1=-sinx`.

Answer 2

use the formula for the cosine of a sum:
`cos(a+b)= cosacosb-sinasinb`

together with the fundamental tigonometric limits:

`lim_(hrarr0)sinx/x = 1` and

`lim_(hrarr0)(cosx-1)/x = 0`

For `f(x)=cosx`, we get:

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

`color(white)"sssss"` `= lim_(hrarr0)(cosxcosh-sinxsinh-cosx)/h`

`color(white)"sssss"` `= lim_(hrarr0)(cosxcosh-cosx-sinxsinh)/h`

`color(white)"sssss"` `= lim_(hrarr0)(cosx(cosh-1)/h-sinxsinh/h)`

`color(white)"sssss"` `= lim_(hrarr0)cosxlim_(hrarr0)(cosh-1)/h-lim_(hrarr0)sinxlim_(hrarr0)sinh/h)`

`color(white)"sssss"` `= (cosx)(0)-(sinx)(1) = -sinx`