Question

Using the limit definition, how do you find the derivative of `f(x)=cosx`?

Answer 1

The classical definition of the derivative is

`(df(x))/dx=lim_(h->0)[f(x+h)-f(x)]/h`

where `f(x)=cosx` and `f(x+h)=cos(x+h)`

hence we have that

`(df(x))/dx=lim_(h->0)[f(x+h)-f(x)]/h=> (df(x))/dx=lim_(h->0)[cos(x+h)-cosx]/h=> (df(x))/dx=lim_(h->0)[[cosx*cosh-sinx*sinh]-cosx]/h=> (df(x))/dx=lim_(h->0)[(cosx(cosh-1))/h-sinx*sinh/h]=> (df(x))/dx=lim_(h->0)[(cosx(cosh-1))/h]-lim_(h->0)[sinx*sinh/h]`

But we know that `lim_(h->0)[((cosh-1))/h]=0` and
`lim_(h->0)(sinh/h)=1`

Hence

`(df(x))/dx=lim_(h->0)[(cosx(cosh-1))/h]-lim_(h->0)[sinx*sinh/h]`

`(df(x))/dx=lim_(h->0)(cosx)*lim_(h->0)(cosh-1)/h-lim_(h->0)(sinx)*lim_(h->0)sinh/h`

`(d(f(x)))/dx=cosx*0-sinx*1=> (d(f(x)))/dx=-sinx`

Finally `(dcosx)/dx=-sinx`