Question

How do you use the limit definition of a derivative to take the derivative of `f(x) = sin(nx)` for a constant n?

Answer 1

It is much like using the limit definition to find the derivative of `sinx`.

Explanation:

Many of the details are like finding the derivative of `sinx`

but instead of `lim_(hrarr0)sinh/h = 1`

we need `lim_(hrarr0)sin(nh)/h = lim_(hrarr0)n(sin(nh)/(nh)) = n`

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

`= lim_(hrarr0)(sinn(x+h)-sinnx)/h`

`= lim_(hrarr0)(sin(nx+nh)-sinnx)/h`

`= lim_(hrarr0)(sin(nx)cos(nh)+cos(nx)sin(nh)-sin(nx))/h`

`= lim_(hrarr0)((sin(nx)cos(nh)-sin(nx))/h+(cos(nx)sin(nh))/h)`

`= lim_(hrarr0)(sin(nx)(cos(nh)-1)/h +cos(nx)sin(nh)/h)`

Now, we need: as `hrarr0`, we get `nh rarr0`, so we have:

`lim_(hrarr0)(cos(nh)-1)/h = nlim_(hrarr0)(cos(nh)-1)/(nh) = n*0 =0`

and
`lim_(hrarr0)sin(nh)/h = n lim_(hrarr0)(sin(nh)/(nh)) = n*1 = n`

So we can continue:

`lim_(hrarr0)(sin(nx)(cos(nh)-1)/h +cos(nx)sin(nh)/h) =sin(nx)*0+cos(nx)*n`

`= ncosnx`