Question

Question #3ebdb

Answer 1

`d^n/(dx^n) cosx = cos(x+(npi)/2)`

Explanation:

We have that:

`d/(dx) cosx = -sinx`

`d^2/(dx^2) cosx = d/(dx) (-sinx) = -cosx`

`d^3/(dx^3) cosx = d/(dx) (-cosx) = sinx`

`d^4/(dx^4) cosx = d/(dx) (sinx) = cosx`

And obviously after the fifth order they start repeating.

We can however find a synthetic expression valid for all orders noting that:

`cos(x+pi/2) =cosxcos(pi/2)- sinxsin(pi/2) = -sinx`

`cos(x+pi) =cosxcos(pi)- sinxsin(pi) = -cosx`

`cos(x+3/2pi) =cosxcos(3/2pi)- sinxsin(3/2pi) = sinx`

`cos(x+2pi) =cosx`

So that:

`d^n/(dx^n) cosx = cos(x+(npi)/2)`