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Answer 1 · 2017-01-09T11:17:44

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

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)#