#cos(x)#
#d/dx(cos(x))=-sin(x)#
#d^2/dx^2(cos(x))=-cos(x)#
#d^3/dx^3(cos(x))=sin(x)#
#d^4/dx^4(cos(x))=cos(x)#
So it seems that the derivatives of #cos(x)# repeat every four times.
So, the #n#th derivative of #cos(x)# should be
- #cos(x)# if #n# is divisible by #4#
- #-sin(x)# if the remainder of #n# divided by #4# is #1#
- #-cos(x)# if the remainder of #n# divided by #4# is #2#
- #sin(x)# if the remainder of #n# divided by #4# is #3#
In other words,
#d^n/dx^n(cos(x))={(cos(x)\ \ \ \ \ \ \ "if"\ n\equiv0\ ("mod"\ 4)),(-sin(x)\ \ \ \ "if"\ n\equiv1\ ("mod"\ 4)),(-cos(x)\ \ \ \ "if"\ n\equiv2\ ("mod"\ 4)),(sin(x)\ \ \ \ \ \ \ \ "if"\ n\equiv3\ ("mod"\ 4)):}#
It's possible to make the expression more compact, as
#d^n/dx^n(cos(x))=1/2(i^n(1+(-1)^n)+i^(n+1)(1+(-1)^(n+1)))cos(x-(pi(1-(-1)^n))/4)#
or
#d^n/dx^n(cos(x))="Re"((1+i)i^n)cos(x-(pi(1-(-1)^n))/4)#
or
#d^n/dx^n(cos(x))=cos((pin)/2)cos(x)-sin((pin)/2)sin(x)#
or
#d^n/dx^n(cos(x))=cos(x+(pin)/2)#
Then, the #125#th derivative of #cos(x)#, or #d^125/dx^125(cos(x))#, is #-sin(x)# since #125\equiv1\ ("mod"\ 4)#.
