Answer: #Sin x#
Explanation:
The derivatives for the #sin# and #cos# functions are interconnected as follows:
#d/dx sin(x) = cos(x)#
#d/dx cos(x) = -sin(x)#
#d/dx -sin(x) = - cos(x)#
#d/dx - cos(x) = sin(x)#
(Also worth noting is that #cos (-x) = cos(x)# and #sin(-x) = -sin(x)#, though those will not have bearing here).
With this in mind, taking the derivative of #f(x) = 1-cos(x)# would proceed as follows:
#d/dx (1 - cos(x)) = d/dx (1) + d/dx (-cos(x))# (Recalling that the derivative of a sum/difference is equal to the sum/difference of the derivatives)
#= 0 + sin (x)# (recalling that the derivative of a constant is 0)
#= sin(x)#
