(If #c# is a constant, then the whole thing is constant and the derivative is #0#)
I will assume that #c# is the independent variable. If we are differentiating with respect to some other variable, then we must apply the chain rule and multiply by the derivative of #c#.
#f(c) = ln((1+sinc)/(1-sinc)) = ln(1+sinc)-ln(1-sinc)#
Now use #d/(dc) (lnu) = 1/u (du)/(dc)# (chain rule) to get
#f'(c) = 1/(1+sinc) * (cosc) - 1/(1-sinc) (-cosc)#
# = cosc/(1+sinc) + cosc/(1-sinc)#
# = (2cosc)/(1-sin^2c)#
# = 2/cosc = 2sec c#.
