How do you differentiate #y=(sinx/(1+cosx))^2#?
1 Answer
May 5, 2017
This can be simplified to
#y = sin^2x/(1 + cosx)^2#
#y = (1 - cos^2x)/(1 + cosx)^2#
#y = ((1 + cosx)(1 - cosx))/(1 + cosx)^2#
#y = (1 - cosx)/(1 + cosx)#
Now use the quotient rule.
#y' = (sinx(1 + cosx) - (1 - cosx)-sinx)/(1 + cosx)^2#
#y' = (sinx + sinxcosx - (-sinx + sinxcosx))/(1 + cosx)^2#
#y' = (sinx + sinxcosx + sinx - sinxcosx)/(1 + cosx)^2#
#y' = (2sinx)/(1 + cosx)^2#
Hopefully this helps!
