Posted by theo on Sunday, July 4, 2010 at 5:43pm.
assume you mean
h = [cos (x)/(1+sin x)]^4
let z = [cos (x)/(1+sin x)]
then
dh/dz = 4 z^3
dh/dx = dh/dz * dz/dx
dh/dx = 4 [cos (x)/(1+sin x)]^3 d/dx[[cos (x)/(1+sin x)]
d/dx[cos (x)/(1+sin x)] = [(1+sin x)(-sin x) -cos^2x] / (1+sin x)^2
= [-sin x -1]/(1+sin x)^2 = -1/(1+sin x)
so
dh/dx = 4 [cos (x)/(1+sin x)]^3 (-1)/(1+sin x)
= -4 cos x/ (1+sin x)^4
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