Sunday

November 23, 2014

November 23, 2014

Posted by **theo** on Sunday, July 4, 2010 at 5:43pm.

- calculus -
**Damon**, Sunday, July 4, 2010 at 6:02pmassume 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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