F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

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- Mar 2nd 2009, 10:39 PM #1

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- Mar 2nd 2009, 10:45 PM #2

- Mar 2nd 2009, 10:51 PM #3

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- Mar 2nd 2009, 11:02 PM #4

- Mar 2nd 2009, 11:37 PM #5

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Keeping the function as a quotient and using quotient rule also works

$\displaystyle u(x) = (1+\cos x)^{\frac{1}{2}}$

$\displaystyle v(x) = (1-\cos x)^{\frac{1}{2}}$

$\displaystyle f'(x) = \frac{vu' - uv'}{v^2}$

Just remember the chain rule when differentiating u and v

- Mar 2nd 2009, 11:40 PM #6

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y=(-(1 - cos x)^(1/2)*1/2*(1 + cos x)^(-1/2)*sin x - (1 + cos x)^(1/2)*1/2*(1 - cos x)^(-1/2)*sin x)/((1 - cos x)^(1/2)^2)

I calculated it using a program. I would use the chain rule and quotient rule. But, I'm not sure if there's an identity you can use...

By the way, how do you guys write the equations using the fancy d/dx and stuff...

- Mar 2nd 2009, 11:47 PM #7

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- Mar 3rd 2009, 02:03 AM #8
Read what's in this subforum: http://www.mathhelpforum.com/math-help/latex-help/

- Mar 3rd 2009, 05:39 AM #9

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