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Math Help - help in differentiatinr :(

  1. #1
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    help in differentiatinr :(

    F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Blush View Post
    F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

    are u asking for F' ?
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  3. #3
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    yeah.. and thanx alot for being concern n for all ur answers
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  4. #4
    MHF Contributor matheagle's Avatar
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    {d\over dx}\biggl((1+\cos x)^{.5} (1-\cos x)^{-.5}\biggr)

    =.5(1+\cos x)^{-.5}(-\sin x) (1-\cos x)^{-.5}+(1+\cos x)^{.5}(-.5)(1-\cos x)^{-1.5}(\sin x)
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  5. #5
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    Keeping the function as a quotient and using quotient rule also works

    u(x) = (1+\cos x)^{\frac{1}{2}}
    v(x) = (1-\cos x)^{\frac{1}{2}}

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

    Just remember the chain rule when differentiating u and v
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  6. #6
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    Quote Originally Posted by Blush View Post
    F(x) = {(1+Cosx)^0.5} / {(1-Cosx)^0.5}

    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...
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  7. #7
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    Quote Originally Posted by econstudent View Post


    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...
    \frac{d}{dx}\left(f(x)\right) just means the differential of f(x) with respect to x
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  8. #8
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    Quote Originally Posted by econstudent View Post
    [snip]

    By the way, how do you guys write the equations using the fancy d/dx and stuff...
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Haha, I completely misinterpreted what he was asking!
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