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Math Help - Differentiate cosecx

  1. #1
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    Differentiate cosecx

    Prove that d/dx (cosecx) = -cosecxcotx
    Last edited by nath_quam; June 16th 2006 at 09:51 PM.
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  2. #2
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    Quote Originally Posted by nath_quam
    Prove that d/dx (cosecx) = -cosecxcotx
    Hint: start with this
    cscx = 1/sinx

    -Dan
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  3. #3
    Grand Panjandrum
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    What would you like us to do with the attachment?

    RonL
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  4. #4
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    For part (a) integrate dI and then using your initial value solve for the constant of integration.
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  5. #5
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    Since,
    \csc x=\frac{1}{\sin x}
    Then, when you take its derivative apply quotient rule,
    (\csc x)'=\frac{(1)'\sin x+1(\sin x)'}{\sin^2 x}
    Then,
    (\csc x)'=\frac{0\cdot \sin x-1\cdot \cos x}{\sin^2 x}=\frac{-\cos x}{\sin ^2x}= \frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot x\sin x
    ---
    For reference,
    If f(x) is a diffrenciable function at x=c and g(x) is a diffrenciable function at x=c such as, g(c)\not = 0, then f(x)/g(x) is diffrenciable at x=c and its derivative at x=cis,
    \frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}
    Last edited by ThePerfectHacker; June 13th 2006 at 01:51 PM.
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  6. #6
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    Quotient Rule

    Perfect Hacker shouldn't the quotient rule have a minus in the middle not a plus as you have it Thanks Nath and how would i go about doing the three parts of the attachment
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  7. #7
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    Quote Originally Posted by nath_quam
    not a plus as you have it
    A plus!
    Where do you see a plus?






    (Okay I admit I edited my message, you are right)
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