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Math Help - I missed a class on one of the Derivative Rules.

  1. #1
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    I missed a class on one of the Derivative Rules.

    Just explain how one would go about doing this:




    What's the rule to this?
    Attached Thumbnails Attached Thumbnails I missed a class on one of the Derivative Rules.-missed-class.jpg  
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    Quote Originally Posted by Zanderist View Post
    Just explain how one would go about doing this:




    What's the rule to this?
    You need to use the Quotient Rule:

    If f(x) = \frac{g(x)}{h(x)} then f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.
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  3. #3
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    to solve that you would use th quotient rules as well as the sin and cos rules.

     <br />
\frac{dy}{dx}\sin{x}=\cos{x}<br />
     <br />
\frac{dy}{dx}\cos{x}=-\sin{x}<br />

    so using the quotient rule, \frac{dy}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^2}

     <br />
u=4\sin{x}<br />
     <br />
v=2+\cos{x}<br />
     <br />
u'=4\cos{x}<br />
     <br />
v'=-\sin{x}<br />

    fill it all in and solve
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Zanderist View Post
    Just explain how one would go about doing this:




    What's the rule to this?
    this would be:

    4cosx(2+cosx) - 4sinx(-sinx)
    ----------------------------
    (2+cosx)^2

    You can simplify even if you miss the entire semester. Go for it!!
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    Quote Originally Posted by harish21 View Post
    this would be:

    4cosx(2+cosx) - 4sinx(-sinx)
    ----------------------------
    (2+cosx)^2

    You can simplify even if you miss the entire semester. Go for it!!
    Yes, now simplify it.
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    tan(x)-5
    --------
    sec(x)

    This some how turns into,

    -(tan(x)^2-5*tan(x)-sec(x)^2)
    ----------------------------------
    sec(x)

    I don't see how the quotient rule gives me that.
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  7. #7
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    Quote Originally Posted by Zanderist View Post
    tan(x)-5
    --------
    sec(x)

    This some how turns into,

    -(tan(x)^2-5*tan(x)-sec(x)^2)
    ----------------------------------
    sec(x)

    I don't see how the quotient rule gives me that.
    I would actually have done:

    \frac{8\cos{x} + 4\cos^2{x} + 4\sin^2{x}}{(2 + \cos{x})^2}

     = \frac{8\cos{x} + 4(\cos^2{x} + \sin^2{x})}{(2 + \cos{x})^2}

     = \frac{8\cos{x} + 4}{(2 + \cos{x})^2}

     = \frac{4(2\cos{x} + 1)}{(2 + \cos{x})^2}.
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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Zanderist View Post
    tan(x)-5
    --------
    sec(x)

    This some how turns into,

    -(tan(x)^2-5*tan(x)-sec(x)^2)
    ----------------------------------
    sec(x)

    I don't see how the quotient rule gives me that.
    tan(x)-5 / secx = (sinx - 5cosx) [ Reason: tanx=sinx/cosx and secx=1/cosx]

    Derivate this and you get cosx + 5 sinx
    Last edited by harish21; February 28th 2010 at 08:56 AM. Reason: mistake in the previous post
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