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Math Help - Need Help Identifying Technique

  1. #1
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    Need Help Identifying Technique

    Hi everyone! I just got out of class and was reviewing my notes. In one of the example exercises, I found a technique that I'm not familiar with. Here's the example question:

     f(x) = \frac{x^8 \cos ^3  x}{\sqrt{x-1}}

     \ln y = \ln (\frac{x^8 \cos ^3  x}{(x-1)^\frac{1}{2}}})

     \ln y = \ln (x^8 \cos ^3x) - \ln(x-1)^\frac{1}{2}

     \ln y = \ln(x^8) + \ln (\cos ^3x) - \frac{1}{2} \ln(x-1)

     \ln y = 8\ln x + 3\ln \cos x - \frac{1}{2} \ln(x-1)

     \frac{1}{y} \frac{dy}{dx} = \frac{8}{x} + \frac{3}{\cos x}  \sin x - \frac{1}{2} \frac{1}{(x-1)}

     \frac{dy}{dx} = \frac{x^8 \cos ^3x}{\sqrt{x-1}} (\frac{8}{x} - \frac{3 \sin x}{\cos x} - \frac{1}{2(x-1)})

    The problem that I have is in the final step (this also isn't fully simplified, but I didn't have time to type the whole thing since I just taught myself LaTeX in the past 20 minutes ). Where does the

     \frac{x^8 \cos ^3x}{\sqrt{x-1}}

    come from? I heard the word "substitution" mentioned, but that's all I know. An explanation would be greatly appreciated! Also, would it be possible for you to show a minor amount of work to help explain the technique?


    Thanks a lot!
    Last edited by Algebrah; October 14th 2011 at 12:10 PM. Reason: Error In Step 2
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  2. #2
    MHF Contributor

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    Tejas
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    Re: Need Help Identifying Technique

    that's y!

    in the step before, on the left, you had:

    (1/y)(dy/dx) = stuff, so...

    dy/dx = (y)(stuff)
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  3. #3
    Junior Member
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    Re: Need Help Identifying Technique

    Oh! That's so obvious!

    Thanks a lot!

    :manythanks:
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