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Math Help - logarithmic differentiation

  1. #1
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    logarithmic differentiation

    use logarithmic differential to find dy/dx, where
    y = \frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    y' = y*________


    sofar:
    ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    = -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3

    = \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4}{cos(x)} -2\frac{2x+2}{x^2+2x} + 3x^2

    not sure what to do next
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    use logarithmic differential to find dy/dx, where
    y = \frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    y' = y*________


    sofar:
    ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    = -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3

    = \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4}{cos(x)} -2\frac{2x+2}{x^2+2x} + 3x^2

    not sure what to do next
    first off, that is incorrect. (your differentiated  \ln (\sin x ) incorrectly)

    when you get the correct solution, just multiply both sides by y. you can replace y with its function of x if you wish

    EDIT: oh, you dont have to write y as a function of x, look at the way they expect the answer, y*_______. so it would be y times whatever you got the derivative to be
    Last edited by Jhevon; May 31st 2007 at 07:12 PM.
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  3. #3
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    i think this is correct so far:

    ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    = -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3

    = \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -2\frac{2x+2}{x^2+2x} + 3x^2

    =  y\frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    i think this is correct so far:

    ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}

    = -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3

    = \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -2\frac{2x+2}{x^2+2x} + 3x^2

    =  y\frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2
    no it isn't. it's y times everything on the other side.

     y \left( \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2 \right)

    we multiplied both sides by y, you can't just multiply one term on the right. what you see in brackets is what should go in the space provided
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  5. #5
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    thanks, forgot to put in the brackets
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