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Math Help - Use logarithmic differentiation to find dy/dx, where

  1. #1
    Junior Member
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    Use logarithmic differentiation to find dy/dx, where

    y = (x-4)^-3(sin(x))^4 / (x^2-2x)^2(e^x^3)
    y'= y*____________

    I set it up as:
    sin^4(x) / (x-4)^3(x^2-2x)^2(e^x^3)

    Take the ln of both sides:
    ln y= 4sin^3(x)cos(x) / ((3(x-4)^2)(2(x^2-2x))(2x-1)(3x))

    But that's wrong. What do I do with this?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Use ln(a/b) = lna - lnb

    and ln(a^b) = bln(a)



    y = (x-4)^-3(sin(x))^4 / (x^2-2x)^2(e^x^3)

    lny = [-3sin(x)] ln(x-4) -[ e^x^3] ln (x^2 -2x)

    Now differentiate:
    y '/y = etc....
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