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Math Help - use logarthmic differntiation to find dy/dx

  1. #1
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    use logarthmic differntiation to find dy/dx

    y=[(x+2)*√(1-x)]/[4x]

    i got as far as

    y'/y=ln(x+2)+1/2ln(1-x)-3ln4x

    im not sure if the part is correct since the cubed is only on the x and not the 4.
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  2. #2
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    Quote Originally Posted by johnnyc418 View Post
    y=[(x+2)*√(1-x)]/[4x]

    i got as far as

    y'/y=ln(x+2)+1/2ln(1-x)-3ln4x

    im not sure if the part is correct since the cubed is only on the x and not the 4.
    y = \frac{(x+2) \sqrt{1-x^2}}{4x^3}

    note that ln(4x^3) can be simplified to ln(4)+ln(x^3) = 2ln(2)+3ln(x)

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

    \frac{y'}{y} = \frac{1}{x+2} - \frac{2x}{2(1-x)^2} - \frac{3}{x}
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  3. #3
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    alright so knowing that, would the answer be

    y'=([(x+2)*√(1-x)]/[4x])*([1/x+2]-[x/1-x]-3/x]

    or is there any simplifying i could do it it. thanks
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by johnnyc418 View Post
    alright so knowing that, would the answer be

    y'=([(x+2)*√(1-x)]/[4x])*([1/x+2]-[x/1-x]-3/x]

    or is there any simplifying i could do it it. thanks
    Indeed it would be, you might be able to work on it but I wouldn't bother
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