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anonymous

  • 5 years ago

what is the derivative of (lnx^2)/(x+1)^(1/2)

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  1. amistre64
    • 5 years ago
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    \[D_x(\frac{\ln(x)}{(x+1)^2})\]

  2. amistre64
    • 5 years ago
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    this thing? and whered my fraction bar end up :)

  3. amistre64
    • 5 years ago
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    use quotient rule; or bring the botom up with a negative exponent and use product rule

  4. amistre64
    • 5 years ago
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    murmuring..... ^(1/2) lol

  5. amistre64
    • 5 years ago
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    bt'-b't f'(x) = ----- b^2

  6. amistre64
    • 5 years ago
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    b = (x+1)^(1/2) ; b' = 1/[2(x+1)^(1/2)] t = ln(x) ; t' = 1/x

  7. amistre64
    • 5 years ago
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    b^2 = (x+1) :) that one was easy lol

  8. amistre64
    • 5 years ago
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    \[\frac{\frac{\sqrt{x+1}}{x} - \frac{\ln(x)}{2 \sqrt{x+1}}}{(x+1)}\] simplify as needed

  9. amistre64
    • 5 years ago
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    2x + 2 - x ln(x) --------------- 2x(x+1) sqrt(x+1)

  10. anonymous
    • 5 years ago
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    amistre64, In the following equation, Ln is spelled Log. \[\int\limits \frac{\frac{\sqrt{1+x}}{x}-\frac{\text{Log}[x]}{2 \sqrt{1+x}}}{1+x} \, dx = \frac{\text{Log}[x]}{\sqrt{1+x}} \] It looks to me like the Ln of x was raised to the second power in the original problem statement above. Perhaps by eyes are failing me in my old age.

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