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anonymous

  • one year ago

IS ANYOME GOOD AT FINDING DERIATIVES FOR LN?! Please I need help as soon as possible! :(

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  1. alekos
    • one year ago
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    post it up

  2. anonymous
    • one year ago
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    Ok

  3. anonymous
    • one year ago
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    y=ln(1-x/(x+2)^5)

  4. alekos
    • one year ago
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    \[y=\ln [1 -\frac{ x }{(x+2)^{5}}]\] Is that it?

  5. anonymous
    • one year ago
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    No sorry my computer is messed up, the 1-x is the numerator, and the denominator is as you out it. Ln is correct too

  6. anonymous
    • one year ago
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    Put*

  7. anonymous
    • one year ago
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    @alekos

  8. IrishBoy123
    • one year ago
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    \(\large y=ln \frac{1-x}{(x+2)^5} \) assuming that you know that \(\large (log \ u(x))' = \frac{1}{u} u'\) then look also at \(\large ln \frac{a}{b} = ln \ a - ln \ b\) and similar properties of logs; and maybe split this up a bit before you start...

  9. alekos
    • one year ago
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    yeah, thanks Irish Boy. I reckon do the properties of logs.... Ln(1-x) - Ln(x+2)^5 = Ln(1-x) - 5Ln(x+2) the rest is easy because the derivative of Lnx is 1/x

  10. IrishBoy123
    • one year ago
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    @alekos look good

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