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hooverst

  • 2 years ago

compute the derivative y=ln((x+1)^2(3x-1)^3)

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  1. hooverst
    • 2 years ago
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    \[y=\ln((x+1)^2(3x-1)^3)\]

  2. slaaibak
    • 2 years ago
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    \[{1 \over (x+1)^2 (3x-1)^3} \times [2\times(x+1)(3x-1)^3 + 3 \times 3 \times (3x-1)^2(x+1)^2] \]

  3. hooverst
    • 2 years ago
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    ok I see

  4. slaaibak
    • 2 years ago
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    or easier: split it up into the following \[2\ln(x+1) + 3\ln(3x-1)\] then derive: \[{2 \over {x+1}} +{ 9 \over 3x-1}\]

  5. hooverst
    • 2 years ago
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    so would that be the answer

  6. slaaibak
    • 2 years ago
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    yes

  7. hooverst
    • 2 years ago
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    do you know how to do \[y=x \sqrt{6}\]

  8. slaaibak
    • 2 years ago
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    yes

  9. hooverst
    • 2 years ago
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    can you show me how to do it? if you have time.

  10. slaaibak
    • 2 years ago
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    well, since \[\sqrt6\] is a constant, the answer would be \[\sqrt6\]

  11. hooverst
    • 2 years ago
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    thanks @slaaibak

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