anonymous
  • anonymous
compute the derivative y=ln((x+1)^2(3x-1)^3)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y=\ln((x+1)^2(3x-1)^3)\]
slaaibak
  • slaaibak
\[{1 \over (x+1)^2 (3x-1)^3} \times [2\times(x+1)(3x-1)^3 + 3 \times 3 \times (3x-1)^2(x+1)^2] \]
anonymous
  • anonymous
ok I see

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slaaibak
  • slaaibak
or easier: split it up into the following \[2\ln(x+1) + 3\ln(3x-1)\] then derive: \[{2 \over {x+1}} +{ 9 \over 3x-1}\]
anonymous
  • anonymous
so would that be the answer
slaaibak
  • slaaibak
yes
anonymous
  • anonymous
do you know how to do \[y=x \sqrt{6}\]
slaaibak
  • slaaibak
yes
anonymous
  • anonymous
can you show me how to do it? if you have time.
slaaibak
  • slaaibak
well, since \[\sqrt6\] is a constant, the answer would be \[\sqrt6\]
anonymous
  • anonymous
thanks @slaaibak

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