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

Mathematics
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\[y=\ln((x+1)^2(3x-1)^3)\]
\[{1 \over (x+1)^2 (3x-1)^3} \times [2\times(x+1)(3x-1)^3 + 3 \times 3 \times (3x-1)^2(x+1)^2] \]
ok I see

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Other answers:

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

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