hooverst
compute the derivative
y=ln((x+1)^2(3x1)^3)



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hooverst
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\[y=\ln((x+1)^2(3x1)^3)\]

slaaibak
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\[{1 \over (x+1)^2 (3x1)^3} \times [2\times(x+1)(3x1)^3 + 3 \times 3 \times (3x1)^2(x+1)^2] \]

hooverst
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ok I see

slaaibak
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or easier: split it up into the following
\[2\ln(x+1) + 3\ln(3x1)\]
then derive:
\[{2 \over {x+1}} +{ 9 \over 3x1}\]

hooverst
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so would that be the answer

slaaibak
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yes

hooverst
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do you know how to do \[y=x \sqrt{6}\]

slaaibak
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yes

hooverst
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can you show me how to do it? if you have time.

slaaibak
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well, since \[\sqrt6\]
is a constant, the answer would be
\[\sqrt6\]

hooverst
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thanks @slaaibak