Question

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

Answer 1

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

Answer 2

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

Answer 3

ok I see

Answer 4

or easier: split it up into the following

\[2\ln(x+1) + 3\ln(3x-1)\]

then derive:

\[{2 \over {x+1}} +{ 9 \over 3x-1}\]

Answer 5

so would that be the answer

Answer 6

yes

Answer 7

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

Answer 8

yes

Answer 9

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

Answer 10

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

Answer 11

thanks @slaaibak