## anonymous 4 years ago compute the derivative y=ln((x+1)^2(3x-1)^3)

1. anonymous

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

2. slaaibak

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

3. anonymous

ok I see

4. slaaibak

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

5. anonymous

so would that be the answer

6. slaaibak

yes

7. anonymous

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

8. slaaibak

yes

9. anonymous

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

10. slaaibak

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

11. anonymous

thanks @slaaibak