## AmTran_Bus one year ago My bus is going to quit if it does not get some help.

1. AmTran_Bus

2. AmTran_Bus

I just need to know which of those choices matches my answer.

3. geerky42

4. geerky42

Do you know how to use logarithmic differentiation?

5. AmTran_Bus

Yes. It always gives it in a different form, but awesome to check your answer.

6. geerky42

You just need to go ahead and apply logarithmic differentiation. $\ln y = 5\ln (3x+1)+ 6\ln(x^4-2)$Now taking derivative of both sides; $\dfrac{y'}{y} = \cdots~?$

7. geerky42

$\dfrac{\mathrm d}{\mathrm dx}~~~5\ln (3x+1)+ 6\ln(x^4-2)=~?$

8. AmTran_Bus

Let me see :)

9. geerky42

Yeah just use WolframAlpha.

10. AmTran_Bus
11. AmTran_Bus

Hello?

12. geerky42

Now multiply both sides by y. Then we can replace y to whether it equals to; $y' = \dfrac{15~y}{1+3 x}+\dfrac{24~ x^3~y}{x^4-2}$Since we are given that $y = (3x+1)^5~(x^4-2)^6$ Using our simple algebra skill, we have $y' = \dfrac{15~\left((3x+1)^5~(x^4-2)^6\right)}{1+3 x}+\dfrac{24~ x^3~\left((3x+1)^5~(x^4-2)^6\right)}{x^4-2}\\~\\~\\\phantom{y'} = \boxed{15~ (3x+1)^4~(x^4-2)^6+24~ x^3~(3x+1)^5~(x^4-2)^5}$

13. AmTran_Bus

Thanks so much you will never know.