Question

My bus is going to quit if it does not get some help.

Answer 1

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

Answer 2

You got your answer using WolframAlpha? lol.

Answer 3

Do you know how to use logarithmic differentiation?

Answer 4

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

Answer 5

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;
\[\dfrac{y'}{y} = \cdots~?\]

Answer 6

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

Answer 7

Let me see :)

Answer 8

Yeah just use WolframAlpha.

Answer 9

ok http://www.wolframalpha.com/input/?i=d%2Fdx+5ln%283x%2B1%29%2B6ln%28x%5E4-2%29

Answer 10

Hello?

Answer 11

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 \[y' = \dfrac{15~\left((3x+1)^5~(x^4-2)^6\right)}{1+3 x}+\dfrac{24~ x^3~\left((3x+1)^5~(x^4-2)^6\right)}{x^4-2}\\~\\~\\\phantom{y'} = \boxed{15~ (3x+1)^4~(x^4-2)^6+24~ x^3~(3x+1)^5~(x^4-2)^5}\]

Answer 12

Thanks so much you will never know.