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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I just need to know which of those choices matches my answer.
You got your answer using WolframAlpha? lol.

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Other answers:

Do you know how to use logarithmic differentiation?
Yes. It always gives it in a different form, but awesome to check your answer.
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; \[\dfrac{y'}{y} = \cdots~?\]
\[\dfrac{\mathrm d}{\mathrm dx}~~~5\ln (3x+1)+ 6\ln(x^4-2)=~?\]
Let me see :)
Yeah just use WolframAlpha.
ok http://www.wolframalpha.com/input/?i=d%2Fdx+5ln%283x%2B1%29%2B6ln%28x%5E4-2%29
Hello?
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 \[y' = \dfrac{15~\left((3x+1)^5~(x^4-2)^6\right)}{1+3 x}+\dfrac{24~ x^3~\left((3x+1)^5~(x^4-2)^6\right)}{x^4-2}\\~\\~\\\phantom{y'} = \boxed{15~ (3x+1)^4~(x^4-2)^6+24~ x^3~(3x+1)^5~(x^4-2)^5}\]
Thanks so much you will never know.

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