AmTran_Bus
  • AmTran_Bus
My bus is going to quit if it does not get some help.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AmTran_Bus
  • AmTran_Bus
AmTran_Bus
  • AmTran_Bus
I just need to know which of those choices matches my answer.
geerky42
  • geerky42
You got your answer using WolframAlpha? lol.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

geerky42
  • geerky42
Do you know how to use logarithmic differentiation?
AmTran_Bus
  • AmTran_Bus
Yes. It always gives it in a different form, but awesome to check your answer.
geerky42
  • 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; \[\dfrac{y'}{y} = \cdots~?\]
geerky42
  • geerky42
\[\dfrac{\mathrm d}{\mathrm dx}~~~5\ln (3x+1)+ 6\ln(x^4-2)=~?\]
AmTran_Bus
  • AmTran_Bus
Let me see :)
geerky42
  • geerky42
Yeah just use WolframAlpha.
AmTran_Bus
  • AmTran_Bus
ok http://www.wolframalpha.com/input/?i=d%2Fdx+5ln%283x%2B1%29%2B6ln%28x%5E4-2%29
AmTran_Bus
  • AmTran_Bus
Hello?
geerky42
  • 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 \[y' = \dfrac{15~\left((3x+1)^5~(x^4-2)^6\right)}{1+3 x}+\dfrac{24~ x^3~\left((3x+1)^5~(x^4-2)^6\right)}{x^4-2}\\~\\~\\\phantom{y'} = \boxed{15~ (3x+1)^4~(x^4-2)^6+24~ x^3~(3x+1)^5~(x^4-2)^5}\]
AmTran_Bus
  • AmTran_Bus
Thanks so much you will never know.

Looking for something else?

Not the answer you are looking for? Search for more explanations.