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AmTran_Bus

  • one year ago

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

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  1. AmTran_Bus
    • one year ago
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  2. AmTran_Bus
    • one year ago
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    I just need to know which of those choices matches my answer.

  3. geerky42
    • one year ago
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    You got your answer using WolframAlpha? lol.

  4. geerky42
    • one year ago
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    Do you know how to use logarithmic differentiation?

  5. AmTran_Bus
    • one year ago
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    Yes. It always gives it in a different form, but awesome to check your answer.

  6. geerky42
    • one year ago
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    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
    • one year ago
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    \[\dfrac{\mathrm d}{\mathrm dx}~~~5\ln (3x+1)+ 6\ln(x^4-2)=~?\]

  8. AmTran_Bus
    • one year ago
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    Let me see :)

  9. geerky42
    • one year ago
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    Yeah just use WolframAlpha.

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

  11. AmTran_Bus
    • one year ago
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    Hello?

  12. geerky42
    • one year ago
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    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
    • one year ago
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    Thanks so much you will never know.

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spraguer (Moderator)
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