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AmTran_Bus

  • one year ago

I disagree with my textbook. Need help

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  1. anonymous
    • one year ago
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    whatever it is, i am on your side

  2. AmTran_Bus
    • one year ago
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  3. AmTran_Bus
    • one year ago
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    So on the pic, under the POWER RULE step

  4. anonymous
    • one year ago
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    can't see it though, screen shot is bad

  5. AmTran_Bus
    • one year ago
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    Where did they take in account for the - 1/2 exponent that was supposed to be there when they took derivative?

  6. AmTran_Bus
    • one year ago
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    Its working for me @satellite73

  7. AmTran_Bus
    • one year ago
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    Can anyone see it?

  8. anonymous
    • one year ago
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    what book is it?

  9. AmTran_Bus
    • one year ago
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    Thomas' calculus Alternete edition

  10. AmTran_Bus
    • one year ago
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    can anyone see the pic?

  11. pooja195
    • one year ago
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    cant find the book sorry!! :( Try taking a better pic.

  12. AmTran_Bus
    • one year ago
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    Gurr. Give me time and I will type it out

  13. AmTran_Bus
    • one year ago
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    \[\ln (x^2+1)+\ln(x+3)^{1/2}-\ln(x-1)\] Then \[\ln (x^2+1)+1/2 \ln (x+3)- \ln (x-1)\]

  14. anonymous
    • one year ago
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    The idea is \[ \ln\left((x+3)^{1/2}\right) = \frac 12 \ln(x+3) \]

  15. AmTran_Bus
    • one year ago
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    How?

  16. AmTran_Bus
    • one year ago
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    http://www.wolframalpha.com/input/?i=ln+*+%28x%2B3%29%5E%281%2F2%29

  17. anonymous
    • one year ago
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    \[\begin{array}{rcl} \ln\left((x+3)^{1/2}\right) &=& y\\ (x+3)^{1/2} &=& e^y\\ x+3 &=& (e^y)^2\\ x+3 &=& e^{2y} \\ \ln(x+3) &=& 2y \\ \frac 12 \ln(x+3) &=& y \\ \ln\left((x+3)^{1/2}\right) &=& \frac 12 \ln(x+3) \end{array} \]

  18. AmTran_Bus
    • one year ago
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    Ok. Thanks. Im satisfied now

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is replying to Can someone tell me what button the professor is hitting...

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