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Help With Derivative

  1. Jun 19, 2006 #1
    Hi, can someone please tell me what I'm doing wrong.

    [tex] y = ln[x^4 +3x +1][/tex]

    [tex] y' = \frac {1}{x^4 +3x +1} 4x^3 + 3[/tex]

    =[tex] \frac {4x^3 + 3}{x^4 +3x +1}[/tex]
     
  2. jcsd
  3. Jun 19, 2006 #2

    StatusX

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    Homework Helper

    Nothing. Except for you forced me to write this sentence because "Nothing" by itself is too short a message to post, but that's not really your fault.
     
  4. Jun 19, 2006 #3
    Hmm, I was told it was wrong. Maybe because I skipped that middle step when I showed it to someone.
     
  5. Jun 19, 2006 #4
    :biggrin: Well, you might want to include some parentheses on that second line:

    [tex]\begin{gathered}
    y = \ln \left( {x^4 + 3x + 1} \right) \hfill \\
    y' = \frac{1}{{x^4 + 3x + 1}}\left( {4x^3 + 3} \right)\;\;\;\;\;\; \leftarrow parentheses\;{\text{:D}} \hfill \\
    y' = \frac{{4x^3 + 3}}{{x^4 + 3x + 1}} \hfill \\
    \end{gathered} [/tex]
     
  6. Jun 19, 2006 #5
    Ok, Thanks Guys.

    Maybe you can check this for me, Since I don't have the answer.

    [tex] xy^3 + x - y + 21 = 0[/tex]

    [tex] 1y^3 + x3y^2 * y' + 1 - y' = 0 [/tex]

    [tex] y'[3xy^2 - 1] = -y^3 - 1 [/tex]

    [tex] y' = \frac {-y^3 - 1}{3xy^2 - 1}[/tex]
     
  7. Jun 20, 2006 #6

    HallsofIvy

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    Looks good to me.
     
  8. Jun 20, 2006 #7
    :biggrin: Just as in my post #4, here's another nitpick:

    ~You can omit some signs in your answer, writing it as:
    [tex] y' = \frac {y^3 + 1}{1-3xy^2}[/tex]

    (Looks cleaner, I think)
     
    Last edited: Jun 20, 2006
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