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anonymous

  • one year ago

Find the derivative

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  1. anonymous
    • one year ago
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    \[f(x)=\sqrt{\left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)}+1}\]

  2. anonymous
    • one year ago
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    It's a nasty one.

  3. zepdrix
    • one year ago
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    yikes 0_o

  4. anonymous
    • one year ago
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    What I did was natural log both sides, but then I get stuck as I can't proceed with the logarithmic differentiation.

  5. anonymous
    • one year ago
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    \[\ln(y)=\frac{ 1 }{ 2 }\ln \left( \left( \frac{ x }{ x^2+3 } \right)^{\arcsin(x^3)} +1\right)\]

  6. anonymous
    • one year ago
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    Now what?

  7. zepdrix
    • one year ago
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    Hmm this might seem strange, but I would recommend isolating the big chunk of x stuff before logging, because ya, that +1 is causing a problem.

  8. zepdrix
    • one year ago
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    \[\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]

  9. anonymous
    • one year ago
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    Wolfram wont even use this method.

  10. anonymous
    • one year ago
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    Ohh okay so you're manipulating the inside. So what?

  11. zepdrix
    • one year ago
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    I'm just thinking that logging from THAT point might be easier:\[\large\rm \ln(y^2-1)=\arcsin(x^3)\cdot \ln\left(\frac{x}{x^2+3}\right)\]Hmm

  12. anonymous
    • one year ago
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    How is that equivalent?

  13. zepdrix
    • one year ago
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    hmm? :o

  14. anonymous
    • one year ago
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    Like how is what I had before and what you just got equivalent?

  15. zepdrix
    • one year ago
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    Did you not see the steps I took? :o I can explain them again maybe :D

  16. anonymous
    • one year ago
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    Now I perfectly get what you did but how is what I had and what you had equivalent?

  17. zepdrix
    • one year ago
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    Hmm that question is confusing... All I did was algebraic manipulation, that shouldn't change anything.

  18. anonymous
    • one year ago
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    So ln(y^2) = ln(y^2-1)?

  19. zepdrix
    • one year ago
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    no?

  20. anonymous
    • one year ago
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    That is what you wrote though?

  21. zepdrix
    • one year ago
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    \[\large\rm \ln(y)=\frac{1}{2}\ln \left(\left( \frac{x}{x^2+3}\right)^{\arcsin(x^3)} +1\right)\qquad \text{your equation}\]\[\large\rm \ln(y^2-1)= \ln\left[\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\right]\qquad \text{my equation}\]You see that the right side is missing the +1, yes?

  22. anonymous
    • one year ago
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    I do. I suppose I am struggling to see the equivalence because I'm used to seeing it without logs.

  23. zepdrix
    • one year ago
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    Well, if you ignore the logs for a sec, you understand how I went from here,\[\large\rm y=\sqrt{\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}+1}\]to here?\[\large\rm y^2-1=\left(\frac{x}{x^2+3}\right)^{\arcsin(x^3)}\]

  24. anonymous
    • one year ago
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    Yep I got that.

  25. anonymous
    • one year ago
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    Ohh then you just logged both sides. Gotcha.

  26. anonymous
    • one year ago
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    ooooo. K I get this now.

  27. anonymous
    • one year ago
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    Thanks!

  28. zepdrix
    • one year ago
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    >.< heh

  29. imqwerty
    • one year ago
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    :)

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