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champ

  • 5 years ago

what is the derivative of x^x2

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  1. anonymous
    • 5 years ago
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    is the second x raised to the 2?

  2. anonymous
    • 5 years ago
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    i think its x(2x)

  3. anonymous
    • 5 years ago
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    hmm, perhaps x^2x..

  4. anonymous
    • 5 years ago
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    This is a bit tricky and requires a little bit of thinking. Firstly, to clarify.. you wand \(\frac{d}{dx}x^{x^2}\)

  5. anonymous
    • 5 years ago
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    Err want.

  6. anonymous
    • 5 years ago
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    Right?

  7. anonymous
    • 5 years ago
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    yeah, i think that's what he means, because had it been multilication he/she would have just put the two in front..

  8. anonymous
    • 5 years ago
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    Are you going to solve this polpak? :)

  9. anonymous
    • 5 years ago
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    Crazy! How's this look? <a href=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d-7291-4b5e-be30-d944f1a0ea34/2011-04-22_1313.png "><img class="embeddedObject" src=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d-7291-4b5e-be30-d944f1a0ea34/2011-04-22_1313.png " width="596" height="653" border="0" /></a>

  10. anonymous
    • 5 years ago
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    Sure. The original poster seems to have gone away, but it's a fun exercise in anycase. Let \(u = x^{x^2} \implies (ln\ u) = x^2(ln\ x)\) \[\frac{d}{dx}(ln\ u) = \frac{d}{dx}x^2(ln\ x)\] \[\implies \frac{d}{du}(ln\ u) * \frac{du}{dx} = x^2*\frac{d}{dx}(ln\ x) + (ln\ x)*\frac{d}{dx}(x^2)\] \[\implies \frac{1}{u} * \frac{du}{dx} = \frac{x^2}{x} + 2x(ln\ x) \] \[\implies \frac{du}{dx} = u(1 + 2x(ln\ x))\] \[ = x^{x^2}(1 + 2x(ln\ x)) = x^{x^2} + 2x^{x^2+1}(ln\ x)\] \[\implies \frac{d}{dx}x^{x^2} = x^{x^2} + 2x^{x^2+1}(ln\ x)\]

  11. anonymous
    • 5 years ago
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    i dont uderstand how you got to ln

  12. anonymous
    • 5 years ago
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    I took the ln(u) to get the x^2 out of the exponent.

  13. anonymous
    • 5 years ago
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    If \(u = x^{x^2}\) then \(ln(u) = x^2ln(x)\)

  14. anonymous
    • 5 years ago
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    did you use a formula on the rest of it am kinda lost

  15. anonymous
    • 5 years ago
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    No formula. Just the chain rule and the product rule.

  16. myininaya
    • 5 years ago
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  17. anonymous
    • 5 years ago
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    The right side of the equation on line 3 is the chain rule, the left is the product rule.

  18. myininaya
    • 5 years ago
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    you made a small error x^2/x=x not 1

  19. anonymous
    • 5 years ago
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    Ah, so I did.

  20. anonymous
    • 5 years ago
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    I had the 1 there because I was gonna factor the x from both terms, then forgot to do it ;p

  21. myininaya
    • 5 years ago
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    the equation still holds if polak takes the natural log of both sides then you use implicit differentiation to solve for y'

  22. anonymous
    • 5 years ago
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    so when trying to deferentiate x^2*lnx you use the product rule

  23. myininaya
    • 5 years ago
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    yes

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