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TomLikesPhysics

  • 3 years ago

I can only find derivative for x>0 , but I don't know why (according to wolfram).

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  1. TomLikesPhysics
    • 3 years ago
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    mom I need to attach the equations

  2. TomLikesPhysics
    • 3 years ago
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    1 Attachment
  3. TomLikesPhysics
    • 3 years ago
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    My f`is apparently only true for x>0 and I have no clue where I make that assumption in my differentiation.

  4. Zekarias
    • 3 years ago
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    Your f' is defined only for |R\{-6^{0.5), 6^(0.5)}, actually

  5. TuringTest
    • 3 years ago
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    how do you figure @Zekarias

  6. TuringTest
    • 3 years ago
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    f' is not defined at +/-6^(0.5), but it is for all other points as far as I see

  7. TuringTest
    • 3 years ago
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    I do see one problem which is that for -6^(0.5)<|x|0 the slope should be negative, but the derivative is positive

  8. TuringTest
    • 3 years ago
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    -6^(0.5)<|x|<0 I meant

  9. TuringTest
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E2-6)%5E(2%2F3)&t=crmtb01 you can see the slope is negative for -6^(1/2)<x<0, but the derivative would be positive as you have it

  10. TomLikesPhysics
    • 3 years ago
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    ok, so how should I take the derivative then? Without rewriting it and using the chainrule over and over?

  11. TuringTest
    • 3 years ago
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    that I'm not so sure about. I'm thinking on it.

  12. TuringTest
    • 3 years ago
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    @amistre64 any ideas here?

  13. amistre64
    • 3 years ago
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    the derivative on the attachment looks fine; im not sure what the question is tho

  14. TuringTest
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=derivative+of+y%3D%28x%5E2-6%29%5E%282%2F3%29 true I don't see wolf giving the condition that x>0

  15. TuringTest
    • 3 years ago
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    but what about the point I brought up? f' for -6^(1/2)<x<0 should be negative, but it's positive

  16. amistre64
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=derivative+cbrt%28%28x%5E2-6%29%5E2%29 i think this has more intricate workings than we think

  17. amistre64
    • 3 years ago
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    \[u=(x^2-6)^2~:~u'=4x(x^2-6)\] \[D[u^{1/3}]=\frac{u^{-1/3}}{3}u'\] \[D[u^{1/3}]=\frac{4x(x^2-6)}{3((x^2-6)^2)^{1/3}}\]

  18. amistre64
    • 3 years ago
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    at -1\[\frac{-*-}{+}=+\]

  19. TuringTest
    • 3 years ago
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    but that should not be if you look at the graph http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E2-6)%5E(2%2F3)&t=crmtb01 should be f'<0 at x=-1

  20. amistre64
    • 3 years ago
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    notice that the function\[\sqrt[3]{((x^2-6)^2)}\ne \left(\sqrt[3]{(x^2-6)}\right)^2\]at all points http://www.wolframalpha.com/input/?i=%28%28x%5E2-6%29%5E2%29%5E%281%2F3%29+-+%28%28x%5E2-6%29%5E%281%2F3%29%29%5E2

  21. amistre64
    • 3 years ago
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    the subtlties are in how we are not using the "correct" use of a derivative

  22. amistre64
    • 3 years ago
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    the way we are used to working with exponents seems to be a misuse of notation and doesnt express the full nature of the problem http://www.wolframalpha.com/input/?i=y%3D%28%28x%5E2-6%29%5E2%29%5E%281%2F3%29%2C+y%3D+%28%28x%5E2-6%29%5E%281%2F3%29%29%5E2

  23. TomLikesPhysics
    • 3 years ago
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    So I should not rewrite the equation and just use the chainrule multiple times then to end up with the same answer as wolfram does?

  24. amistre64
    • 3 years ago
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    correct, I would simplify it by making a substitution; then replacing those values in the end

  25. TomLikesPhysics
    • 3 years ago
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    k, thx. I will try that now.

  26. TomLikesPhysics
    • 3 years ago
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    Damn. That went well and it was pretty quick and easy too. So the first attempt did not work because I messed the exponents up?

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