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joshuanoble2000

Find a function whose derivative is 3x^2(x^3 + 2)^12

  • 2 years ago
  • 2 years ago

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  1. AnwarA
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    Do you know what integration or an antiderivative is?

    • 2 years ago
  2. joshuanoble2000
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    No. I think we're meant to be doing this by figuring one out that would work by simply working backwards...we're working on the chain rule now.

    • 2 years ago
  3. dave444
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    Try letting \(u=x^3\) and go from there.

    • 2 years ago
  4. AnwarA
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    I see, we can do that. Actually try putting \(u=x^3+2\), that would work better.

    • 2 years ago
  5. dave444
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    True.

    • 2 years ago
  6. AnwarA
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    You will have then something like \(u'u^{12}\). Now try to relate this to the product rule, can you do that?

    • 2 years ago
  7. AnwarA
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    Oh sorry, I mean relate it ot the chain rule.

    • 2 years ago
  8. dave444
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    Or note that if \(u=x^3+2\), then \( du = 3x^2\;dx\) and so \[\int 3x^2(x^3+2)^{12}\;dx = \int u^{12} du = \frac{u^{13}}{13}+C = \frac{(x^3+2)^{13}}{13} +C \]

    • 2 years ago
  9. joshuanoble2000
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    Sorry, I'm going to have to go through this rather painstakingly, I fear. U is x^3 + 2. So I say dy/du is (dy/(u^3 + 2) * (u^3 +2) / dx) to begin with?

    • 2 years ago
  10. joshuanoble2000
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    Sorry, du, so 1...

    • 2 years ago
  11. joshuanoble2000
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    3x^2, I mean

    • 2 years ago
  12. AnwarA
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    I can't read what you wrote well. But what I did is that I just found the derivative of \(u\) w.r.t \(x\). \(u=x^3 + 2 \implies \frac{du}{dx}=3x^2\). So the expression we have can then be written as \(\large u'u^{12}\). So far so good?

    • 2 years ago
  13. joshuanoble2000
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    Yes.

    • 2 years ago
  14. AnwarA
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    Good. Now you know from your study of derivatives, a method for finding derivatives of composite functions called the Chain rule. Can you state it for me?

    • 2 years ago
  15. joshuanoble2000
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    f'(x) = f'(g(x)) * g'(x)

    • 2 years ago
  16. joshuanoble2000
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    (assuming the original functions were stated f(g(x))

    • 2 years ago
  17. AnwarA
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    True, you mean \([f(g(x))]'\), or \(\frac{d}{dx}[f(g(x))]\)

    • 2 years ago
  18. joshuanoble2000
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    Right.

    • 2 years ago
  19. AnwarA
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    Now we want to relate this formula you wrote with the expression we have. Consider \(g'(x)=u' \implies g(x)=u,\) and \(f'(g(x))=u^{12}=g^{12}.\) Can you then find \(f(g(x))\)?

    • 2 years ago
  20. AnwarA
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    Simply you have \(f'=u^{12}\), so \(f\)=..?!

    • 2 years ago
  21. AnwarA
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    Are you still there?

    • 2 years ago
  22. joshuanoble2000
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    Yeah.

    • 2 years ago
  23. AnwarA
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    All you need here is to apply the power rule. You have f'(u)=u^12, so f(u)=..?

    • 2 years ago
  24. AnwarA
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    The symbols I just used are not really accurate, but I used them for simplification.

    • 2 years ago
  25. AnwarA
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    You can see that \(\large f(u)=\frac{u^{13}}{13}\)?

    • 2 years ago
  26. joshuanoble2000
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    If you had 1/13 U^13

    • 2 years ago
  27. AnwarA
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    Very good. So we're saying u^13/13 is a function that has a derivative \(u'u^{12}\). Now just resubstitute for \(x\) is \(\large \frac{u^{13}}{13}\), and this should be your final answer. Don't forget that \(u=x^3+2\).

    • 2 years ago
  28. AnwarA
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    *for x in*

    • 2 years ago
  29. joshuanoble2000
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    Oh my, I feel silly now. It should be 1/13(x^3+2)^13?

    • 2 years ago
  30. AnwarA
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    Exactly! :)

    • 2 years ago
  31. joshuanoble2000
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    This whole trouble was because the 3x^2 was moved to the front of the equation...quite dense on my part. Thanks so much for being patient.

    • 2 years ago
  32. AnwarA
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    So we can write a general rule from what we just did, that states: An anti derivative of \(f'(x)[f(x)]^n\) is \(\large \frac{[f(x)]^{n+1}}{n+1}\). I hope this doesn't make it any more complicated. You're welcome!

    • 2 years ago
  33. AnwarA
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    In our case f(x) is obviously x^3+2.

    • 2 years ago
  34. joshuanoble2000
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    No, you're thoroughly helpful and when I'm done with my homework I'll more carefully look over what you've written.

    • 2 years ago
  35. AnwarA
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    Good luck then!

    • 2 years ago
  36. joshuanoble2000
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    Thanks!

    • 2 years ago
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