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TuringTest

  • 4 years ago

Let\[y=x^p(1+x)^q\]find\[y^{(p+q)}\]

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  1. TuringTest
    • 4 years ago
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    that's the (p+q)th derivative by the way Sorry that I have to go soon, but I look forward to seeing your responses when I return.

  2. amistre64
    • 4 years ago
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    ln(y) = ln(x^p) + ln(1+x)^q y'/y = p/x + q/(1+x) hmm

  3. joemath314159
    • 4 years ago
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    i think the answer is just (p+q)!

  4. joemath314159
    • 4 years ago
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    since the highest power of x will be x^(p+q), and all the other powers of x will become 0 after P+q derivatives.

  5. amistre64
    • 4 years ago
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    sounds plausible to me

  6. TuringTest
    • 4 years ago
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    hint:(Leibniz' theorem/ binomial stuff)

  7. TuringTest
    • 4 years ago
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    at least in MIT's solution I'd like to see others

  8. amistre64
    • 4 years ago
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    lol, id like to see mits solution :)

  9. TuringTest
    • 4 years ago
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    I'll send it to you....

  10. AccessDenied
    • 4 years ago
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    im curious, what class is this problem from?

  11. TuringTest
    • 4 years ago
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    OCW calc I

  12. AccessDenied
    • 4 years ago
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    Ahh, okay. Thank you. :)

  13. Mr.Math
    • 4 years ago
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    http://en.wikipedia.org/wiki/Leibniz_rule_(generalized_product_rule)

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