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TechnoFiction

  • 3 years ago

Can anyone help me with the derivative of: y=x^11 e^x

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  1. TechnoFiction
    • 3 years ago
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    \[y=x ^{11}e ^{x}\] Much easier to look at.

  2. Mimi_x3
    • 3 years ago
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    do you know the product rule

  3. TechnoFiction
    • 3 years ago
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    Yeah, I somehow worked out the problem by myself. Just want to make sure if I got it right or not.

  4. Mimi_x3
    • 3 years ago
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    well tell me what you got :)

  5. TechnoFiction
    • 3 years ago
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    I look really ridiculous, but: \[11x ^{10}e ^{x}+x ^{11}e ^{x}\] Here you go.

  6. Mimi_x3
    • 3 years ago
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    looks right to me

  7. TechnoFiction
    • 3 years ago
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    hm....ok, if you say so. I just don't feel right somehow. It look too long and weird.

  8. Mimi_x3
    • 3 years ago
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    It's right! \(u=x^{11} => u' = 11x^{10}\) \(v=e^{x} => v' = e^{x}\) \(=> 11x^{10}*e^{x}+ x^{11}e^{x}\)

  9. Mimi_x3
    • 3 years ago
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    You can also write it like: \(e^{x} (11x^{10}+x^{11})\) if it looks too "weird" :)

  10. TechnoFiction
    • 3 years ago
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    ok, thanks :D

  11. calculusfunctions
    • 3 years ago
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    You could also factor the \[x ^{10}\]and express the derivative as\[\frac{ dy }{ dx }=x ^{10}e ^{x}(11+x)\]

  12. TechnoFiction
    • 3 years ago
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    Cool, thanks

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