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anonymous

  • 5 years ago

the derivative of x^(-2x) is -2x^(-2x) (log(x)+1) why is there that +1?

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  1. mathteacher1729
    • 5 years ago
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    See image. 1) Go to http://www.wolframalpha.com 2) type "derivative x^(-2x) Hope this helps. :)

  2. anonymous
    • 5 years ago
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    \[x^{-2x}\] take the log get \[-2x\ln(x)\] take the derivative using product rule get \[-2x\frac{1}{x} -2ln(x)\]

  3. anonymous
    • 5 years ago
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    now factor out the 2 get \[2(\ln(x)-1)\]

  4. anonymous
    • 5 years ago
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    of course the x's in the first term cancel. then multiply by the original function to get your answer: \[x^{-2x}(2(ln(x)-1)=2\times x^{-2x}(ln(x)-1)\]

  5. anonymous
    • 5 years ago
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    i made a mistake is see. start with \[-2-2ln(x)\] gives \[-2(ln(x)+1)\]

  6. anonymous
    • 5 years ago
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    sorry about that.

  7. anonymous
    • 5 years ago
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    but that is where the -2 comes from

  8. anonymous
    • 5 years ago
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    you can do this because ln(x) '=1/x so if you take the log and at the end you multiply by the function than you get the derrivate

  9. anonymous
    • 5 years ago
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    this is a question :)

  10. anonymous
    • 5 years ago
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    oh yes. if you want to take the derivative of a function with the variable in the exponent the easiest thing to do is first take the log.

  11. anonymous
    • 5 years ago
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    I have never seen this before, but it is a clever idea

  12. anonymous
    • 5 years ago
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    you know that the derivative of \[ln(f(x))\] is \[\frac{f'(x)}{f(x)}\]

  13. anonymous
    • 5 years ago
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    yep, I see

  14. anonymous
    • 5 years ago
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    which means you take the log, take the derivative, and then multiply by original function.

  15. anonymous
    • 5 years ago
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    yes yes, cool technique

  16. anonymous
    • 5 years ago
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    you can also view it this way: \[x^{2x}=e^{-2x\ln(x)}\]

  17. anonymous
    • 5 years ago
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    then take the derivative using the chain rule. the derivative of e to the something is just e to the something times the derivative of something. you will see that the work is identical

  18. anonymous
    • 5 years ago
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    thanks, I see why this helps

  19. anonymous
    • 5 years ago
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    in any case all the work is taking the derivative of the log of that thing, either way. hope this helps

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