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anonymous

  • 5 years ago

Find the derivative of the function.. g(x)= ln[(e^x)/(1+e^x)]

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  1. anonymous
    • 5 years ago
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    1/(e^x+1)

  2. anonymous
    • 5 years ago
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    how did you get that?

  3. anonymous
    • 5 years ago
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    is it because ln(e^x) = x and the derivative of x = 1??

  4. anonymous
    • 5 years ago
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    Wolfram told me so. I really don't feel like doing that derivative. If you want to see the steps type it in yourself, it shows you the steps.

  5. anonymous
    • 5 years ago
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    oh okay.. thanks!

  6. anonymous
    • 5 years ago
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    Wolfram do it in an ugly way. In general: \[\frac{\mathbb{d}}{\mathbb{d}x} \ln f(x) = \frac{f'(x)}{f(x)} \] Make it a bit slicker by splitting it up using Ln(a/b) = Ln a - Ln b before differentiating

  7. anonymous
    • 5 years ago
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    that's what i've been trying to work with because i realized that the function is f'(x)/f(x)!

  8. anonymous
    • 5 years ago
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    i.e \[ \frac{\mathbb{d}}{\mathbb{d}x} \ln \frac{e^x}{1+e^x} = \frac{\mathbb{d}}{\mathbb{d}x} \ln e^x - \frac{\mathbb{d}}{\mathbb{d}x}\ln 1+ e^x = \text{blah...} \] Unfortunately, I think that is nicer than something involving using the fact it is on the form ln[f'(x)/f(x)] , but I could be missing something.

  9. anonymous
    • 5 years ago
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    ohh i see what you're saying... so then the derivative of ln(e^x) will be 1

  10. anonymous
    • 5 years ago
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    and the derivative of ln(1+e^x) is e^x/1+e^x?

  11. anonymous
    • 5 years ago
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    Indeed it is

  12. anonymous
    • 5 years ago
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    fantastic! thanks so much.. i'll definitely be posting more up soon.. im studying for a calc 2 final!

  13. anonymous
    • 5 years ago
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    :)

  14. anonymous
    • 5 years ago
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    Ouch, sorry. Calc 2 was easy, but I absolutely hated series haha. Easy while you are doing them, but even easier to forget :). Vector calculus is so much more fun.. (*sarcasm*)

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