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anonymous

  • 4 years ago

Integral(e^(ab))da= e^(ab)*b^-1 Why?

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  1. anonymous
    • 4 years ago
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    As, going the other way: if I differentiate the right equation with respect to a, b doesn't dissapear, as it is a constant, then why does a b^-1 materialise in the antiderivative?

  2. amistre64
    • 4 years ago
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    i take it thats a partial and not b(a)?

  3. amistre64
    • 4 years ago
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    \[D_a(exp({ab}))=exp(ab)*(ab)'\] \[D_a(exp({ab}))=exp(ab)(a'b+ab')\] or maybe if bs a constant \[D_a(exp({ab}))=exp(ab)*b\]

  4. amistre64
    • 4 years ago
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    we need the 1/b to catch it; 1/b * b = 1

  5. amistre64
    • 4 years ago
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    in other words, if we dont put something there to catch the "b" that flies out we dont get the derivative we are looking for

  6. amistre64
    • 4 years ago
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    Dx is a way to notate a derivative operation on something.

  7. amistre64
    • 4 years ago
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    Da means take the derivative of this stuff with respect to a

  8. amistre64
    • 4 years ago
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    go ahead and replace b with your favorite constant thats not 0

  9. amistre64
    • 4 years ago
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    youll see that if we dont have a 1/2 in the antiderivative that we end up with : exp(2a)*2 which is NOT what we are trying to undo.

  10. amistre64
    • 4 years ago
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    so we apply a useful form of 1 into the antiderivative to help us out; since 1* anything doesnt change its value; we need a "2", or a "b" as the case may be, so lets use 2/2 and just pull out the 1/2 for later

  11. anonymous
    • 4 years ago
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    Thanks, after writing that out it clicked.

  12. amistre64
    • 4 years ago
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    :) youre welcome

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