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mathivh

  • one year ago

Partial derivative in terms of x and y

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  1. mathivh
    • one year ago
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    What is the partial derivate in terms of x and in terms of y? \[\int\limits_{x}^{y}e^{{-t}^{2}}dt\] The solutions are: \[\frac{ df }{ dx }=-e^{{-x}^{2}}\] and \[\frac{ df }{ dy }=e^{{-y}^{2}}\]

  2. Loser66
    • one year ago
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    What don't you understand?

  3. mathivh
    • one year ago
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    I don't know how to get the 't' to terms of x and y. I first tried the derivative in terms of t under the integral sign but that still contains \[e^{{-t}^2}\] which I shouldn't be integrating I guess (becomes it would to some erf function)

  4. Loser66
    • one year ago
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    Partial of x means y is a constant, right?

  5. Loser66
    • one year ago
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    \[\int_x^y e^{-t^2}dt=-\int_y^x e^{-t^2}dt\]

  6. Loser66
    • one year ago
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    By Fundamental theorem, its derivative is just replace t by x it is = \(-e^{-x^2}\)

  7. Loser66
    • one year ago
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    partial w.r.t. y is just as it is and consider x as constant. YOu replace t by y and get the solution as given. \(e^{-y^2}\)

  8. mathivh
    • one year ago
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    Ok, but why is it that partial w.r.t. x is taken after switching the limits of the integral?

  9. Loser66
    • one year ago
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    because the Fundamental theorem of calculus just apply to lower limit is a constant.

  10. mathivh
    • one year ago
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    Ok I understand it now, thank you for your time!

  11. Loser66
    • one year ago
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    http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

  12. mathivh
    • one year ago
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    Interesting read, strange that my textbook didn't mention anything of this theorem.

  13. Loser66
    • one year ago
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    Actually, I learn it from this site, not from my school/books That's why I love this site. :)

  14. Loser66
    • one year ago
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    @mathivh if you want to take derivative of the integral w.r.t.y, you consider x as constant. \[\int_x^y e^{-t^2}dt= e^{-y^2} \] Actually, it is \(e^{-y^2}*y'\) but \(y'=1\), they don't put it there. In the new case, \[\int_x^{xy} e^{-t^2}dt \] The upper limit is a function of y, you must take its derivative, hence it becomes \[= e^{-y^2} (xy)' = e^{-y^2}*x\] That is why x is there.

  15. Loser66
    • one year ago
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    If the upper limit is y^2, then your result will be \(e^{(-y^2)^2}*(y^2)'= 2y e^{-y^4}\) Is it clear??

  16. mathivh
    • one year ago
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    It is clear now, thanks again! ;)

  17. mathivh
    • one year ago
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    In the second last post you made, I would think the last equation would be: \[= e^{-x^{2}y^{2}}(xy)'=e^{-x^{2}y^{2}}*x\] Because you mentioned to replace t by the upper limit of the integral?

  18. Loser66
    • one year ago
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    What is the original problem? I don't get what you meant on the last post

  19. Loser66
    • one year ago
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    oh, yeah, typo there. I see it.

  20. mathivh
    • one year ago
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    Or does the lower limit changes something about this? Because \[\int\limits_{0}^{xy}e^{{-t}^{2}}= e^{-x^{2}y^{2}}(xy)'\]

  21. Loser66
    • one year ago
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    Don't forget that is the PARTIAL DERIVATIVE w.r.t.y , as long as the lower limit is not y or function of y, you are ok.

  22. mathivh
    • one year ago
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    So it should be \[e^{-x^{2}y^{2}}*x\] instead of \[e^{-y^{2}}*x\] right?

  23. Loser66
    • one year ago
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    yes, you are right. My bad.

  24. mathivh
    • one year ago
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    It's ok :) What would change if the lower limit were to contain y aswell?

  25. Loser66
    • one year ago
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    In that case, you have to take integral, then derivative of the result. No choice.

  26. Loser66
    • one year ago
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    oh, you can break it. Give me an example, I will show you how

  27. Loser66
    • one year ago
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    brb

  28. mathivh
    • one year ago
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    I don't have any exercise like that, I was just curious. An example would be: \[\int\limits_{y}^{xy} e^{{-t}^{2}}dt\] You say one should integrate it first, but that would give the erf function which we don't want to use most likely.

  29. Loser66
    • one year ago
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    |dw:1439212948147:dw|

  30. mathivh
    • one year ago
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    Right, stupid of me not to think it about like that :) Now I should be able to solve al similar exercises :p Am I right to say that the link you posted doesn't mention to multiply the anti-derivative of f(t) by the upper limit?

  31. Loser66
    • one year ago
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    Actually they have. Search.

  32. Loser66
    • one year ago
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    They have all the cases, the page I gave you just the common mistake page.

  33. mathivh
    • one year ago
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    I found some examples using a function of x in the upper limit. Capiche!

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