Partial derivative in terms of x and y

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Partial derivative in terms of x and y

Mathematics
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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 \[\frac{ df }{ dy }=e^{{-y}^{2}}\]
What don't you understand?
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)

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Partial of x means y is a constant, right?
\[\int_x^y e^{-t^2}dt=-\int_y^x e^{-t^2}dt\]
By Fundamental theorem, its derivative is just replace t by x it is = \(-e^{-x^2}\)
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. \(e^{-y^2}\)
Ok, but why is it that partial w.r.t. x is taken after switching the limits of the integral?
because the Fundamental theorem of calculus just apply to lower limit is a constant.
Ok I understand it now, thank you for your time!
http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html
Interesting read, strange that my textbook didn't mention anything of this theorem.
Actually, I learn it from this site, not from my school/books That's why I love this site. :)
@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.
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??
It is clear now, thanks again! ;)
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?
What is the original problem? I don't get what you meant on the last post
oh, yeah, typo there. I see it.
Or does the lower limit changes something about this? Because \[\int\limits_{0}^{xy}e^{{-t}^{2}}= e^{-x^{2}y^{2}}(xy)'\]
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.
So it should be \[e^{-x^{2}y^{2}}*x\] instead of \[e^{-y^{2}}*x\] right?
yes, you are right. My bad.
It's ok :) What would change if the lower limit were to contain y aswell?
In that case, you have to take integral, then derivative of the result. No choice.
oh, you can break it. Give me an example, I will show you how
brb
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.
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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?
Actually they have. Search.
They have all the cases, the page I gave you just the common mistake page.
I found some examples using a function of x in the upper limit. Capiche!

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