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mathivh
 one year ago
Partial derivative in terms of x and y
mathivh
 one year ago
Partial derivative in terms of x and y

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mathivh
 one year ago
Best ResponseYou've already chosen the best response.0What 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 \[\frac{ df }{ dy }=e^{{y}^{2}}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4What don't you understand?

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0I 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)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4Partial of x means y is a constant, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4\[\int_x^y e^{t^2}dt=\int_y^x e^{t^2}dt\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4By Fundamental theorem, its derivative is just replace t by x it is = \(e^{x^2}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4partial 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. \(e^{y^2}\)

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0Ok, but why is it that partial w.r.t. x is taken after switching the limits of the integral?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4because the Fundamental theorem of calculus just apply to lower limit is a constant.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0Ok I understand it now, thank you for your time!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0Interesting read, strange that my textbook didn't mention anything of this theorem.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4Actually, I learn it from this site, not from my school/books That's why I love this site. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4@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.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4If the upper limit is y^2, then your result will be \(e^{(y^2)^2}*(y^2)'= 2y e^{y^4}\) Is it clear??

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0It is clear now, thanks again! ;)

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0In 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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4What is the original problem? I don't get what you meant on the last post

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4oh, yeah, typo there. I see it.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0Or does the lower limit changes something about this? Because \[\int\limits_{0}^{xy}e^{{t}^{2}}= e^{x^{2}y^{2}}(xy)'\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4Don'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.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0So it should be \[e^{x^{2}y^{2}}*x\] instead of \[e^{y^{2}}*x\] right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4yes, you are right. My bad.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0It's ok :) What would change if the lower limit were to contain y aswell?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4In that case, you have to take integral, then derivative of the result. No choice.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4oh, you can break it. Give me an example, I will show you how

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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 antiderivative of f(t) by the upper limit?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4Actually they have. Search.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.4They have all the cases, the page I gave you just the common mistake page.

mathivh
 one year ago
Best ResponseYou've already chosen the best response.0I found some examples using a function of x in the upper limit. Capiche!
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