Partial derivative in terms of x and y

- mathivh

Partial derivative in terms of x and y

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- mathivh

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
\[\frac{ df }{ dy }=e^{{-y}^{2}}\]

- Loser66

What don't you understand?

- mathivh

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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## More answers

- Loser66

Partial of x means y is a constant, right?

- Loser66

\[\int_x^y e^{-t^2}dt=-\int_y^x e^{-t^2}dt\]

- Loser66

By Fundamental theorem, its derivative is just replace t by x
it is = \(-e^{-x^2}\)

- Loser66

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. \(e^{-y^2}\)

- mathivh

Ok, but why is it that partial w.r.t. x is taken after switching the limits of the integral?

- Loser66

because the Fundamental theorem of calculus just apply to lower limit is a constant.

- mathivh

Ok I understand it now, thank you for your time!

- Loser66

http://www.mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

- mathivh

Interesting read, strange that my textbook didn't mention anything of this theorem.

- Loser66

Actually, I learn it from this site, not from my school/books
That's why I love this site. :)

- Loser66

@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

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??

- mathivh

It is clear now, thanks again! ;)

- mathivh

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?

- Loser66

What is the original problem? I don't get what you meant on the last post

- Loser66

oh, yeah, typo there. I see it.

- mathivh

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

- Loser66

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.

- mathivh

So it should be \[e^{-x^{2}y^{2}}*x\]
instead of
\[e^{-y^{2}}*x\]
right?

- Loser66

yes, you are right. My bad.

- mathivh

It's ok :) What would change if the lower limit were to contain y aswell?

- Loser66

In that case, you have to take integral, then derivative of the result. No choice.

- Loser66

oh, you can break it. Give me an example, I will show you how

- Loser66

brb

- mathivh

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.

- Loser66

|dw:1439212948147:dw|

- mathivh

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?

- Loser66

Actually they have. Search.

- Loser66

They have all the cases, the page I gave you just the common mistake page.

- mathivh

I found some examples using a function of x in the upper limit. Capiche!

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