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

1. 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 $\frac{ df }{ dy }=e^{{-y}^{2}}$

2. Loser66

What don't you understand?

3. 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)

4. Loser66

Partial of x means y is a constant, right?

5. Loser66

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

6. Loser66

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

7. 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. $$e^{-y^2}$$

8. mathivh

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

9. Loser66

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

10. mathivh

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

11. Loser66
12. mathivh

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

13. Loser66

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

14. 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.

15. 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??

16. mathivh

It is clear now, thanks again! ;)

17. 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?

18. Loser66

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

19. Loser66

oh, yeah, typo there. I see it.

20. mathivh

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

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

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

23. Loser66

yes, you are right. My bad.

24. mathivh

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

25. Loser66

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

26. Loser66

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

27. Loser66

brb

28. 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.

29. Loser66

|dw:1439212948147:dw|

30. 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?

31. Loser66

Actually they have. Search.

32. Loser66

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

33. mathivh

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