Question

Find
d/dx (integral[1,xy] (e^(t^2))dt)

Answer 1

I have the answer, y e^(x^2 y^2).
I just don't understand why it is this.

Answer 2

I cannot even understand your function.

Answer 3

http://www.wolframalpha.com/input/?i=d%2Fdx+%28integral%5B1%2Cxy%5D+%28e%5E%28t%5E2%29%29dt%29

Answer 4

\[\huge \frac{d}{dx}\int\limits_1^{xy}e^{t^2}\;dt\]
We need to apply the `Fundamental Theorem of Calculus, Part 1`:\[\large \frac{d}{dx}\int\limits_a^x f(t)\;dt \qquad=\qquad f(x)\]

It'll be a little bit trickier since we have two variables, hmm.

Answer 5

I guess we're treating this derivative operator as partial with respect to x?
Otherwise we don't get the answer they indicate :o hmm.

Answer 6

First, integrate \(\int e^{t^2}dt\) then apply the FToC.

Answer 7

So what happens is...
~We're integrating,
~Then we're taking a derivative, which gives us back the thing we started with.

Between those two steps, we plugged in the upper and lower limits of integration, which causes the argument of our function to change.

When we apply that second step, since the argument of our function is more than just x, we have to apply the chain rule.

\[\large \frac{d}{dx}\int\limits\limits_1^{xy}e^{t^2}\;dt\qquad=\qquad e^{(xy)^2}\frac{d}{dx}(xy)\]

Answer 8

It's a little tricky if you're rusty on the fundamental theorem :(

Answer 9

@abb0t you don't need to integrate... and good luck integrating e^t^2 if you did.

Answer 10

I would of used polar coordinates to integrate.

Answer 11

If you know partial deriv's you should know double integrals.

Answer 12

You don't need to... FTC makes it a shortcut. The d/dx essentially "cancels out" the integral, and you plug in the limits, differentiate the limits.

Answer 13

thanks, that helps, I'm good with the fundamental th, it's just the partial ferivative.

Answer 14

Treating y as a constant, d/dx(xy) is just y.

Answer 15

Ah, is that the way they presented it? With the curly d's in front? :)

Answer 16

\[\Large \frac{ \delta }{\delta x }(xy) = y\]

Answer 17

Yah, you can kind of think of it like this if it helps at all.
\[\large \frac{\partial}{\partial x}(yx) \qquad\to\qquad \frac{d}{dx}(cx)\]
As agent indicated, the y is being held constant when we take our partial.

Answer 18

thx for helping guys.