Find the partial derivative of a function whih is the integral of another function I've got an exercise and I am not really sure which is the correct answer.
Exercise
$f$ is R-integrable. Find the partial derivatives for function $g(x,y) = \int_0^{xy} \! f(t) \, \mathrm{d}t.$
End
I've found out that I should use The Second Fundamental Theorem of Calculus (from Michael Spivak - Calculus) which says:
If $f$ is integrable on $[a,b]$ and $f=g'$ for some function $g$ then
$\int_a^b \ = g(b) - g(a)$
I didn't found out how to apply it for my function $g(x,y)$.
I wanted to use The First Fundamental Theorem of Calculus, but function $f$ needs to be continuous and I don't have this information.
My Solution
$f = g'$ which means that:
$f(x) = \frac{\partial g}{\partial x} (x,y) = \frac{\partial g}{\partial x} \int_0^{xy} \! f(t) \, \mathrm{d}t.$
$f(y) = \frac{\partial g}{\partial y} (x,y) = \frac{\partial g}{\partial y} \int_0^{xy} \! f(t) \, \mathrm{d}t.$
What now? In my opinion the partial derivatives are $f(x)$ and $f(y)$, but I have no proof for that.
 A: There's some confusion here.  You're using $g$ in two ways.  The second way, as an antiderivative of $f$, needs to be scrapped. In problems like this, it can help to imagine that you have an antiderivative of $f$, but let's call it $F(t)$, because it can't be a function of two variables (so it can't be $g$.)  Then:
$$g(x,y) = \int_0^{xy} f(t) \; dt = F(t)\mid_0^{xy} =F(xy)-F(0).$$
Then by chain rule:
$$g_x(x,y) = \frac{\partial}{\partial x} (F(xy) - F(0)) = F'(xy)y - 0 = yf(xy).$$
Using the fact that $F' = f$ and the chain rule.
Edit:  To mop up the details:  By the definition:
$$g_x = \lim_{h \to 0} \frac{1}{h}\left( \int_0^{(x+h)y} f(t) \; dt - \int_0^{xy} f(t) \; dt\right) =  \lim_{h \to 0} \frac{1}{h} \int_{xy}^{xy+hy} f(t) \; dt.$$
Let $u=xy$ and $w=hy$:
$$=\lim_{h \to 0} \frac{y}{hy} \int_{xy}^{xy+hy} f(t) \; dt = y\lim_{w \to 0} \frac{1}{w} \int_{u}^{u+w} f(t) \; dt =y \lim_{w\to 0} f(c) $$
by the Mean Value Theorem (which needs only $f$ to be R-integrable) for some $c$ between $u$ and $u+w$.  So if we believe this last limit is $f(u)$ we're done.
Hand-wavy argument that the last limit is $f(u)$:  Let $w_n = 1/n$ and let $c_n$ be the corresponding value of $c$.  If $f(c_n)$ and $f(w)$ remain far apart as $n$ increases, then so do the upper and lower Reimann sums, which contradicts that $f$ is R-integrable.  So $f(c_n) \to f(w).$
