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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.

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    I am not sure that the exercise is well posed. In general, $g$ is differentiable almost everywhere only. $f$ needs to be continuous almost everywhere. So you can apply chain rule and FTC almost everywhere. – 2017-01-08
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    @user251257 Like I said you don't know if $f$ is continuous and I didn't find any theorem for that. We knot that if $f$ is continuous than $f$ is integrable, but opposite isn't true. – 2017-01-08
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    And The Second Theorem of Calculus doesn't need $f$ to be continuous. It is not necessary. – 2017-01-08
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    teachers are prone to error. Just ask if yours forgot some details. It happens all the time. Or have you only defined integral for continuous functions? – 2017-01-08
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    Our teacher is not that kind of person. I don't think he made a mistake. We used Calculus from Michael Spivak. – 2017-01-08
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    If you need an example: let $f(0) = 1$ and $f(x) = 0$ for $x\ne 0$. What is $F(x) = \int_0^x f(t) dt$ and what is $F'(0)$? – 2017-01-08
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    @user251257 Function $f$ cannot be a derivative – 2017-01-08
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    at least not at $x=0$, yes. Just ask your teacher. Teachers do make error and it happens more often than students think. Btw. the idea of @B.Goddard's answer is clearly correct. – 2017-01-08
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    @user251257 So what is wrong with the exercise? – 2017-01-08
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    just the assumption that $f$ is continuous is missing. – 2017-01-08
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    @user251257 I will ask him about it. Anyway I still think that he wants me to use STC. – 2017-01-08
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    like I wrote, @B.Goddard's correct and probably the intended answer, assuming $f$ is continuous. – 2017-01-08
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    But in the STC it isn't neccessary to be continuous. $f$ only needs to be integrable. That's why I don't understand why $f$ should be continuous. STC doesn't need this condition. – 2017-01-08
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    in your notation of STC, if $f=g'$ then $\int_a^b f = g(b) - g(a)$. But it doesn't say that $g(x) = \int_a^x f$ or claim $g'=f$ (it's an assumption!). – 2017-01-08
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    @user251257 I think I get it. In my example I don't know if $f = g'$, so I can't use STC (I can only assume that $f=g'$. Am I right? – 2017-01-08
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    yes, exactly. For the step $f=g'$ you need the first fundamental theorem. – 2017-01-08
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    @user251257 I talked with my teacher and he said that $f$ must be continuous. BUT, the task is GOOD. He didn't write that f is continuous, but you can assume it and solve the problem. So I can solve the exercise and after that I can claim that I have a result, but I assume that $f$ is continuous. – 2017-01-10

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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 $f(c_n) \to f(w).$

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    $F′(xy)y−0=yf(xy).$ I don't get this party. – 2017-01-08
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    @AlexM. Thanks. – 2017-01-08
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    $F' =f$ holds if $f$ is continuous. Otherwise it is in general not true. – 2017-01-08
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    @user251257 It's just a way of thinking it through. $F$ is, in some sense, just a dummy function. – 2017-01-08
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    @user251257 I only know that $f$ is R-integrable. Nothing about continuity :( – 2017-01-08
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    @ValentinEmilCudelcu First, $F' = f$. Second, by the chain rule, the derivative of $F(xy) = F'(xy)$(derivative of the inside with respect to $x$) $=F'(xy)y.$ – 2017-01-08
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    @B.Goddard you may call $F$ what ever you want. Just apply FTC correctly. – 2017-01-08
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    @user251257 When you're walking students through new concepts, the important thing is that they make it to the other side, not that they're in awe of your pedantry. – 2017-01-08