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Calculate the partial derivatives of the function $ \displaystyle F(x,y)=\int_{x}^{ \displaystyle\int_{0}^{y} g(s) ds} f(t)dt $ where $f,g$ are continuous from $\mathbb{R}$ to $\mathbb{R}$.

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Try the fundamental theorem of calculus and the chain rule. –  mrf Jan 14 '13 at 22:11
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Let's define

$$ \displaystyle H(x,y)=\int_{x}^{ \displaystyle\int_{0}^{y} ds \: g(s)} f(t)dt$$

$$ \displaystyle H(x,y) = F \left [ \displaystyle\int_{0}^{y} ds \: g(s) \right ] - F(x) $$

where $F$ is the antiderivative of $f$. Then

$$\frac{\partial H}{\partial x} = -F'(x) = -f(x) $$

$$\frac{\partial H}{\partial y} = f \left [ \displaystyle\int_{0}^{y} ds \: g(s) \right ] g(y) $$

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