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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
Hi and welcome to the site. Both you and the site would benefit if you would consider registering your account. Also, if you find answers helpful, up vote them. If you find a "best answer", accept it by clicking on the check mark under the question. – JohnD Jan 14 '13 at 22:17
up vote 2 down vote accepted

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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