How do I find $\frac{\partial F}{\partial y}$ and $\frac{\partial F}{\partial z}$ with regards to differentiating under the integral sign?

Question

I have come across a corollary of Leibniz' Rule which says for an integral: $$F(x,y,z)=\int_{y}^{z}f(x,t) \cdot dt$$ which satisfies Leibniz' Rule we should be able to obtain the following: $$\frac{\partial F}{\partial x} = \int_{y}^{z}f'_{1}(x,t) \cdot dt$$ $$\frac{\partial F}{\partial y} = -f(x,y)$$ $$\frac{\partial F}{\partial z} = f(x,z)$$ I have proved the first partial derivative by simply going through the same procedures as Leibniz' Rule, but I am not sure how to proceed proving the bottom two partial derivatives? The book gives a hint which is: 'Use the Fundamental Theorem of Calculus' but I am unsure how to apply it.

Any help would be greatly appreciated.

Answer 1

When you differentiate partially w.r.t. $y$ the variable $x$ and $z$ act as constants. So you are simply integrating a function from $y$ to a constant $z$ and differentiating it. The result is minus the value of the integrand at $y$, i.e. $-f(x,y)$. [Note that $\int_y^{z} f(x,t)dt=-\int_z^{y} f(x,t)dt$]. This is just the fundamental theorem of calculus. Similarly the partial derivative w.r.t. $z$ is $f(x,z)$.

Answer 2

I'll show the partial derivative with respect to $z$; the derivative with respect to $y$ is similar enough.

We're treating $x$ and $y$ as constants. To clarify that, define a new function $g_x(t) = f(x,t)$. This is a single-variable function. It's potentially a different function for each $x$, which is why we put that subscript in the notation, but it's still a function of only one argument. Now, the integral we're trying to differentiate is $$F = \int_y^z g_x(t)\,dt$$ By the Fundamental theorem (derivative of an integral form), $$\frac{\partial F}{\partial z} = g_x(z)=f(x,z)$$ There it is.