Question

Let \(z: \mathbb{R}^2 \to \mathbb{R}\) be a continuous mapping and with respect to its second variable partially differentiable. Define the following mapping \[f: \begin{cases} \mathbb{R}^2 &\longrightarrow \mathbb{R} \\ (x,y) &\longmapsto \displaystyle \int_0^x z(s,y)ds \end{cases} \] and show that \(f\) is differentiable with respect to \(x\) and \(y\) (managed to show it with respect to \(y\) but stuck with respect to \(x\) in a rigorous approach)

Answer 1

My approach: (Succeeded) with respect to y:
Define rectancgle \(R:= \lbrace (x,y) \in \mathbb{R}^2 \mid a \leq x \leq b, \ c \leq y \leq d \rbrace \subset \mathbb{R}^2\) and define the following mapping: \[F: \begin{cases} [c,d] & \longrightarrow \mathbb{R} \\ (x,y) & \longmapsto \displaystyle \int_0^x z(s,y)ds \end{cases} \]
Computing the definition of differentiability in higher dimensions \[\large \left| F(y_0+h)-F(y_0) - \left( \int \partial_y z(s,y)ds \right) h\right| \] and using the Mean Value theorem to show that it's convergent to zero by setting \[\large \partial_yz(s, \gamma) h = z(s,y_0+h)-z(s,y_0) \text{ for } \gamma \in ]y_0,y_0+h[ \]

Answer 2

Professor @mathmale

Answer 3

I'd love to help, but am afraid that this sort of problem has not been part of my experience. So sorry. Hope someone else will step in. Best of luck to you, and bravo for taking such a challenging course.

Answer 4

@mathmale, thanks a lot for your kind words, appreciate it.

Answer 5

@eliassaab

Answer 6

two differentiate it w.r.t 'x', use fundamental theorem of calculus.
to differentiate it w.r.t. 'y' use lebnitz differentiation under integral sign.

Answer 7

that is
\[ \large{|} F(x+h, y) - F(x,y) - h \left (z(x,y) - z(0,y)\right )\large{|} < \epsilon \]
as \( h\to 0 \)

Answer 8

*leibniz integration rule.