Suppose that $f(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$are continuous in a nbhd of a point $(a,b) \in \mathbb{R}^{2}$. Show that:
$g(x,y) = \int_{a}^{x}f(t,y) dt $ is $\mathcal{C}^{1}$ in a neighbourhood of $(a,b)$
The equation $y = b + \int_{a}^{x}f(t,y)dt$ can be solved for $y$ as a $\mathcal{C}^{1}$ function of $x$ in some neighbourhood of $a$
To do 1) this I need to show two things:
i) that $g(x,y)$ is continuous
ii) that $Dg(x,y)$ is continuous.
For i):
Consider the function $G: \mathbb{R} \to \mathbb{R}$, defined as $G(\xi) = \int_{a}^{\xi}f(t,y)dt$. As well define the function $\sigma : \mathbb{R}^{2} \to \mathbb{R}$, defined as $\sigma(x,y) = x$.
Therefore:
$$G(\sigma(x,y)) = \int_{a}^{\sigma(x,y)}f(t,y)dt = \int_{a}^{x}f(t,y)dt = g(x,y)$$
One can see that $g$ is a composition of continuous functions and as such $g$ is also continuous.
ii) This is the one place I'm having trouble with.
I'm not sure if the composition I chose actually works out for this situation because I have an inclination based on what we've been learning to this part to apply Leibniz integral rule.
and differentiate with regards to $y$, that is taking $\frac{\partial f}{\partial y}$ and then somehow use the hypothesis of continuity, but based on how I defied $G(x,y)$ that would turn out to just be $0$, which I don't want.
Question 2
I know this is a direct application of the Implicit function theorem, but maybe because I'm seeing the integral I'm not sure how to actually do the problem.
Could I get some guiding steps to address these problems?