Let $g: \mathbb R \rightarrow \mathbb R$ a differentiable function in $\mathbb R$. $f(x,y) = \frac{g(y)}{1+g^2(x)}$ is differentiable in its domain? If $g: \mathbb R \rightarrow \mathbb{R}$ is differentiable in $\mathbb{R}$ I am trying to prove that:
$f(x,y) = \frac{g(y)}{1+g^2(x)}$ is differentiable in its domain.
Here is my reasoning:
$g: \mathbb{R} \rightarrow \mathbb{R}$, is defined in $\mathbb{R}$. Then, if $x, y \in \mathbb{R}$, $g(x)$ are $g(y)$ defined. Therefore, $D_{f}= \mathbb{R} \times \mathbb{R} = \mathbb{R}^2$.
A function $f:U\subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ defined in $U$ in $\mathbb{R}^2$, is differentiable at $p=(x_{0}, y_{0})\in U$ if partial derivatives of $f$ exist at $p$.
\begin{align}
   A_{1} = \frac{\partial f}{\partial x} (x_{0}, y_{0})\\
   A_{2} = \frac{\partial f}{\partial y} (x_{0}, y_{0})
\end{align}
and $r(h_{1}, h_{2})$ is defined in:
\begin{align}
                 f((x_{0}, y_{0}) + (h_{1}, h_{1})) = f(x_{0}, y_{0}) + A_{1}h_{1} + A_{2}h_{2} + r(h_{1}, h_{2})
             \end{align}
with the following property
\begin{align}
   \lim_{(h_{1}, h_{2}) \rightarrow (0,0)} \frac{r(h_{1}, h_{2})}{\left \lVert (h_{1}, h_{2}) \right \rVert} = 0
\end{align}
Assuming $f(x,y)$, then partial derivatives should exist:
\begin{align}
A_{1} &= \frac{\partial }{\partial x}\left( \frac{g(y)}{1+g^{2}(x)}\right) \nonumber \\
             &= g(y)\frac{\partial}{\partial x} \left( \frac{1}{1+g^2(x)} \right) \nonumber \\
             &= g(y) \frac{\partial}{\partial x} \left( (1+g^2(x))^{-1} \right)\nonumber \\
             &=g(y)\left[ (-1)(1+g^{2}(x))^{-2}(2g(x)) g_{x}\right] \nonumber\\
             &=\frac{g(y) (-1)(2g(x))(g_{x})}{[1+g^{2}(x)]^{2}}\nonumber \\
             &=\frac{-2 g(y) g(x) g^{'}_{x}}{[1+g^{2}(x)]^{2}}
\end{align}
Similarly:
\begin{align}
A_{2} &= \frac{\partial }{\partial y}\left( \frac{g(y)}{1+g^{2}(x)}\right) \nonumber \\
             &= \frac{1}{1+g^2(x)}\frac{\partial}{\partial y} \left( g(y) \right) \nonumber \\
             &= \frac{g^{'}_{y}}{1+g^2(x)}
\end{align}
As $g(x)$ y $g(y)$ are defined, $x,y \in \mathbb{R}$, then partial derivatives are combination of $g(x)$, $g(y)$ continuous functions (because $g$ is differentiable). Then, if $g^{'}_{x}$ and $g^{'}_{y}$ are also continuous, then the entire partial derivatives will also be continuous, which will prove that $f(x,y)$ is differentiable. But, I am not sure if it is possible with the info of the problem.
My questions are:

*

*Is this strategy adequate to solve the problem? If not, how do you solve the problem?

Thanks,
This is my first post, and I am very new to maths, so I will be happy to edit the post to clarify it.
 A: It seems that you know some theorems about differentiability, for example that if the partial derivatives of $f$ exist and are continuous, then $f$ is differentiable. Unfortunately this does not help you because you do not know whether $A_1, A_2$ are continuous. They are continuous iff $g$ is continuously  differentiable, but that is not required.
In my opinion you need some more theorems to prove that $f$ is differentiable, using the definition itself will probably be not very helpful. I suggest to proceed as follows:

*

*Show that the coordinate projections $p_1 : \mathbb R^2 \to \mathbb R, p_1(x,y) = x$,  and $p_2 : \mathbb R^2 \to \mathbb R, p_1(x,y) = y$, are differentiable. Here you can easily use the definition.


*Show that sum, product and quotient of differentiable functions $u, v: \mathbb R^2 \to \mathbb R$ are differentiable (for the quotient we need that the function in the denominator has no zeros).


*Prove the chain rule: If $u : \mathbb R^2 \to \mathbb R$ and $v : \mathbb R \to \mathbb R$  are differentiable, then $v \circ u$ is differentiable.
Perhaps you already know some of these facts.
Then everything is easy: The functions $g \circ p_1$ and $g \circ p_2$ are differentiable, thus $(g \circ p_1)^2$ and $1 + (g \circ p_1)^2$ are differentiable and so is $f = \dfrac{g \circ p_2}{1 + (g \circ p_1)^2}$.
