Differentiability of a mutlivariable series We have the following function:
$$f:(0,\infty)\times(0,\infty)\rightarrow\mathbb{R}, \ f(x,y)=\sum_{n=0}^{\infty}cos^2(nx)e^{-nxy}$$

*

*Show that f is partially differentiable and calculate $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$.

*Show that f is Frechet-Differentiable.


My thoughts so far:

*

*I first tried to directly calcualte the Limit but coudln't do it. Maybe with uniform convergence and it would be possible but I have no idea how.
Calculating the derivatives is not a problem.

*I havent tried a lot here, maybe I am missing something from the first task.

Thanks in advance for the help.
 A: Here's a solution to the first part. I'm a bit rusty so I might have made a mistake somewhere but I'm pretty sure this is a valid solution.
Let $f_k(x,y) = \sum_{n = 0}^k \cos^2(nx) e^{-nxy}$. Fix $y_1 >0$
To calculate the partial derivate $\frac {\partial }{\partial x}$ we first show that the sequence of functions $\tilde f_k : (0,\infty) \rightarrow \mathbb R$ defined by
$$ \tilde f_k(x) := f_k(x,y_1)$$
converges uniformly on $[a,b] \subset (0,\infty)$ for all $a<b$ by using the Cauchy criterion for uniform convergence. Let $x \in [a,b]$ and  $k< l $
\begin{align*}
\vert \tilde f_l(x) - \tilde f_k(x) \vert &= \vert \sum_{n = k}^l \cos^2(nx) e^{-nxy_1}\vert  \\
& \leq \sum_{n = k}^l e^{-nxy_1} \\
&= \sum_{n = k}^l (e^{-xy_1})^n \\
& \leq \sum_{n = k}^l (e^{-by_1})^n
\to 0
\end{align*}
This means that the sequence $ \big( \tilde{f}_k \big)_{k \in \mathbb N} $ converges uniformly on all $[a,b] \subset (0,\infty)$.
Let $x_1 > 0 $ and take $\epsilon > 0 $ such that $[x_1-\epsilon,x_1 + \epsilon] \subset (0,\infty)$. By uniform convergences of $\tilde f_k$ on $[x_1-\epsilon,x_1 + \epsilon]$ we have that $f(x,y_1) = \lim \tilde f_k(x)$ is differentiable on $(x_1-\epsilon,x_1 + \epsilon)$ and that
$$ \frac {\partial f}{ \partial x}(x_1,y_1) := \lim_{k \to \infty} \frac{\partial \tilde f_k}{\partial x}(x_1,y_1) $$
This means that we can simply differentiate term by term.
By similar arguments you should be able to find the partial derivative with respect to $y.$
