Proving that $F(x,y)=\sum_{n\ge 1} \sin (ny)e^{-n(x+y)}$ is continuously differentiable Let $$F(x,y)=\sum_{n\ge 1} \sin (ny)e^{-n(x+y)}.$$ Prove that there is $\delta > 0$ and a unique differentiable function $y=\phi(x)$ on $I=(1-\delta,1+\delta)$ such that $\phi(1)=0, F(x,\phi(x))=0$ for all $x\in I$.
This obviously boils down to the Implicit function theorem. But before I need to show that $F$ is $C^1$. For the one-variable case to show $\sum f_n(x)$ is $C^1$ it suffices to show that $\sum f_n$ converges at some point and $\sum f'_n$ converges uniformly. In our case there are two variables. But how to show that $F$ is $C^1$ in this case?
 A: First off, let's define a neighborhood of $(1,0)$ where the series converges. How about $U=(3/4,5/4)\times (-1/4,1/4)?$ If $(x,y)\in U,$ then $x+y > 3/4 -1/4 = 1/2.$ It follows that the given series converges uniformly on $U$ by Weierstrass M. Thus we can define $F(x,y)$ to be the sum of this series for $(x,y)\in U.$ Since each summand is continuous on $U,$ so is $F.$
Does this $F(x,y)$ have partial derivatives in $U?$ Let's think about $\partial F/\partial y.$ What happens if we just blatantly consider
$$\tag 1 \sum_{n\ge 1} \partial (\sin (ny)e^{-n(x+y)})/\partial y.$$
This series converges uniformly in $U,$ which you should verify. Now we're set up to use one variable results. Fix $x$ and look at the corresponding function of $y.$ The corresponding series converges uniformly on $(-1/4,1/4).$ From one variable it follows that $\partial F/\partial y$ exists everywhere in $U,$ and equals the sum in $(1).$ Furthermore, since each summand is continuous in $U,$ so is $\partial F/\partial y.$ Now repeat this process for $\partial F/\partial x.$ This gives $F\in C^1(U).$
