Question: Suppose that we have a function $f(A) = Ab$ for $A \in \mathbb{R}^{m\times n}$ and $b \in \mathbb{R}^{n}$. Is this correct? $$\frac{df}{dA} = \begin{bmatrix} b^T\\ \vdots \\ b^T \end{bmatrix}$$

Answer: 
Hint: If the series were uniformly convergent on some $[0,a],$ then for large $N,$

$$\sum\_{n=N}^{2N}\frac{\sin x}{1+n^2x^2}$$

would be less than $1/10$ for $x\in [0,a].$ Try $x=1/N$ to see this doesn't happen.
Thanks @Jack D'Aurizio pointing out my error. Following the hint given by @zhw. we have
\begin{align\*}
\sum\_{n=N}^{2N}\dfrac{\sin x}{1+n^{2}x^{2}}<\dfrac{1}{2}(\tan^{-1}2-\tan^{-1}1)
\end{align\*}
for large $N$ and $x\in[0,a]$. Putting $x=1/N$, 
\begin{align\*}
\sin(1/N)\sum\_{n=N}^{2N}\dfrac{1}{1+(n/N)^{2}}&\geq\sin(1/N)\int\_{N}^{2N}\dfrac{1}{1+(x/N)^{2}}dx\\
&=N\sin(1/N)\int\_{1}^{2}\dfrac{1}{1+u^{2}}du\\
&=\dfrac{\sin(1/N)}{1/N}(\tan^{-1}2-\tan^{-1}1),
\end{align\*}
taking $N\rightarrow\infty$ will violate the inequality.