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: 
Following the hint of zhw,

If $S\_{n}(x)=\sum\_{k=0}^{n}\frac{\sin{x}}{n^2 x^2+1}$ convereges uniformly to some $S(x)$ on the interval $[-1,1]$ then the tail sequence $T\_{n}(x)=\sum\_{k=n}^{\infty}\frac{\sin{x}}{k^2 x^2+1}$ should converge to zero
for every $x\in [-1,1]$.

But for $n\sim \frac{1}{\delta}$ $$ | T\_{n}(\delta)|= \sin{(\delta)}\,\sum\_{n\sim \frac{1}{\delta}} \frac{1}{1+n^2 \delta^2}\sim \frac{\sin{\delta}}{\delta} \sim 1 $$

So there does not exist any $N$ however large such that the tail $T\_{n}(x)$ is arbitrarily small for all (and this is key) $x$, when $n\geq N$.
If such $N$ exists we would have $T\_{N}(x)$ arbitrarily small for all $x\in [-1,1]$ but have just seen $T\_{N}(\frac{1}{N})\gtrsim 1$.
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.