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Derivative of $x^TA$

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Let be $A$ an $n\times n$ matrix and $x\in\mathbb{R}^n$. Then wikipedia and the books say that the derivative of $f(x):=x^TA$ is $A^T$.

I was trying to apply this result to the official definition of differentiability i.e.

$f$ is differentiable in $a$ if there exists a linear mapping $B$ such that$$\lim\limits_{x\to a}\frac{\Vert x^T-a^TA-B(x-a)\Vert}{\Vert x-a\Vert}=0.$$


If I try $B=A^T$ then the nomiator is not well defined due to different dimensions. However, if I take the rows of $A^T$ and convert them into a $1\times n$ vector where the $i$-th row is located in the $i$-th entry of the $1\times n$ vector, then it works.

So my question is: Instead of saying that $A^T$ is the dervitative would it be "more" correct to say that the above constructed $1\times n$ vector is the derivative? Or how do I make sense of the commonly known statement that $A^T$ is the derivative?


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