I am not too familiar with matrix calculus, how can we go about taking the derivative of this quantity: $$ \nabla_{x}Ax\operatorname{det}(2\pi\boldsymbol\Sigma)^{-\frac{1}{2}} \, e^{ -\frac{1}{2}(\mathbf{x} - \boldsymbol\mu)'\boldsymbol\Sigma^{-1}(\mathbf{x} - \boldsymbol\mu)}, $$ where $A,\Sigma$ are $n\times n$ dimensional matrices and $\mu$ is an element of $\mathbb{R}^n$; $n\geq 1$?
What I've done so far.... the 1-dimensional Case
I've computed the $(n=1)$ dimensional case to be $$ \frac{e^{\frac{-(x - \mu)^2}{\Sigma^2}} (\frac{-(2 x^2)}{\Sigma^2} + \frac{(2 \mu x)}{\Sigma^2} + 1)}{ \sqrt{2\pi \Sigma} } . $$
Define some new variables to reduce the visual clutter $$\eqalign{ M &= M^T=\Sigma^{-1}\,\,\,\, &w = (x-\mu) \cr \lambda &= \det(2\pi M)^{1/2}\,\,\, &\phi = \exp\Big(\frac{-w^TMw}{2}\Big) \cr }$$ So the function of interest is the vector $$\eqalign{f &= \lambda\phi Ax}$$ Find its differential and gradient, noting that $\{A,\lambda\}$ are constants wrt $x$. $$\eqalign{ df &= \lambda\phi A\,dx + \lambda Ax\,d\phi \cr &= \lambda\phi A\,dx - \lambda Ax(\phi w^TM\,dw) \cr &= \lambda\phi\,(A-Axw^TM)\,dx \cr \frac{\partial f}{\partial x} &= \lambda\phi\,(A-Axw^TM) \cr }$$