Right now I'm in an analysis course and we are on differentiation of multivariable functions. I know that if we have a set of functions f: $\mathbb{R^{n}}$ to $\mathbb{R^{m}}$, the derivative of a function at a point is defined as m x n matrix of partial derivatives. 
However, this matrix is written in terms of the standard basis for $\mathbb{R^{n}}$ and the standard basis for $\mathbb{R^{m}}$. I want to rewrite this matrix in terms of a standard basis for $\mathbb{R^{n}}$ and the standard basis for $\mathbb{R^{n}}$. I.e. I want to find:
, where B is a basis given for $\mathbb{R^{n}}$.
My linear algebra is pretty rusty and I'm not sure even if the matrix would change at all. Any help, ideas would be very much appreciated. Thank you!
Let $\varphi \colon \Bbb{R}^n \to \Bbb{R}^n$ be a change of coordinates given by $\varphi(x_1,\dots,x_n) = (y_1,\dots,y_n)$, where each $y_i$ is a function of $x_1,\dots,x_n$. Then you can express your function $f \colon \Bbb{R}^n \to \Bbb{R}^m$ in the $y$-coordinates by composing with $\varphi^{-1}$. That is, the composition $f \circ \varphi^{-1}$. This is because $\varphi^{-1}$ first changes back to the original coordinates, and then applies the map $f$.
The chain rule says that the derivative of a composion is the composition of the derivatives. Using the notation you've written above, let's call $a=(x_1,\dots,x_n)$ a point in the original coordinates, and let's call $b = \varphi(a) = (y_1,\dots,y_n)$ the representation of the same point in the new coordinates. Then the chain rule says that $D(f \circ \varphi^{-1})(b) = Df(a) \cdot D\varphi^{-1}(b)$.
As a simple example, consider changing to polar coordinates in $\Bbb{R}^2$. Then $\varphi(x,y) = (r,\theta)$, and the inverse transformation is given by $\varphi^{-1}(r,\theta) = (x,y) = (r\cos(\theta), \, r\sin(\theta))$. Then
$$ D\varphi^{-1}(r,\theta) = \left( \begin {array}{cc} \cos(\theta) & -r\sin(\theta) \\[1.2ex] \sin(\theta) & r\cos(\theta) \end{array} \right) $$
Then if you take any function $f \colon \Bbb{R}^2 \to \Bbb{R}^m$ given in the original $x,y$-coordinates, then the expression of the Jacobian matrix of partial derivatives in polar coordinates is given by
$$ \begin {align*} Df(x,y) \cdot D\varphi^{-1}(r,\theta) &= \left(\begin{array}{cc} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \vdots & \vdots \\ \frac{\partial f_m}{\partial x} & \frac{\partial f_m}{\partial y} \end{array} \right) \left( \begin{array}{cc} \cos(\theta) & -r\sin(\theta) \\[1.2ex] \sin(\theta) & r\cos(\theta) \end{array} \right) \\ &= \left( \begin{array}{cc} \cos(\theta) \frac{\partial f_1}{\partial x} + \sin(\theta) \frac{\partial f_1}{\partial y} & r \cos(\theta) \frac{\partial f_1}{\partial y} - r \sin(\theta) \frac{\partial f_1}{\partial x} \\ \vdots & \vdots \\ \cos(\theta) \frac{\partial f_m}{\partial x} + \sin(\theta) \frac{\partial f_m}{\partial y} & r \cos(\theta) \frac{\partial f_m}{\partial y} - r \sin(\theta) \frac{\partial f_m}{\partial x} \end{array} \right) \end {align*} $$