Matrix calculus in multiple linear regression OLS estimate derivation The steps of the following derivation are from here
Starting from $y= Xb +\epsilon $, which really is just the same as
$\begin{bmatrix}
y_{1} \\
y_{2} \\
\vdots \\
y_{N}
\end{bmatrix}
=
\begin{bmatrix}
1 & x_{21} & \cdots & x_{K1} \\
1 & x_{22} & \cdots & x_{K2} \\
\vdots & \ddots & \ddots & \vdots \\
1 & x_{2N} & \cdots & x_{KN}
\end{bmatrix}
*
\begin{bmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{K}
\end{bmatrix}
+
\begin{bmatrix}
\epsilon_{1} \\
\epsilon_{2} \\
\vdots \\
\epsilon_{N}
\end{bmatrix} $
it all comes down to minimzing $e'e$:
$\epsilon'\epsilon = \begin{bmatrix}
e_{1} & e_{2} & \cdots & e_{N} \\
\end{bmatrix}
\begin{bmatrix}
e_{1} \\
e_{2} \\
\vdots \\
e_{N}
\end{bmatrix} = \sum_{i=1}^{N}e_{i}^{2}
$
So minimizing $e'e'$ gives us:
$min_{b}$ $e'e = (y-Xb)'(y-Xb)$
$min_{b}$ $e'e = y'y - 2b'X'y + b'X'Xb$
(*) $\frac{\partial(e'e)}{\partial b} = -2X'y + 2X'Xb \stackrel{!}{=} 0$
$X'Xb=X'y$
$b=(X'X)^{-1}X'y$
I'm pretty new to matrix calculus, so I was a bit confused about (*).
In step (*), $\frac{\partial(y'y)}{\partial b} = 0$, which makes sense. And then $\frac{\partial(-2b'X'y)}{\partial b} = -2X'y$, but why exactly is this true? If it were $\frac{\partial(-2b'X'y)}{\partial b'}$, then that would make perfect sense to me. Is taking the partial derivative with respect to $b$ the same as taking the partial derivative with respect to $b'$? 
Similarly, $\frac{\partial(b'X'Xb)}{\partial b} = X'Xb$ Why is this true? Shouldn't it be $= b'X'X$? 

This is not exaclty a proof but rather a way to think about it.
You are trying to minimize a scalar function $F(b)$. Now use the implicit derivative:
$$dF=d(y'y)-2d(b'X'y)+d(b'X'Xb)=-2db'X'y+db'X'Xb+b'X'Xdb.$$
Now transpose the last expression (which is a scalar) and factor $db'$.
$$dF=2db'(-X'y+X'Xb)$$
So the gradient of $F(b)$ is $2(-X'y+X'Xb)$. Set this to zero and solve for $b$. This procedure is sometimes also called the external definition of the gradient.
