update

doc/src/week35/week35.do.txt

@@ -315,16 +315,16 @@ $\bm{z}^T=\bm{y}^T\bm{A}$, a vector of length $n$. We have then, using the defin
 \alpha = \bm{z}^T\bm{x},
 \]
 !et
-which means that (using our previous example) we have
+which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have
 !bt
 \[
-\frac{\partial \alpha}{\partial \bm{x}} = \bm{z}=\bm{A}^T\bm{y}.
+\frac{\partial \alpha}{\partial \bm{x}} = \frac{\partial \bm{z}^T\bm{x}}{\partial \bm{x}}=\bm{z}^T=\bm{A}^T\bm{y}.
 \]
 !et
 
 Note that the resulting vector elements are the same for $\bm{z}^T$ and $\bm{z}$, the only difference is that one is just the transpose of the other.
 
-Since $\alpha$ is a scalar we have $\alpha =\alpha^T=\bm{x}^T\bm{A}^T\bm{y}$. Defining now $\bm{z}=\bm{x}^T\bm{A}^T$ we find that
+Since $\alpha$ is a scalar we have $\alpha =\alpha^T=\bm{x}^T\bm{A}^T\bm{y}$. Defining now $\bm{z}^T=\bm{x}^T\bm{A}^T$ we find that
 !bt
 \[
 \frac{\partial \alpha}{\partial \bm{y}} = \bm{z}^T=\bm{x}^T\bm{A}^T.
@@ -477,9 +477,15 @@ or as
 We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)
 !bt
 \[
-\frac{\partial (\bm{b}^T\bm{a})}{\partial \bm{a}} = \bm{b},
+\frac{\partial (\bm{x}^T\bm{a})}{\partial \bm{x}} = \bm{a}^T,
 \]
 !et
+!bt
+\[
+\frac{\partial (\bm{a}^T\bm{x})}{\partial \bm{x}} = \bm{a}^T,
+\]
+!et
+
 !bt
 \[
 \frac{\partial tr(\bm{B}\bm{A})}{\partial \bm{A}} = \bm{B}^T,