update

doc/pub/week35/html/._week35-bs010.html

@@ -379,14 +379,14 @@ $$
 \alpha = \boldsymbol{z}^T\boldsymbol{x},
 $$
 
-<p>which means that (using our previous example) we have</p>
+<p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
 <p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
 
-<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
+<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$
 \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T.
 $$

doc/pub/week35/html/._week35-bs014.html

@@ -364,9 +364,14 @@ MathJax.Hub.Config({
 
 <p>We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)</p>
 $$
-\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
 $$
 
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
+$$
+
+
 $$
 \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
 $$

doc/pub/week35/html/week35-reveal.html

@@ -537,16 +537,16 @@ $$
 $$
 <p>&nbsp;<br>
 
-<p>which means that (using our previous example) we have</p>
+<p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 <p>&nbsp;<br>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 <p>&nbsp;<br>
 
 <p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
 
-<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
+<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 <p>&nbsp;<br>
 $$
 \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T.
@@ -715,7 +715,13 @@ $$
 <p>We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)</p>
 <p>&nbsp;<br>
 $$
-\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
+$$
+<p>&nbsp;<br>
+
+<p>&nbsp;<br>
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
 $$
 <p>&nbsp;<br>
 

doc/pub/week35/html/week35-solarized.html

@@ -602,14 +602,14 @@ $$
 \alpha = \boldsymbol{z}^T\boldsymbol{x},
 $$
 
-<p>which means that (using our previous example) we have</p>
+<p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
 <p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
 
-<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
+<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$
 \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T.
 $$
@@ -739,9 +739,14 @@ $$
 
 <p>We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)</p>
 $$
-\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
 $$
 
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
+$$
+
+
 $$
 \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
 $$

doc/pub/week35/html/week35.html

@@ -679,14 +679,14 @@ $$
 \alpha = \boldsymbol{z}^T\boldsymbol{x},
 $$
 
-<p>which means that (using our previous example) we have</p>
+<p>which means that (using our previous example and keeping track of our definition of the derivative of a scalar) we have</p>
 $$
-\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{z}=\boldsymbol{A}^T\boldsymbol{y}.
+\frac{\partial \alpha}{\partial \boldsymbol{x}} = \frac{\partial \boldsymbol{z}^T\boldsymbol{x}}{\partial \boldsymbol{x}}=\boldsymbol{z}^T=\boldsymbol{A}^T\boldsymbol{y}.
 $$
 
 <p>Note that the resulting vector elements are the same for \( \boldsymbol{z}^T \) and \( \boldsymbol{z} \), the only difference is that one is just the transpose of the other.</p>
 
-<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
+<p>Since \( \alpha \) is a scalar we have \( \alpha =\alpha^T=\boldsymbol{x}^T\boldsymbol{A}^T\boldsymbol{y} \). Defining now \( \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T \) we find that</p>
 $$
 \frac{\partial \alpha}{\partial \boldsymbol{y}} = \boldsymbol{z}^T=\boldsymbol{x}^T\boldsymbol{A}^T.
 $$
@@ -816,9 +816,14 @@ $$
 
 <p>We list here some other useful relations we may encounter (recall that vectors are defined by boldfaced low-key letters)</p>
 $$
-\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+\frac{\partial (\boldsymbol{x}^T\boldsymbol{a})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
 $$
 
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{x})}{\partial \boldsymbol{x}} = \boldsymbol{a}^T,
+$$
+
+
 $$
 \frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
 $$

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,