orthogonal; work around plotly()

quarto/differentiable_vector_calculus/matrix_calculus_notes.qmd

@@ -355,9 +355,15 @@ The operator $f'(x)= g'(h(x)) h'(x)$ is a product of matrices.
 
 ### Computational differences with expressions from the chain rule
 
-Of note here is the application of the chain rule to three (or more compositions):
+Of note here is the application of the chain rule to three (or more compositions) where $c:R^n \rightarrow R^j$, $b:R^j \rightarrow R^k$, and $a:R^k \rightarrow R^m$:
 
-The derivative of $f(x) = a(x) b(x) c(x)$  can be expressed as
+If  $f(x) = a(b(c(x)))$ then the derivative is:
+
+$$
+f'(x) = a'(b(c(x))) b'(c(x)) c'(x),
+$$
+
+which can be expressed as three matrix multiplications two ways:
 
 $$
 f' = (a'b')c'  \text{ or } f' = a'(b'c')
@@ -380,7 +386,7 @@ Whereas computing from the right to left is first $jk1$ operations leaving a $j
 * left to right is $njk + nk$ = $nk \cdot (j + 1)$.
 * right to left is $jk + jn = j\cdot (k+n)$.
 
-When $j=k$, say, we can compare and see the second is a factor less in terms of operations. This can be quite significant in higher dimensions, whereas the dimensions of calculus (where $n$ and $m$ are $3$ or less) it is not an issue.
+When $j=k$, say, we can compare and see the second is a factor less in terms of operations. This can be quite significant in higher dimensions.
 
 
 ##### Example
@@ -405,7 +411,7 @@ Whereas the relationship is changed when the first matrix is skinny and the last
 
 ----
 
-In calculus, we have $n$ and $m$ are $1$,$2$,or $3$. But that need not be the case, especially if differentiation is over a parameter space.
+In calculus, we typically have $n$ and $m$ are $1$, $2$,or $3$. But that need not be the case, especially if differentiation is over a parameter space.
 
 ## Derivatives of matrix functions
 
@@ -546,7 +552,7 @@ Appropriate sizes for $A$, $B$, and $C$ are determined by the various products i
 
 If $A$ is $m \times n$ and $B$ is $r \times s$, then since $BC$ is defined, $C$ has $s$ rows, and since $CA^T$ is defined, $C$ must have $n$ columns, as $A^T$ is $n \times m$, so $C$ must be $s\times n$. Checking this is correct on the other side, $A \times B$ would be size $mr \times ns$ and $\vec{C}$ would be size $sn$, so that product works, size wise.
 
-The referred to notes have an explanation for this formula, but we confirm with an example with $m=n=2$ and $r=s=3$:
+The referred to notes have an explanation for this formula, but we only confirm it with an example using $m=n=2$ and $r=s=3$:
 
 ```{julia}
 @syms A[1:2, 1:2]::real B[1:3, 1:3]::real C[1:3, 1:2]::real