formatting edits

quarto/derivatives/derivatives.qmd

@@ -660,12 +660,20 @@ $$
 There are $n$ terms, each where one of the $f_i$s have a derivative. Were we to multiply top and bottom by $f_i$, we would get each term looks like: $f \cdot f_i'/f_i$.
 
 
-With this, we can proceed. Each term $x-i$ has derivative $1$, so the answer to $f'(x)$, with $f$ as above, is $f'(x) = f(x)/(x-1) + f(x)/(x-2) + f(x)/(x-3) + f(x)/(x-4) + f(x)/(x-5)$, that is:
+With this, we can proceed. Each term $x-i$ has derivative $1$, so the answer to $f'(x)$, with $f$ as above, is
 
+\begin{align*}
+f'(x) &= f(x)/(x-1) + f(x)/(x-2) + f(x)/(x-3)\\
+      &+ f(x)/(x-4) + f(x)/(x-5),
+\end{align*}
 
-$$
-f'(x) = (x-2)(x-3)(x-4)(x-5) + (x-1)(x-3)(x-4)(x-5) + (x-1)(x-2)(x-4)(x-5) + (x-1)(x-2)(x-3)(x-5) + (x-1)(x-2)(x-3)(x-4).
-$$
+That is
+
+\begin{align*}
+f'(x) &= (x-2)(x-3)(x-4)(x-5) + (x-1)(x-3)(x-4)(x-5)\\
+      &+ (x-1)(x-2)(x-4)(x-5) + (x-1)(x-2)(x-3)(x-5) \\
+	  &+ (x-1)(x-2)(x-3)(x-4).
+\end{align*}
 
 ---
 
@@ -680,10 +688,12 @@ $$
 ### Chain rule
 
 
-Finally, the derivative of a composition of functions can be computed using pieces of each function. This gives a rule called the *chain rule*. Before deriving, let's give a slight motivation.
+Finally, the derivative of a composition of functions can be computed using pieces of each function. This gives a rule called the *chain rule*. Before deriving, let's give a slight motivation through two examples.
+
+The first involves working out on a treadmill. For this example, there is a presumed linear relationship between miles run and calories burned. With that, the rate of calories burned per hour would be proportional to the miles per hours
 
 
-Consider the output of a factory for some widget. It depends on two steps: an initial manufacturing step and a finishing step. The number of employees is important in how much is initially manufactured. Suppose $x$ is the number of employees and $g(x)$ is the amount initially manufactured. Adding more employees increases the amount made by the made-up rule $g(x) = \sqrt{x}$. The finishing step depends on how much is made by the employees. If $y$ is the amount made, then $f(y)$ is the number of widgets finished. Suppose for some reason that $f(y) = y^2.$
+Now consider the output of a factory for some widget. It depends on two steps: an initial manufacturing step and a finishing step. The number of employees is important in how much is initially manufactured. Suppose $x$ is the number of employees and $g(x)$ is the amount initially manufactured. Adding more employees increases the amount made by the made-up rule $g(x) = \sqrt{x}$. The finishing step depends on how much is made by the employees. If $y$ is the amount made, then $f(y)$ is the number of widgets finished. Suppose for some reason that $f(y) = y^2.$
 
 
 How many widgets are made as a function of employees? The composition $u(x) = f(g(x))$ would provide that. Changes in the initial manufacturing step lead to changes in how much is initially made; changes in the initial amount made leads to changes in the finished products. Each change contributes to the overall change.
@@ -806,11 +816,10 @@ This is a useful rule to remember for expressions involving exponentials.
 Find the derivative of $\sin(x)\cos(2x)$ at $x=\pi$.
 
 
-$$
-[\sin(x)\cos(2x)]'\big|_{x=\pi} =
-(\cos(x)\cos(2x) + \sin(x)(-\sin(2x)\cdot 2))\big|_{x=\pi} =
-((-1)(1) + (0)(-0)(2)) = -1.
-$$
+\begin{align*}
+[\sin(x)\cos(2x)]'\big|_{x=\pi} &=(\cos(x)\cos(2x) + \sin(x)(-\sin(2x)\cdot 2))\big|_{x=\pi} \\
+& =((-1)(1) + (0)(-0)(2)) = -1.
+\end{align*}
 
 ##### Proof of the Chain Rule