daviddoji/CalculusWithJuliaNotes.jl
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quarto/derivatives/derivatives.qmd
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@@ -688,12 +688,9 @@ $$
### 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 through two examples.
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 an example.
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
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.$
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.
@@ -824,7 +821,11 @@ Find the derivative of $\sin(x)\cos(2x)$ at $x=\pi$.
##### Proof of the Chain Rule
A function is *differentiable* at $a$ if the following limit exists $\lim_{h \rightarrow 0}(f(a+h)-f(a))/h$. Reexpressing this as: $f(a+h) - f(a) - f'(a)h = \epsilon_f(h) h$ where as $h\rightarrow 0$, $\epsilon_f(h) \rightarrow 0$. Then, we have:
A function is *differentiable* at $a$ if the following limit exists $\lim_{h \rightarrow 0}(f(a+h)-f(a))/h$.
This is reexpressed as: $f(a+h) - f(a) - f'(a)h = \epsilon_f(h) h$ where as $h\rightarrow 0$, $\epsilon_f(h) \rightarrow 0$.
With that in mind, we have:
$$
@@ -843,11 +844,12 @@ f(g(a) + g'(a)h + \epsilon_g(h)h) - f(g(a)) \\
Rearranging:
$$
f(g(a+h)) - f(g(a)) - f'(g(a)) g'(a) h = f'(g(a))\epsilon_g(h)h + \epsilon_f(h')(h') =
(f'(g(a)) \epsilon_g(h) + \epsilon_f(h') (g'(a) + \epsilon_g(h)))h =
\epsilon(h)h,
$$
\begin{align*}
f(g(a+h)) &- f(g(a)) - f'(g(a)) g'(a) h\\
&= f'(g(a))\epsilon_g(h)h + \epsilon_f(h')(h')\\
&=(f'(g(a)) \epsilon_g(h) + \epsilon_f(h') (g'(a) + \epsilon_g(h)))h \\
&=\epsilon(h)h,
\end{align*}
where $\epsilon(h)$ combines the above terms which go to zero as $h\rightarrow 0$ into one. This is the alternative definition of the derivative, showing $(f\circ g)'(a) = f'(g(a)) g'(a)$ when $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$.