From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Tue, 2 Apr 2019 11:10:15 +0000 (+1100)
Subject: [spec] clarify stationary points & chain rule
X-Git-Tag: yr12~170
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/cc220a67dfb1dfe7cdfbcb695761406a5e402ade

[spec] clarify stationary points & chain rule
---

diff --git a/spec/calculus.md b/spec/calculus.md
index 5c3b3fc..a085f3b 100644
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -95,12 +95,14 @@ $$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$
 
 ## Chain rule for $(f\circ g)$
 
-$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
-$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$
+If $f(x) = h(g(x)) = (h \circ g)(x)$:
+
+$$f^\prime(x) = h^\prime(g(x)) \cdot g^\prime(x)$$
 
-Function notation:
+If $y=h(u)$ and $u=g(x)$:
 
-$$(f\circ g)^\prime(x)=f^\prime(g(x))g^\prime(x),\quad \mathbb{where}\hspace{0.3em} (f\circ g)(x)=f(g(x))$$
+$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
+$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$
 
 Used with only one expression.
 
@@ -110,7 +112,6 @@ ${du \over dx} = 2x$
 $y=u^7$  
 ${dy \over du} = 7u^6$  
 
-
 ## Product rule for $y=uv$
 
 $${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
@@ -196,7 +197,8 @@ Order of polynomial $n$th derivative decrements each time the derivative is take
 
 ### Points of Inflection
 
-*Point of inflection* - point of maximum gradient (either +ve or -ve). Occurs where $f^{\prime\prime} = 0$
+*Stationary point* - point of zero gradient (i.e. $f^\prime(x)=0$)  
+*Point of inflection* - point of maximum $|$gradient$|$ (i.e.  $f^{\prime\prime} = 0$)
 
 - if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up)
 - if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down)