From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Thu, 16 Aug 2018 10:57:25 +0000 (+1000)
Subject: clarify chain/product/quotient rules
X-Git-Tag: yr11~67
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/becf5ca1430ee99bdbe44b482b7c57a81789632c

clarify chain/product/quotient rules
---

diff --git a/spec/calculus.md b/spec/calculus.md
index c0056c4..4c33318 100644
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -70,12 +70,12 @@ Given point $P(a, b)$ and function $f(x)$, the gradient is $f^\prime(a)$
 
 ## Derivatives of $x^n$
 
-For $f: \mathbb{R} \rightarrow \mathbb{R}$ where $f(x)=x^n, x \in \mathbb{N}$
-
-Derivative is $f^\prime(x) = nx^{n-1}$
+$${d(ax^n) \over dx}=anx^{n-1}$$
 
 If $x=$ constant, derivative is $0$
 
+If $y=ax^n$, derivative is $a\times nx^{n-1}$
+
 If $f(x)={1 \over x}=x^{-1}, \quad f^\prime(x)=-1x^{-2}={-1 \over x^2}$
 
 If $f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}$
@@ -84,15 +84,22 @@ If $f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)$
 
 $$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$
 
+## Derivatives of $u \pm v$
+
+$${dy \over dx}={du \over dx} \pm {dv \over dx}$$
+where $u$ and $v$ are functions of $x$
+
 ## Euler's number as a limit
 
 $$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$
 
 ## Chain rule
 
+$$(f \circ g)^\prime = (f^\prime \circ g) \cdot g^\prime$$
+
 Leibniz notation:
 
-$${dy \over dx} = {dy \over du} \times {du \over dx}$$
+$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
 
 Function notation:
 
@@ -109,15 +116,15 @@ ${dy \over du} = 7u^6$
 
 $7u^6 \times$
 
-## Product rule
-
-If $f(x)=u(x) \times v(x)$, then $f^\prime (x) = u(x) \times v^\prime(x) + v(x)\times u^\prime(x)$
+## Product rule for $y=uv$
 
-If $y=uv$, then derivative ${dy \over dx} = u{dv \over dx} + v{du \over dx}$
+$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
 
 Surds can be left on denomintaors.
 
-## Quotient rule
+## Quotient rule for $y={u \over v}$
+
+$${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$$
 
 If $f(x)={u(x) \over v(x)}$, then $f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$