## 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 $f(x) = h(g(x)) = (h \circ g)(x)$:
-Function notation:
+$$f^\prime(x) = h^\prime(g(x)) \cdot g^\prime(x)$$
-$$(f\circ g)^\prime(x)=f^\prime(g(x))g^\prime(x),\quad \mathbb{where}\hspace{0.3em} (f\circ g)(x)=f(g(x))$$
+If $y=h(u)$ and $u=g(x)$:
+
+$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
+$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$
Used with only one expression.
$y=u^7$
${dy \over du} = 7u^6$
-
## Product rule for $y=uv$
$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
### 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)
$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
-# Rational functions
+## Rational functions
$$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}$$
-## Addition of ordinates
+### Addition of ordinates
- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
+
+## Implicit Differentiation
+
+On CAS: Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$.
+
+Used for differentiating circles etc.
+
+If $p$ and $q$ are expressions in $x$ and $y$ such that $p=q$, for all $x$ nd $y$, then:
+
+$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$
+
+