$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
-# 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:
+
+$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$
+
+