+Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}}$.
+
+$${dy \over dx} = {1 \over {dx \over dy}}$$
+
+## Second derivative
+
+$$f(x) \implies f^\prime (x) \implies f^{\prime\prime}(x)$$
+
+$$\therefore y \implies {dy \over dx} \implies {d({dy \over dx}) \over dx} \implies {d^2 y \over dx^2}$$
+
+Order of polynomial $n$th derivative decrements each time the derivative is taken
+
+### Maxima and minima
+
+- 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)
+- if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection
+- - if also $f^\prime(a)=0$, then it is a stationary point of inflection
+
+*Point of inflection* - point of maximum gradient (either +ve or -ve)