${dy \over du} = 7u^6$
-$7u^6 \times$
-
## Product rule for $y=uv$
$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
> the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$, must be raised, to produce that number $x$
-### Logarithmic identities
+### Logarithmic identities
+
$\log_b (xy)=\log_b x + \log_b y$
$\log_b x^n = n \log_b x$
$\log_b y^{x^n} = x^n \log_b y$
### Index identities
+
$b^{m+n}=b^m \cdot b^n$
$(b^m)^n=b^{m \cdot n}$
$(b \cdot c)^n = b^n \cdot c^n$
### Differentiating logarithms
$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
-## Solving $e^x$ etc
+## Derivative rules
| $f(x)$ | $f^\prime(x)$ |xs
| ------ | ------------- |
## Differentiating $x=f(y)$
-Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$
+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)
## Antidifferentiation
To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
+## Rates
+
+### Related rates
+
+$${da \over db} \quad \text{change in } a \text{ with respect to } b$$
+
+#### Gradient at a point on parametric curve
+
+$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
+