## Chain rule for $(f\circ g)$
-$$(f \circ g)^\prime = (f^\prime \circ g) \cdot g^\prime$$
-
-Leibniz notation:
-
$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
+$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$
Function notation:
${dy \over du} = 7u^6$
-$7u^6 \times$
-
## 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
+
+$b^{m+n}=b^m \cdot b^n$
+$(b^m)^n=b^{m \cdot n}$
+$(b \cdot c)^n = b^n \cdot c^n$
+${a^m \div a^n} = {a^{m-n}}$
+
### $e$ as a logarithm
-$$\log_e e = 1$$
+$$\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y$$
$$\ln x = \log_e x$$
### Differentiating logarithms
-$${d \over dx} \log_b x = {1 \over x \ln b}$$
+$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
-## Solving $e^x$
+## Derivative rules
-| $f(x)$ | $f^\prime(x)$ |
+| $f(x)$ | $f^\prime(x)$ |xs
| ------ | ------------- |
| $\sin x$ | $\cos x$ |
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
+| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $\log_e x$ | $1 \over x$ |
| $\log_e {ax}$ | $1 \over x$ |
+| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
+| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
+
+<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+
+Reciprocal derivatives:
+
+$${{dy \over dx} \over 1} = dx \over dy$$
+
+## Differentiating $x=f(y)$
+
+Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}}$.
+
+$${dy \over dx} = {1 \over {dx \over dy}}$$
+
+## 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
+
+$$y={x^{n+1} \over n+1} + c$$
+
+## Integration
+
+$$\int f(x) dx = F(x) + c$$
+
+- area enclosed by curves
+- $+c$ should be shown on each step without $\int$
+
+$$\int x^n = {x^{n+1} \over n+1} + c$$
+
+### Integral laws
+
+$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$
+$\int k f(x) dx = k \int f(x) dx$
+
+| $f(x)$ | $\int f(x) \cdot dx$ |
+| ------------------------------- | ---------------------------- |
+| $k$ (constant) | $kx + c$ |
+| $x^n$ | ${1 \over {n+1}}x^{n+1} + c$ |
+| $a x^{-n}$ | $a \cdot \log_e x + c$ |
+| $e^{kx}$ | ${1 \over k} e^{kx} + c$ |
+| $e^k$ | $e^kx + c$ |
+| $\sin kx$ | $-{1 \over k} \cos (kx) + c$ |
+| $\cos kx$ | ${1 \over k} \sin (kx) + c$ |
+| ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ |
+| $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)|
+| $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ |
+| ${1 \over {ax+b}}$ | ${1 \over a} \log_e (ax+b) + c$ |
+| $(ax+b)^n$ | ${1 \over {a(n+1)}}(ax+b)^{n-1} + c$ |
+
+## Applications of antidifferentiation
+
+- $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$
+- the nature of any stationary point of $y=F(x)$ is determined by the way the sign of the graph of $y=f(x)$ changes about its $x$-intercepts
+- if $f(x)$ is a polynomial of degree $n$, then $F(x)$ has degree $n+1$
+
+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
+
+$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$