## Derivatives of $x^n$
-For $f: \mathbb{R} \rightarrow \mathbb{R}$ where $f(x)=x^n, x \in \mathbb{N}$
-
-Derivative is $f^\prime(x) = nx^{n-1}$
+$${d(ax^n) \over dx}=anx^{n-1}$$
If $x=$ constant, derivative is $0$
+If $y=ax^n$, derivative is $a\times nx^{n-1}$
+
If $f(x)={1 \over x}=x^{-1}, \quad f^\prime(x)=-1x^{-2}={-1 \over x^2}$
If $f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}$
$$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$
+## Derivatives of $u \pm v$
+
+$${dy \over dx}={du \over dx} \pm {dv \over dx}$$
+where $u$ and $v$ are functions of $x$
+
## Euler's number as a limit
$$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$
-## Chain rule
-
-Leibniz notation:
+## Chain rule for $(f\circ g)$
-$${dy \over dx} = {dy \over du} \times {du \over dx}$$
+$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
+$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$
Function notation:
$7u^6 \times$
+## Product rule for $y=uv$
+
+$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
+
+Surds can be left on denomintaors.
+
+## Quotient rule for $y={u \over v}$
+
+$${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$$
+
+If $f(x)={u(x) \over v(x)}$, then $f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$
+
+If $y={u(x) \over v(x)}$, then derivative ${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$
+
+## Logarithms
+
+$$\log_b (x) = n \quad \operatorname{where} \hspace{0.5em} b^n=x$$
+
+Wikipedia:
+
+> 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
+$\log_b (xy)=\log_b x + \log_b y$
+$\log_b x^n = n \log_b x$
+$\log_b y^{x^n} = x^n \log_b y$
+
+### 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
+
+$$\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y$$
+$$\ln x = \log_e x$$
+
+### Differentiating logarithms
+$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
+
+## Solving $e^x$ etc
+
+| $f(x)$ | $f^\prime(x)$ |xs
+| ------ | ------------- |
+| $\sin x$ | $\cos x$ |
+| $\sin ax$ | $a\cos ax$ |
+| $\cos x$ | $-\sin x$ |
+| $\cos ax$ | $-a \sin ax$ |
+| $e^x$ | $e^x$ |
+| $e^{ax}$ | $ae^{ax}$ |
+| $ax^{nx}$ | $an \cdot e^{nx}$ |
+| $\log_e x$ | $1 \over x$ |
+| $\log_e {ax}$ | $1 \over x$ |
+| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
+| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+
+<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+
+## 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.
+