$b^{m+n}=b^m \cdot b^n$
$(b^m)^n=b^{m \cdot n}$
$(b \cdot c)^n = b^n \cdot c^n$
+${a^m \div a^n} = {a^{m-n}}$
### $e$ as a logarithm
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
+Reciprocal derivatives:
+
+$${{dy \over dx} \over 1} = dx \over dy$$
+
+## Differentiating $x=f(y)$
+
+Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$
+
## Antidifferentiation
$$y={x^{n+1} \over n+1} + c$$
- area enclosed by curves
- $+c$ should be shown on each step without $\int$
-$$\int xn = {x^{n+1} \over n+1} + c$$
+$$\int x^n = {x^{n+1} \over n+1} + c$$
### 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) | $kc + c$ |
-| $x^n (n \in J\\\{-1\})$ | ${1 \over {n+1}}x^{n+1} + c$ |
+| $f(x)$ | $\int f(x) \cdot dx$ |
+| ------------------------------- | ---------------------------- |
+| $k$ (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.