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# Differential calculus
## Limits
$$f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}$$
-$$m_{\operatorname{tangent}}=\lim_{h \rightarrow 0}f^\prime(x)$$
+$$m_{\tan}=\lim_{h \rightarrow 0}f^\prime(x)$$
-$$m_{\operatorname{chord PQ}}=f^\prime(x)$$
+$$m_{\vec{PQ}}=f^\prime(x)$$
first principles derivative:
-$${m_{\operatorname{tangent at P}} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$
+$${m_{\text{tangent at }P} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$
## Gradient at a point
$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
-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 ${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$
+$$f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$$
## Logarithms
## Derivative rules
-| $f(x)$ | $f^\prime(x)$ |xs
+| $f(x)$ | $f^\prime(x)$ |
| ------ | ------------- |
| $\sin x$ | $\cos x$ |
| $\sin ax$ | $a\cos ax$ |
| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
-<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->
-
-Reciprocal derivatives:
+## Reciprocal derivatives
-$${{dy \over dx} \over 1} = dx \over dy$$
+$${1 \over {dy \over dx}} = {dx \over dy}$$
## Differentiating $x=f(y)$
-Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}}$.
+Find $dx \over dy$. Then ${dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}$.
$${dy \over dx} = {1 \over {dx \over dy}}$$
## Second derivative
-$$f(x) \implies f^\prime (x) \implies f^{\prime\prime}(x)$$
+$$f(x) \longrightarrow f^\prime (x) \longrightarrow f^{\prime\prime}(x)$$
-$$\therefore y \implies {dy \over dx} \implies {d({dy \over dx}) \over dx} \implies {d^2 y \over dx^2}$$
+$$\therefore y \longrightarrow {dy \over dx} \longrightarrow {d({dy \over dx}) \over dx} \longrightarrow {d^2 y \over dx^2}$$
Order of polynomial $n$th derivative decrements each time the derivative is taken
*Stationary point* - point of zero gradient (i.e. $f^\prime(x)=0$)
*Point of inflection* - point of maximum $|$gradient$|$ (i.e. $f^{\prime\prime} = 0$)
-- 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
+* 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
-## Antidifferentiation
+## Implicit Differentiation
-$$y={x^{n+1} \over n+1} + c$$
+**On CAS:** Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$.
+
+Used for differentiating circles etc.
+
+If $p$ and $q$ are expressions in $x$ and $y$ such that $p=q$, for all $x$ nd $y$, then:
+
+$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$
## Integration
-$$\int f(x) dx = F(x) + c$$
+$$\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$
+
+$$\int x^n \cdot dx = {x^{n+1} \over n+1} + c$$
- 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
$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$
| $e^k$ | $e^kx + c$ |
| $\sin kx$ | $-{1 \over k} \cos (kx) + c$ |
| $\cos kx$ | ${1 \over k} \sin (kx) + c$ |
+| $\sec^2 kx$ | ${1 \over k} \tan(kx) + c$ |
+| $1 \over \sqrt{a^2-x^2}$ | $\sin^{-1} {x \over a} + c \>\vert\> a>0$ |
+| $-1 \over \sqrt{a^2-x^2}$ | $\cos^{-1} {x \over a} + c \>\vert\> a>0$ |
+| $a \over {a^2-x^2}$ | $\tan^{-1} {x \over a} + c$ |
| ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ |
| $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)|
| $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ |
+Note $\sin^{-1} {x \over a} + \cos^{-1} {x \over a}$ is constant for all $x \in (-a, a)$.
+
+### Definite integrals
+
+$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$
+
+- Signed area enclosed by: $\> y=f(x), \quad y=0, \quad x=a, \quad x=b$.
+- *Integrand* is $f$.
+- $F(x)$ may be any integral, i.e. $c$ is inconsequential
+
+### Integration by substitution
+
+$$\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du$$
+
+Note $f(u)$ must be one-to-one $\implies$ one $x$ value for each $y$ value
+
+e.g. for $y=\int(2x+1)\sqrt{x+4} \cdot dx$:
+let $u=x+4$
+$\implies {du \over dx} = 1$
+$\implies x = u - 4$
+then $y=\int (2(u-4)+1)u^{1 \over 2} \cdot du$
+Solve as a normal integral
+
+#### Definite integrals by substitution
+
+For $\int^b_a f(x) {du \over dx} \cdot dx$, evaluate new $a$ and $b$ for $f(u) \cdot du$.
+
+### Trigonometric integration
+
+$$\sin^m x \cos^n x \cdot dx$$
+
+**$m$ is odd:**
+$m=2k+1$ where $k \in \mathbb{Z}$
+$\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x$
+Substitute $u=\cos x$
+
+**$n$ is odd:**
+$n=2k+1$ where $k \in \mathbb{Z}$
+$\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x$
+Subbstitute $u=\sin x$
+
+**$m$ and $n$ are even:**
+Use identities:
+
+- $\sin^2x={1 \over 2}(1-\cos 2x)$
+- $\cos^2x={1 \over 2}(1+\cos 2x)$
+- $\sin 2x = 2 \sin x \cos x$
+
+### Partial fractions
+
+On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`
+
+
## 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
+- nature of stationary points is determined by sign of $y=f(x)$ on either side of 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.
### Related rates
-$${da \over db} \quad \text{change in } a \text{ with respect to } b$$
+$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
#### Gradient at a point on parametric curve
-$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
-$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
-# Rational functions
+## Rational functions
$$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}$$
-## Addition of ordinates
+### Addition of ordinates
- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
-