----
-geometry: margin=2cm
-columns: 2
-graphics: yes
-tables: yes
-author: Andrew Lorimer
-classoption: twocolumn
----
-
# 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
## Implicit Differentiation
-On CAS: Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$.
+**On CAS:** Action $\rightarrow$ Calculation $\rightarrow$ `impDiff(y^2+ax=5, x, y)`. Returns $y^\prime= \dots$.
Used for differentiating circles etc.
$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$
-## Antidifferentiation
-
-$$y={x^{n+1} \over n+1} + c$$
-
## Integration
-$$\int f(x) dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$
+$$\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$
+
+$$\int x^n \cdot dx = {x^{n+1} \over n+1} + c$$
- 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
-$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)_{}$$
+$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$
+
+- 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
## 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