# Differential calculus

## Limits

$$\lim_{x \rightarrow a}f(x)$$

$L^-$ - limit from below

$L^+$ - limit from above

$\lim_{x \to a} f(x)$ - limit of a point  

- Limit exists if $L^-=L^+$
- If limit exists, point does not.

Limits can be solved using normal techniques (if div 0, factorise)

## Limit theorems

1. For constant function $f(x)=k$, $\lim_{x \rightarrow a} f(x) = k$
2. $\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G$
3. $\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G$
4. ${\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0$

Corollary: $\lim_{x \rightarrow a} c \times f(x)=cF$ where $c=$ constant

## Solving limits for $x\rightarrow\infty$

Factorise so that all values of $x$ are in denominators.

e.g.

$$\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2$$


## Continuous functions

A function is continuous if $L^-=L^+=f(x)$ for all values of $x$.

## Gradients of secants and tangents

Secant (chord) - line joining two points on curve

Tangent - line that intersects curve at one point

given $P(x,y) \quad Q(x+\delta x, y + \delta y)$:
gradient of chord joining $P$ and $Q$ is ${m_{PQ}}={\operatorname{rise} \over \operatorname{run}} = {\delta y \over \delta x}$

As $Q \rightarrow P, \delta x \rightarrow 0$. Chord becomes tangent (two infinitesimal points are equal).

Can also be used with functions, where $h=\delta x$.

## First principles derivative

$$f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}$$

$$m_{\operatorname{tangent}}=\lim_{h \rightarrow 0}f^\prime(x)$$



$$m_{\operatorname{chord PQ}}=f^\prime(x)$$

first principles derivative:
$${m_{\operatorname{tangent at P}} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$

## Gradient at a point

Given point $P(a, b)$ and function $f(x)$, the gradient is $f^\prime(a)$


## Derivatives of $x^n$

$${d(ax^n) \over dx}=anx^{n-1}$$

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 $f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}$

If $f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)$

$$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$

## 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 for $(f\circ g)$

$${dy \over dx} = {dy \over du} \cdot {du \over dx}$$
$${d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}$$

Function notation:

$$(f\circ g)^\prime(x)=f^\prime(g(x))g^\prime(x),\quad \mathbb{where}\hspace{0.3em} (f\circ g)(x)=f(g(x))$$

Used with only one expression.

e.g. $y=(x^2+5)^7$ - Cannot reasonably expand  
Let $u-x^2+5$ (inner expression)  
${du \over dx} = 2x$  
$y=u^7$  
${dy \over du} = 7u^6$  


$7u^6 \times$

## 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
| ------ | ------------- |
| $\sin x$ | $\cos x$ |
| $\sin ax$ | $a\cos ax$ |
| $\cos x$ | $-\sin x$ |
| $\cos ax$ | $-a \sin ax$ |
| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ax^{nx}$ | $an \cdot e^{nx}$ |
| $\log_e x$ | $1 \over x$ |
| $\log_e {ax}$ | $1 \over x$ |
| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |

<!-- $${d(ax^{nx}) \over dx} = an \cdot e^nx$$ -->

Reciprocal derivatives:

$${{dy \over dx} \over 1} = dx \over dy$$

## Differentiating $x=f(y)$

Find $dx \over dy$. Then $dx \over dy = {1 \over {dy \over dx}} \therefore {dy \over dx} = {1 \over {dx \over dy}$

## 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.