# User:Saul/calculus

## Contents

## Differentiation

We will use the function notation **ƒ(x)** which just applies some action to **x** like this function will double x: **ƒ(x) = 2x**

Differentiation of a function is taking a function (usually a curve) and finding the gradient at the single instant of **x**.

For example the function **ƒ(x) = x ^{2}** will represent a 'U' shaped curve, the gradient at point

**x**will be:

**ƒ**

^{′}(x) = 2xA derivative of a function will be notated with

**ƒ**, second derivatives are marked

^{′}**ƒ**and so on.

^{′′}### Various Notations

The following are various notations for derivatives.

Note that: **y = ƒ(x) = ƒ**

**dy / dx**

**dƒ / dx**

**dƒ(x) / dx**

**y ^{′}**

**[y]**

^{′}**ƒ**

^{′}**[ƒ]**

^{′}Notation for second derivative:

**ƒ**

^{′′}**d**

^{2}y / dx^{2}### Various Rules

Note: **n** represents a *real* number, and **a** represents a constant.

**ƒ(x) = x ^{n}**

**ƒ**

^{′}(x) = nx^{n-1}**ƒ(x) = ax**

^{n}**ƒ**

^{′}(x) = anx^{n-1}**ƒ(x) = a**

**ƒ**

^{′}(x) = 0**ƒ(x) = x**

**ƒ**

^{′}(x) = 1**ƒ(x) = fg**

**ƒ**

^{′}(x) = f^{′}g + g^{′}f**ƒ(x) = f/g**

**ƒ**

^{′}(x) = (f^{′}g - g^{′}f) / g^{2}**ƒ(x) = sin(x)**

**ƒ**

^{′}(x) = cos(x)**ƒ(x) = cos(x)**

**ƒ**

^{′}(x) = -sin(x)**ƒ(x) = tan(x)**

**ƒ**

^{′}(x) = 1 / cos^{2}(x) = sec^{2}(x)### Uses

#### Finding x-Intercept

Differentiation can be used to find the x-intercepts of a line.

The following formulae is used to find this:

**x _{n+1} = x_{n} - ƒ(x_{n}) / ƒ^{′}(x_{n})**

Applying this function until

**ƒ(x) ≈ 0**

The initial

**x**is just a guess.

Note: if this results in divergence (

**x**does not converge at

**0**or

**x → ∞**) then

**x**was a bad guess.

_{1}#### Approximating Roots and Powers

Differentiation can be used to approximate roots.

For example approximating the answer to:

^{3}√(120)

First round to the nearest known value:

^{3}√(125) = 5

Differentiate the original equation:

**ƒ(x) = ^{3}√x = x^{1/3}**

**ƒ**

^{′}(x) = x^{-2/3}/ 3Substitute the approximation:

**ƒ**

^{′}(120) = 120^{-2/3}/ 3**120**

^{-2/3}/ 3 = 0.014Place this into the equation of gradient:

**(y - 5) / (x - 125) = 0.014**

Prearrange to the formulae of a line:

**y = 0.014x + 3.29**

Substitute for the number to approximate with

**x**:

**y = 0.014(120) + 3.29**

**y = 4.934**

Therefore:

^{3}√(120) ≈ 4.934**(calculated)**

^{3}√(120) = 4.932#### Turning Points

A turning point is the point where the gradient is **0**, an equation may have multiple, quadratics usually have one, cubic functions usually have 2.

##### Locating

The turning point can be found by obtaining the first derivative (**ƒ ^{′}(x)**).

Then solve for when

**ƒ**.

^{′}(x) = 0The results of this are the

**x**values of the turning points, the

**y**values can be found by substituting the

**x**values into the original equation.

##### Minimum and Maximum

A **Minimum** is the the turning point that is **Concave Up**.

A **Maximum** is the the turning point that is **Concave Down**.

The **Minimum** can be found where the first derivative (**ƒ ^{′}(x)**) is

**0**and the second derivative (

**ƒ**) is

^{′′}(x)**positive**.

**ƒ**and

^{′}(x) = 0**ƒ**

^{′′}(x) > 0The

**Maximum**can be found where the first derivative (

**ƒ**) is

^{′}(x)**0**and the second derivative (

**ƒ**) is

^{′′}(x)**negative**.

**ƒ**and

^{′}(x) = 0**ƒ**

^{′′}(x) < 0##### Concave Up and Down

Concave Up is the part of the graph that is **U** shape.

Concave Down is the part of the graph that is **n** shape.

Cubic functions have one of each, quadratics have one or the other.

To find out if a point is Concave Up or Down, find the second derivative (**ƒ ^{′′}(x)**).

If

**ƒ**, then the point is Concave Up.

^{′′}(x) > 0If

**ƒ**, then the point is Concave Down.

^{′′}(x) < 0##### Point of Inflection

The Point of Inflection (P.O.I.) is the point on the graph where Concave Up changes to Concave Down.

The P.O.I. is when the second derivative (**ƒ ^{′′}(x)**) is equal to

**0**and it changes from a negative value to a positive one immediately beforehand (or vice versa).

**If**:

**ƒ**

^{′′}(x) = 0**ƒ**

^{′′}(x - 0.1) < 0**ƒ**

^{′′}(x + 0.1) > 0**Or**:

**ƒ**

^{′′}(x) = 0**ƒ**

^{′′}(x - 0.1) > 0**ƒ**

^{′′}(x + 0.1) < 0**Then**:

**x = P.O.I.**

## Integration

Integration is the reverse of differentiation, sometimes known as anti-derivative.

Integration is usually noted like so:

**∫ƒ ^{′}(x)dx = ƒ(x) + c**

Where

**dx**shows that it will be the integral relative to

**x**(there may be other variables in the equation).

**c**represents a unknown constant - this must exist because when a function is derived (differentiation) some information is lost so

**c**is to make up for that loss.

### Various Rules

**k** represents a constant.

**∫k dx = k + c**

**∫kx dx = k∫x dx**

**∫x ^{n} dx = x^{n + 1} / (n + 1)**

**∫(x**

^{n}+ x^{m}) dx = ∫x^{n}dx + ∫x^{m}dx**∫( ƒ(x) ± g(x) )dx = ∫ƒ(x)dx ± ∫g(x)dx**

**∫(1 / x) dx = Ln(x) + c**

Note:

**Ln(x) = Log**

_{e}(x)### Uses

#### Area

A use of integration is to find the area between a line and the **x** axis between two points on a graph.

The area between the two points **a** and **b** (where **x = 0**) can be found like so:

First find the derivative:

_{a}∫^{b}ƒ^{′}(x) dx = ƒ(x)

Say that the derivative is **F**, then find the difference of the values for **F** for each limit like so:

**F(b) - F(a) = Area**

Note that the area will be negative if it is under the **x** axis and positive if above.

Also if the line crosses the **x** axis more than twice then there will be more than two values where **x = 0** and at least one positive area and a negative area, these will at least partially cancel each other out if only the left most and right most **x** values are taken, so to effectively calculate the area the absolute areas between every set of points (where **x = 0**) muse be calculated.

#### Solid of Revolution

A cool use for integration is to calculate a solid of revolution - the volume of a shape that is made from the area of two points spun around a axis, for example a triangle could create a cone.

The formulae for this is as follows:

**V = Π _{a}∫^{b} ƒ^{2}(x) dx**

Note:

**ƒ**

^{2}(x) = ƒ(x) * ƒ(x)