Differential calculation of function many the variables. Derivate of part.

Part and complete differentials. Application complete a differential.

Making most of your lectures

This is a revision lecture so you should know “at least” the basics already.

DO NOTCopy down word for word what is written in this PowerPoint, notes are already in your subject booklets.

DOListen, and make additions to your notes. Ask questions

CALCULUS

Calculus is the branch of maths concerned with the study of change.It comprises of two chief branches, differential calculus and Integral calculus.They collectively focus on infinite series, functions, derivatives, integrals and limits.

(don’t worry about infinite series, not apart of methods)

Reaching your LIMIT

“The limit of a function is the value that the function approaches as x approaches a given value” – Maths Quest

Approach meaning becoming close to an extent where the approaching value basically equals the given value.

But it does not equal that value.

Limits, continued

Mathematically this can be stated as follows.As, x given value “c”

f(x) f(c)

Therefore, “Δx” or “h” 0

Because Δx is infinitesimal, we can neglect it and use direct substitution to solve limit problems.

i.e, [ lim (5x - 4) ] = [ 5(1) – 4 ] = [ 1 ]x1

Limits, a revised definition

A function does not need to be continuous at a point In order for its limit to be defined. For a limit to be defined than the left limit must equal the right limit. Take for example the following scenario.

let, f(x)=(x3/x) then, f(0) = 0/0 = (UNDEFINED) therefore discontinuous at “x=0”

take lim(x3/x) = lim(x2) = 0

lim(x3/x) = 0 because

lim(x3/x) = 0 = lim(x3/x)

Notice in this example how lim(x3/x)

exists even though it is discontinuous

at x=0.

x0

f(x) = x3/xx0-

x0

x0

x0+

x0

x0-x0+

Limit example

(x2-x-6) (x-3)(x+2)

(x-2) (x-2) lim = lim = lim (x-3) = -1

x2 x2 x2

Direct substitution is impossible as denominater = “0”

Simplify expression by factorising and cancelling (x-2)

Direct substitution is possible

Final answer

Derivatives and differentiation

The derivative of a function gives the gradient function.

i.e. the instantaneous rate of change of function f(x).

Differentiation is the process of determining the derivative from a given function.

Thus, this implies that…

the derivative f’(x) is found by the differentiation of function f’(x)

Differentiation – first principles

Linear approximation

Find square root of 4.1

Remember: f(x+h) = f(x) + hf`(x)

x=4 and h = 0.1

Let f(x) = x ………… f`(x) =

f(4) = 2 ………….f`(x) = ¼

f(4+0.1) = 2 + 0.1 x ¼ = 2.025

1 r2 x

Differentiation Rules

If f(x) = x6 +4x2 – 18x + 90

f’(x) = 6x5 + 8x – 18

*multiply by the power, than subtract one from the power.

Differentiation of transcendental functions

Transcendental functions are those that cannot be expressed by a polynomial equation. Thus they include exponential, logarithmic and trigonometric functions.

Below are the transcendental functions and their respective derivatives.

f(x) = sin (x) f`(x) = cos (x)

f(x) = cos (x) f`(x) = -sin (x)

f(x) = tan (x) f`(x) = sec2 (x)

f(x) = ex f`(x) = ex f(x) = loge(x) f`(x) = 1/x

Chain rule

The chain rule is useful in many situations, usually when you have a function inside of another function.

i.e. y=(1+2x)15 ……..Let (1+2x) = u

y=u15 ….dy/du = 15u14 and du/dx = 2

Use above formula gives,

dy/dx=15x2xu14 = 30(1+2x)14

Chain rule in transcendental

Take for example

y=esin(x)….let sin(x) = u gives; du/dx = cos(x)

y=eu….dy/du =eu

Use chain rule formula

dy/dx= eu.cos(x)

= cos(x).esin(x)

Thus, assign the function that is inside of another function “u”, in this case sin(x) in inside the exponential.

Pruduct Rule

The product Rule is stated as follows; If f(x) = u.v than….. f`(x) = vu’ +uv’

example,

Quotient Rule

The Quotient rule is stated as follows;

If f(x) = than f`(x) =vu’-uv’ nnv2

u v

Tangent and Normal Equations

The gradient of line at A = the derivative at A

To find the equation of the tangent above:

1. Find ‘m’ using dy/dx

2. Find the y-value at A if not given

3. Substitute all (x-, y-value at A, and m) into y = mx + c

The normal (thin line) is perpendicular (90° to) to the tangent.

To find the equation of the tangent above:

1. Find ‘mT’ using dy/dx

2. Find ‘mN’ using mT x mN = -1

3. Find the y-value of A if not given

4. Substitute all into y = mx + c

Stationary Points

A

B

C

x

y

Point A: Local Maximum

Point B: Local minimum

Point C: Stationary Point of Inflection

Integration

Anti-differentiation is known as integration

The general indefinite formula is shown below,

Integration

FORMULAS FOR INTEGRATIONGENERAL Formulae

Trigonometric Formulae

Exponential and Logarithmic Formulae

Linear bracket Formula

Indefinite integrals Examples

• ∫ x5 + 3x2 dx = x6/6 + x3 + c

• ∫ 2sin (x/3) dx = 2 ∫ sin(x/3) dx = -2x3cos(x/3) + c

• ∫ x-2 dx = -x-1 + c

• ∫ e2x dx = ½ e2x + c

• ∫ 20 dx = 20x + c

Definite integrals

1 3 x

yy = x2 – 2x + 5 Area under curve = A

A = ∫1 (x2-2x+5) dx = [x3/3 – x2 + 5x]1

= (15) – (4 1/3) = 10 2/3 units2

3

3

Area under curves – signed area

Area Between 2 curves

Area Between two curves is found by subtracting the Area of the upper curve by Area of the lower curve.

This can be simplified into

Area = ∫ (upper curve – lower curve) dx

A = ∫-5 25-x2-(x2-25) dx

OR

A = 2 ∫0 25-x2-(x2-25) dx

OR

A = 4 ∫0 25-x2 dx

A = 83 1/3 units2

y = x2 -25

y = 25 - x2

5

5

5

Area Between 2 curves continued…

If 2 curves pass through eachother multiple times than you must split up the integrands.

y1y2

C D

A = ∫C(y1-y2)dx + ∫0(y1-y2)dx D0A1 A2

Let A be total bounded by the curves y1 and y2 area,

thus;

A = A1 +A2

Integration – Area Approximation

The area under a curve can be estimated by dividing the area into rectangles.

Two types of which is the Left endpoint and right endpoint approximations.

The average of the left and right end point methods gives the trapezoidal estimate.

y

y = x2 – 2x + 5

x

y = x2 – 2x + 5

x

LEFT

RIGHT