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 NOT Copy down word for word what is written in this PowerPoint, notes are already in your subject booklets. DO Listen, 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 ] x1x1

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)=(x 3 /x) then, f(0) = 0/0 = (UNDEFINED) therefore discontinuous at “x=0” take lim(x 3 /x) = lim(x 2 ) = 0 lim(x3/x) = 0 because lim(x3/x) = 0 = lim(x3/x) Notice in this example how lim(x 3 /x) exists even though it is discontinuous at x=0. x0x0 f(x) = x 3 /x x0-x0- x0x0 x0x0 x0+x0+ x0x0 x0-x0- x0+x0+

Limit example (x 2 -x-6) (x-3)(x+2) (x-2) (x-2) lim = lim = lim (x-3) = -1 x2x2x2x2x2x2 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) = x ¼ = r 2 x

Differentiation Rules If f(x) = x 6 +4x 2 – 18x + 90 f’(x) = 6x 5 + 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) = sec 2 (x) f(x) = e x  f`(x) = e x f(x) = log e (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=u 15 ….dy/du = 15u 14 and du/dx = 2 Use above formula gives, dy/dx=15x2xu 14 = 30(1+2x) 14

Chain rule in transcendental Take for example y=e sin(x) ….let sin(x) = u gives; du/dx = cos(x) y=e u ….dy/du =e u Use chain rule formula dy/dx= e u.cos(x) = cos(x).e sin(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’ nnv 2 uvuv

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 ‘m T ’ using dy/dx 2.Find ‘m N ’ using m T x m N = -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 INTEGRATION GENERAL Formulae Trigonometric Formulae Exponential and Logarithmic Formulae Linear bracket Formula

Indefinite integrals Examples ∫ x 5 + 3x 2 dx = x 6 /6 + x 3 + c ∫ 2sin (x/3) dx = 2 ∫ sin(x/3) dx = -2 x 3cos(x/3) + c ∫ x -2 dx = -x -1 + c ∫ e 2x dx = ½ e 2x + c ∫ 20 dx = 20x + c

Definite integrals 1 3 x y y = x 2 – 2x + 5 Area under curve = A A = ∫ 1 (x 2 -2x+5) dx = [x 3 /3 – x 2 + 5x] 1 = (15) – (4 1/3) = 10 2/3 units 2 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 = ∫ x 2 -(x 2 -25) dx OR A = 2 ∫ 0 25-x 2 -(x 2 -25) dx OR A = 4 ∫ 0 25-x 2 dx A = 83 1/3 units 2 y = x y = 25 - x

Area Between 2 curves continued… If 2 curves pass through eachother multiple times than you must split up the integrands. y1y1 y2y2 C D A = ∫ C (y 1 -y 2 )dx + ∫ 0 (y 1 -y 2 )dx D 0 A1A1 A2A2 Let A be total bounded by the curves y 1 and y 2 area, thus; A = A 1 +A 2

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 = x 2 – 2x + 5 x x LEFT RIGHT