Calculus can be divided into two ke areas: INTRODUCTION TO CALCULUS Differential Calculus dealing wit its, rates of cange, tangents and normals to curves, curve sketcing, and applications to maima and minima problems. Integral Calculus, dealing wit areas and volumes, and approimate areas under and between curves. LIMITS Consider te function f( ). Te it of te function is defined as te value of (ordinate) tat is approaced b te curve, as te value of (or abscissa) approaces a particular number, sa of value ' a '. For eample, if we approac an value of 3 from te left or te rigt, on te curve f ( ) we see tat in bot cases, te iting value is 9. i.e. 9. 3 f ( ) 1 8 f (3) (3) 9 6 4-4 -3 - -1 1 3 4 As a furter eample, te diagram below is a grap of a piecemeal function. Note tat one end point is open, and te oter closed, above 3. f ( ) f( ) + 1 18 16 14 1 1 8 6 4-4 -3 - -1 1 3 4 5 6 7 8 Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 17
Regardless, as we approac 3 from te left, te iting value is 9, but as we approac 3 from te rigt, te iting value is 13. Hence we sa tat te it ere does not eist. i.e. f ( ) does not eist. Limits (and ence tangents) can onl be found on curves tat 3 are continuous at te point in question. Note: Do not confuse tis wit te functional value f (3), wic in tis case is clearl defined as 13. EVALUATING LIMITS In eac of te cases below, ou sould initiall make te substitution of te given value, in an attempt to evaluate te it. Depending on wat results from tis ten determines te necessar strateg. Note tat in all cases, te use of a calculator to substitute a number ver close to te given value will give ou a ver good idea of wat to epect as a final answer ver and in an eam if ou forget te formal metod! CASE I: You substituted into te epression, and te result was well defined. In oter words, te resulting epression could be evaluated do so! QUESTION 18 Find (4 1). (4 1) 4() 1 7 QUESTION 19 Find 1 1. 5 ( 1) 5 5 ( 1) 1 6 Te last epression was a "Quotient". Quotients of epressions often create difficult. Imagine ou substituted into suc an epression, and te result was one of te following... Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 18
CASE II: Zero Constant ie.. c Te it is. QUESTION 5 Find 5 + 4. 5 5 (5) 5 + 4 (5) + 4 9 CASE III: Constant Zero c ie.. Te it is undefined since it approaces ±. QUESTION 1 Find. + ( ) + ( ) + 4 ± Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 19
CASE IV: Infinit Constant ie.. c Te it is undefined since it approaces ±. QUESTION 9 + Find 7. 9 + 7 9 + ( ) 7 7 CASE V: Constant Infinit ie.. c Te it is. QUESTION 3 8 Find. 8 8 ( ) Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page
CASE VI: Zero Zero ie.. You sould simplif te algebraic fraction b first factoring, and ten cancelling, before re-substituting te given value. QUESTION 4 3 4 Find. 1 1 + 1 1 3 4 1 + 1 CASE VII: Infinit Infinit ie.. Divide ever term in te epression b te igest power of present in te denominator, simplif eac term, ten use te rule for Case V. QUESTION 5 5 Find 8+ + 3 5 8+ + 3. ( is te igest power of in te denominator). Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 1
WATCH OUT! If ever in doubt about te correct tecnique, or in doubt of our answer, simpl substitute a value for tat is close to te given value, into te given formula, on our calculator. For eample, for sa, use.99 or 3.1, and for 3, use some large value like 1,,. THE DERIVATIVE OF A FUNCTION In graping, ou ave used tables of values, to plot points (along wit oter algebraic tools suc as finding an aes intercepts etc) and to draw sketces of functions. Te original function f( ) is used to find te values or ordinates of points, on te grap, so tat we can see its sape and position. Te "derivative", or derived function is written as d or '( ) d f. As we are assuming ere tat te independent variable in te function is, ten te derivative is found "wit respect to ". Te derivative is related to, and found from, te original function f( ). Te derivative is used to find a different feature of te grap it is used to find te gradient. Te process of finding te derivative is called "differentiation". Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page
QUESTION 6 Sketc te derived function d d for te given function,. + 1 d d Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 3
QUESTION 7 Sketc te derived function d for te given function,. d d d Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 4
QUESTION 8 Sketc te derived function d d for te given function,. 3 d d DIFFERENTIABILITY A function is said to be "differentiable" at a particular point if it satisfies two conditions te grap of te function at te point in question must be continuous (te curve is not broken at te point), and te grap at te point must also be smoot (te curve must ave te same slope on te left and approac to te point as it does on te rigt and approac). Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 5
ESTIMATING THE GRADIENT OF THE TANGENT If we ad an accurate grap of a function, we could draw a tangent at te point in question, judging tis purel b ee. B coosing two points on te tangent and reading off te coordinates of te two points, we 1 would use te gradient formula m to calculate te gradient of te tangent at tat 1 particular point. If we ad a rule for te function, witout an accurate grap, we could use te following metod. Sa we wanted te gradient of f ( ) at te point were 3. We could use a second point on te curve, ver close to te first and ten appl te gradient formula. Let's use te two points at 3 and 3.1. Note tat tis cannot give te eact gradient at 3 - just someting close. f ( ) f (3.1) (3.1) 9.61 f (3) (3) 9 3 3.1 It is te gradient of tis secant tat we are using. Tis is wat we reall want te gradient of te tangent. Tis is because te slope of te tangent and te slope of te curve are te same. m 1 1 9.61 9 3.1 3.61 6.1.1 Tis value represents te gradient of te secant joining te two points, and not of te curve or tangent. But we migt guess tat te gradient is around 6, and in fact we will sortl find out it is eactl 6. Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 6
DIFFERENTIATION USING FIRST PRINCIPLES f( ) f ( + ) f ( ) + Te gradient of te secant is given b: m 1 1 f ( + ) f( ) ( + ) Hence te gradient of te tangent (and ence te curve) is: d d f '( ) f ( + ) f( ) METHOD: Step 1: Write te epressions for f () and f ( + ). Step : Substitute te epressions for f () and f ( + ) into te it teorem. Step 3: Epand and collect like terms. d [ f ( + ) f ( )] d Step 4: Remove as a common factor and simplif. Step 5: Substitute. Note: To find te gradient at a specific point, we substitute te value of into te derivative. Note: Te gradient at is denoted as f '(). Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 7
QUESTION 9 Find te derivative, using first principles, of f '( ) f + f ( ) ( ) ( + ) + + + + ( + ) f ( ). Tis answer is called te derivative or derived function or gradient function of f ( ). Note tat at 3, f '(3) (3) 6. Tis is te same answer as tat guessed in te earlier eample. QUESTION 3 Find, using first principles, te gradient function of f( ) 5+. WATCH OUT! Wen te function as two terms or more, make sure tat te function is written inside of brackets beind te minus sign. f ( ) 5 + f ( + ) ( + ) 5( + ) + + + + 5 5 f '( ) f ( + ) f( ) ( + + 5 5+ ) ( 5+ ) + + 5 5 + + 5 (+ 5) + 5 (+ 5) + () 5 5 Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 8
DIFFERENTIATION RULES Te derivative of an algebraic term is obtained b multipling te term b te power and ten lowering te power on tat term b one. Tis rule applies for all algebraic epressions, including rational functions, providing tat n. If For eample: If n a ten d d 6 ten an n 1 d 6 1 5 6 d 1 Note also tat if a ten d a d. Te derivative of a constant (a term tat does not contain an variables) is equal to zero. d If c (were c is a constant) ten d Te derivative of a sum (or of a difference) is te sum (or differences) of te individual derivatives. d If u ( ) ± v ( ) ten u'( ) v'( ) d ± QUESTION 31 Find te derivative of 5 +. WATCH OUT! "d" for differentiate, "d" for decrease te power. (Please note tat tis is onl true for a power wic is a constant onl) d d d d + d d d d ( ) (5 ) () 5+ 5 Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 9
Step 1: Rewrite all terms as powers of. FINDING DERIVATIVES 1 1 1 q 3 3 p ( q ) p q ( ) p Step : Bring terms involving in te denominator (bottom of a fraction) to te top, b canging te sign on te power. For eample: 1 Note: 1 6 6 Step 3: Simplif epressions so tat terms are separated b addition and subtraction and ten differentiate eac term individuall. Alternativel, reduce epressions down to 1 term onl (use log and inde laws). Products: Epand simple products rater tan appling te Product Rule. Quotients: Remove common factor(s) and simplif. Factorise and einate terms b cancellation. If tere is onl one term in te denominator, write eac term in te numerator over te denominator so tat individual fractions are formed. Ten simplif eac term b cancellation. For eample: + 1 1 + 1 + (Don t use te Quotient Rule). Step 4: Differentiate. Step 5: Re-write te answer using positive powers. Bring terms wit negative powers in te numerator (top of a fraction) to te bottom, b canging te sign on eac power. Step 6: State restrictions on te values of. Te Scool For Ecellence 16 Trial Eam Revision Lectures Matematics Book 1 Page 3