Differential Calculus (The basics) Prepared by Mr. C. Hull

Transcription

1 Differential Calculus Te basics)

2 A : Limits In tis work on limits, we will deal only wit functions i.e. tose relationsips in wic an input variable ) defines a unique output variable y). Wen we work wit limits, we are concerned wit wat te function values are ONLY as we get closer and closer to a particular input value i.e. a particular -value. We are not necessarily concerned wit wat te function value is AT te particular -value. Tis is an important point always to bear in mind.

3 A : Limits 0; 0) f ) 1; 0) f ) =!1) = 2! Consider wat appens as we let get closer and closer to 1 i.e. let get arbitrarily close to 1 from te left and te rigt it doesn t matter). ½; ¼) Intuitively, we see tat te function values get arbitrarily close to 0. 0 ½ 0,9 0,99 0, ,001 1,01 1,1 2 f) 0 ¼ 0,09 0,0099 0, , ,11 2 We write tis as follows: lim!1 2 " ) = 0 eists, because its value can be calculated. and we say tat te limit Note tat lim 2 ".!1 ) = f 1 ) = 0

4 A : Limits f ) f ) =!1) = 2! if "1 1 2 if =1 1; ½) As before, lim 2 ".!1 ) = 0 However, we note tat ½; ¼) 1; 0) ) = 1 2! 0 f 1. In tis case, te limit, altoug it eists, is not equal to te function value at te -value in question i.e. lim!1 2 " ) f 1 ). Remember, we are not focused so muc on te function value itself at = 2 for now).

5 A : Limits f ) f ) =!1) = 2! ½; ¼) 1; 0) Consider wat appens as as gets larger and larger positive and negative). Te function values temselves get arbitrarily large positive. Tere is no value towards wic te function values approac i.e. te limit does not eist. We write tis as lim 2. Note tat we don t speak of.!±" ) = " f! ) =!

6 A : Limits f ) ½; ¼) f ) =!1) if " 1 2!1) if > 1 2 Consider wat appens as as gets closer and closer to ½ from te rigt. Te function values temselves get arbitrarily close to 0. However, as gets arbitrarily close to ½ from te left, te function values get closer and closer to ¼. In tis case, we write lim and. We say tat + 2 " " 2 "! 1 2 ) = 0 lim! 1 2 ) = " 1 4 lim! 1 2 f ) does not eist even toug f! " 1 = 1 2 4

7 A : Limits Epsilon-delta definition of a limit: )! a A number L is a limit for a sequence of function values f as if, for some arbitrarily cosen i.e. tink very small ) interval ε) from L i.e L - ε and L + ε ), tere is an interval δ) on eiter side of = a suc ) <! 0 <! a <! tat L! f wenever.

8 A : Limits f ) L+ε L L - ε - δ a + δ

9 B : Limits Calculate te following limit: 1. lim!" f ) = f ) = lim!"2!"2 ) + 2) 2 " = lim 2 " 2 + 4) =12 We find te limit by straigtforward substitution because g ) = 2! is continuous for all. Note tat f for all. )! g )

10 B : Limits Calculate te following limit: 2. lim!5 5" " 5 f ) = lim!5 ) " 5 " " 5 = lim!5 = "1 "1) f ) = 5!! 5 Note: It is tempting to tink tat g ) =!1 0;!1) f ) = 5!! 5 is te same grap as but tere is a fundamental difference in teir domains. ) = 5! We write f =!1, for all " R, 5.! 5

11 B : Limits a) b) f!2) =1 lim!"2 f ) = 4 f ) = " + 6 <!2 1 =!2 2! 2 < <1 1 >1 f ) c) Removable discontinuity at = 2 d) f 1) =1 e) lim!1 f ) =1 f) Removable discontinuity at = 1.

12 B : Limits Calculate te following limit: 3. lim!1 "1 "1 f ) = lim!1 "1 ) +1) "1 ) ) = f "!1!1 = lim! ) ) Note: f is defined only for > 0 and 1. = 1 2 We can substitute = 1 because is continuous for all > 0. g ) = 1 +1

13 B : Limits Calculate te following limit: 4. lim 2 + " 2 f ) = lim 2 + " 2 ) f ) = 2 +! 2 = lim 2 + " ) = lim = ) ) Note: f is defined only for > 2 and 0. We can substitute = 0 because is continuous for all > 2. g ) = )

14 B : Continuity A function is said to be continuous at = a under te following conditions: 1. f a is defined ) lim!a lim!a f ) eists i.e. lim f ) = lim f )!a +!a " f ) = f a) Wen a function is not continuous at = a, it is typically because tere is a removable discontinuity or a jump discontinuity.

15 C : Average rate of cange between two points Consider te grap below. We say tat it as an average rate of cange between A and B calculated as follows: f ) Average rate of cange B + ; f + ) ) = = BC AC )! f ) + )! f + A ; f ) ) C = )! f ) f +

16 C : Average rate of cange between two points If f ) = 2! 2, calculate te average rate of cange between = 1 and = 3. Average gradient f ) = f!1!1 )! f!3) )!!3) A!3; 15) m AB =!6 = 3 )! 15) =! 6 2 B!1; 3)

17 C : Rate of cange at a point f ) B + ; f + ) ) ; f ) ) y = f ) We say tat te rate of cange of y = f ) at A is understood to be te limit of a sequence of gradients. A ; f ) ) As! 0, so will te successive average gradients get closer and closer to te gradient of te tangent line to y = f at A. ) Te instantaneous rate of cange at A is calculated as follows: ) = lim f ) " f ) f +

18 C : Rate of cange at a point ) = lim f ; f ) ) ) " f ) f + Important considerations and vocabulary: ) f ) ) f ) 1. f is called te derivative of. Te resulting epression for f is derived from te original function. ) 2. f itself is generally a function of. Tis is easily understood if you consider tat, in general, tangents will ave different gradients at different points. DON T FORGET WHAT f IS! 3. Finding an epression for f using tis metod is called FIRST PRINCIPLES. Tere is a sorter metod called te POWER RULE and bot are etremely important to your work in differential calculus. ) 4. f does not necessarily always eist!. mmmm. ) )

19 C : Rate of cange at a point ) = lim f ; f ) ) ) " f ) f + Furter important considerations and vocabulary: 5. f is differentiable at = a if: i) ii) ) f ) is continuous at = a. makes sense, no?) lim f ) = lim f )!a "!a +. tis says tat te slope of te tangent line must be te same weter you approac = a from te left or te rigt. ) = f ) = 6. If f., ten. f dased of dy If y =., ten. dy by d d = We also ave D!" f ) =... te derivative wit respect to of f)

20 C : Rate of cange at a point: ) = 3 ) ) = lim f ) " f ) f + Given f. Find f by First Principles. ) = lim f ) " f ) f + y = m + c f ) ) = lim f 3 + " 3 Te tangent line as a gradient of! 4 at tis point. 3 ) In general, m tangent = f.! A 3 " 2 ; 2 f ) = 3 ) = lim f 3 " 3 " 3 + ) f ) =!3 2 ) = lim f f ) = lim "3 + ) "3 + ) = "3 2 We use te derivative to find te rate of cange of f at A: )! 3 f = 3 2 = 3 " 2! 3 9 " 2 4 = 12 9 = 4 3

21 C : Rate of cange at a point: ) = lim f ) " f ) f + Given y = 2 + 3, use te definition of te derivative to calculate te gradient of te tangent at = 3. dy d = lim 2 3+ ) + 3 " 2 3) + 3 Te tangent line f ) dy d = lim " 3 ) as a gradient of 1 3 at tis point y = dy d = lim " ) dy d = lim ) "! 3 2 ; 0 dy d = 1 3

22 C : Rate of cange at a point: ) = lim f ) " f ) f + Finding derivatives by using te Power Rule We ave seen te following two cases: ) = 3 = 3!1 f ) =!3 1. f. 2 =!3!2 ) = 2 f ) = 2 = f. ) = a n f ) = a! n n"1 In general, if f, ten. Tis is given witout proof at tis stage. Wait for your tertiary studies if you can?

23 C : Rate of cange at a point: Furter considerations: ) = lim f ) " f ) f + ) + g ) ) + g ) 7. D!" f = f. Te derivative of a sum of terms is equal to te sum of te individual derivatives using te power rule). 8. Te derivative instantaneous rate of cange) of a constant function is zero. a orizontal line as a gradient equal to zero). 9. Te derivative instantaneous rate of cange) of a linear function is a constant. a straigt line as a constant gradient) 10. Te derivative instantaneous rate of cange) of a quadratic function is a linear function a parabola canges at a rate equal to te slope of a tangent line).

24 C : Rate of cange at a point: Find te derivative in eac case: ) = lim f ) " f ) f + e.g. D!" 2 = 2 e.g. 1 ) = f ) = f ) e.g. y = 3 3! 2 y = 3 3! 2!1 dy d = 9 2! 2 "!1"!1!1 dy d = !2 ) f ) = ! 1 + 3" 1 2 " 2 ) 1 2! 1 ) = 1 4! f f ) = 1 f ) = )! " 2 + OR ) =

25 C : Rate of cange at a point: ) = lim f ) " f ) f + e.g. Find te equation of te tangent line to te grap of f at = 3. i.e. 3; 20) ) = ! 7 Te tangent line y = m + c ) will generally ave a non-zero gradient. Tis gradient will be te value of te derivative at te point were = 3. Remember, te derivative allows us to calculate te value of te instantaneous) rate of cange at ANY point on f. ) = ! 7 ) = ) = 4 3) + 3 ) =15 f f 3 f 3 Find te derivative by using te Power Rule. Calculate te rate of cange at =3. Tis will be te slope of te tangent line at tat point y =15 + c 20 =15 3 ) + c!25 = c " y =15! 25 Need c. Substitute te point 3; 20)

26 C : Rate of cange at a point: ) = lim f ) " f ) f + f ) = ! 7 f ) y =15! 25 3 ; 20) Te parabola as a rate of cange of 15 at tis point. 0 ; -25) Were does te parabola ave rate of cange = 0? Wat are te coordinates of tis point? Wat is te equation of te tangent line at tis point?

27 C : Rate of cange at a point: ) = lim f ) " f ) f + Reminder: A function is differentiable at = a if: i) ii) f ) is continuous at = a lim f ) = lim f )!a "!a +. tis says tat te slope of te tangent line must be te same weter you approac = a from te left or te rigt. Don t forget too tat if a grap is differentiable at = a, it means tat te rate of cange at tat point can be calculated!

28 C : Rate of cange at a point: ) = lim f ) " f ) f +! a " a f ) a +! f ) = ! 3 Successive slopes of te blue tangent lines will coincide wit te slope of te red tangent line at = a as approaces a from te rigt. Successive slopes of te purple tangent lines will coincide wit te slope of te red tangent line at = a as approaces a from te left. f ) wen wen = a is differentiable at = a lim f ) = lim f )!a "!a + f ) is continuous at

29 C : Rate of cange at a point: ) = lim f ) " f ) f + f ) =!1 f ) f ) =! f ) =1 A 4; 2) ) f is not differentiable at A In tis eample, we can see tat lim f ) = lim f )!4 "!4 +. In fact, it sould be clear tat, since tere are infinitely many tangent lines to f at = 4, te derivative does not eist at A, even toug f is continuous at A. ) )

30 C : Rate of cange at a point: f ) ) = lim f ) " f ) f + ) = f "!1!1 F 1; ½) ) f is not differentiable at F because it is not continuous at F Even toug lim f ) = lim f )!1 "!1 +, te derivative does not eist at F because tere is a removable discontinuity at F. f1) is not even defined. It stands to reason tat tere cannot be a rate of cange at F.

31 C : Rate of cange at a point: ) = lim f ) " f ) f +