MATHEMATICS XII. Topic. Revision of Derivatives Presented By. Avtar Singh Lecturer Paramjit Singh Sidhu June 19,2009
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1 MATHEMATICS XII 1 Topic Revision of Derivatives Presented By Avtar Singh Lecturer Paramjit Singh Sidhu June 19, June 2009 Punjab EDUSAT Society (PES) 1
2 Continuity 2 Def. In simple words, a function is continuous at a fixed point if we can draw graph of the function at around that point without lifting the pen from the plane of the paper. Another Def. : A function is continuous at x=a if the function is defined at x=a and if the value of the function at x=a is equal the limit of the function at x=a. 19 June 2009 Punjab EDUSAT Society (PES) 2
3 Discontinuity 3 Note 1 : If f is discontinuous at x=a then a is called point of discontinuity. Note 2 : A function f is discontinuous at x=a in following cases : (i) f is not defined at x=a i.e. f(a) does not exist. (ii) Limit of f(x) at x=a does not exit. (iii) Limit of f(x) at x=a exits but not equal to f(a). 19 June 2009 Punjab EDUSAT Society (PES) 3
4 Cases of Discontinuity Limit of f(x) at x=a does not exit. This happens in following cases : 4 Case I Case II Case III 19 June 2009 Punjab EDUSAT Society (PES) 4
5 Kinds of Discontinuity 1. Removable Discontinuity : Some times a function f is not defined at x=a or f(a) is defined in such a way that it is not equal to limit of f(x) at x=a, then this discontinuity can be removed by defining f(a) in such a way that it may equal to limit of f(x) at x=a. 2. Non Removable Discontinuity : This is of two kinds. (i) Discontinuity of first kind (ii) Discontinuity of second kind 19 June 2009 Punjab EDUSAT Society (PES) 5 5
6 Non Removable Discontinuity 6 (i)discontinuity of first kind : (ii) Discontinuity of second kind : or 19 June 2009 Punjab EDUSAT Society (PES) 6
7 Continuity in interval 7 (i) Open interval : A real valued function f defined on open interval (a,b)is said to be continuous in on (a,b) if it is continuous at x=c for all c (a,b) (ii) Closed interval : A real valued function f defined on closed interval [a,b]is said to be continuous in on [a,b] if (i) f is right continuous at x=a. (ii) f is left continuous at x=b. (iii) f is continuous at x=c for all c (a,b) Such have continuous graph on [a,b] 19 June 2009 Punjab EDUSAT Society (PES) 7
8 Algebra of Continuity 8 If f, g are two continuous functions at x=a then (i) kf is continuous function at x=a,k R. (ii) (f ±g) is continuous function at x=a. (iii) fg is continuous function at x=a. (iv) f/g is continuous function at x=a, g(a) 19 June 2009 Punjab EDUSAT Society (PES) 8
9 Prove that A constant function is continuous everywhere. 9 Let f(x) =c be a constant function x R. Let a be any real number. Now Also f(a) =c. f is continuous everywhere. 19 June 2009 Punjab EDUSAT Society (PES) 9
10 Is the function f(x)=x 2 -sin x+5 continuous at x=π 10 We have f(x)=x 2 -sin x+5. = π 2 -sin π+5 = π = π 2 +5 Also f(π)=π 2 -sin π+5 = π = π 2 +5 f is continuous at π 19 June 2009 Punjab EDUSAT Society (PES) 10
11 Board Questions 11 Q. 1 If f(x) is continuous at x=0, 2008) find the value of k. (Mar 19 June 2009 Punjab EDUSAT Society (PES) 11
12 Board Questions 12 Q. 1 (Mar 2008) If f(x) is continuous at x=0, find the value of k. Sol. 19 June 2009 Punjab EDUSAT Society (PES) 12
13 Board Questions 13 Sol. Now f(0) = k. (given) Because function is continuous at x=0 So Limiting value of f(x) is same as f(0). Ans. So k=2 19 June 2009 Punjab EDUSAT Society (PES) 13
14 Board Questions 14 Q. 1 Discuss continuity of f(x) at x=0 (Mar 2007) 19 June 2009 Punjab EDUSAT Society (PES) 14
15 Board Questions 15 Q. 1 Discuss continuity of f(x) at x=0 (Mar 2007) Sol. 19 June 2009 Punjab EDUSAT Society (PES) 15
16 Board Questions 16 Sol. 19 June 2009 Punjab EDUSAT Society (PES) 16
17 Board Questions 17 Sol. 19 June 2009 Punjab EDUSAT Society (PES) 17
18 Board Questions 18 Sol. 19 June 2009 Punjab EDUSAT Society (PES) 18
19 Board Questions 19 Discuss the continuity of f(x) at x=0, where 2003) (Mar Solution :- Here function contain modulus of x hence we have find its limit from both sides because I x I=x if x 0 and I x I =-x if x<0. 19 June 2009 Punjab EDUSAT Society (PES) 19
20 Continued Sol June 2009 Punjab EDUSAT Society (PES) 20
21 Continued Right limit June 2009 Punjab EDUSAT Society (PES) 21
22 Continued So L.H.L. R.H.L. Limit at x=0 does not exist. Hence from def. of continuity f(x) is discontinuous at x=0. Here we need not to find f(0) June 2009 Punjab EDUSAT Society (PES) 22
23 Question Q.No. 1 Find K so that function is continuous at point x= π/2 f(x)= K cos x if x π/2 Π-2x 3 if x= π/2 Q. No.2 Examine the continuity at x=0 23 f(x)= Sin x cos x if x 0-1 if x=0 19 June 2009 Punjab EDUSAT Society (PES) 23
24 Question Q.No Find the all points of discontinuity of function f(x) = 2x+3 if x 0 { 2x-3 if x > 0 Q. No.2Find the value of a and b if function is continuous. f(x)={ 5 If x 2 ax + b if 2< x <10 21 If x June 2009 Punjab EDUSAT Society (PES) 24
25 Derivative Differentiation Introduction Increment Differential Co-efficient Notation 19 June 2009 Punjab EDUSAT Society (PES) 25
26 Definition Derivative Let f(x) be function defined then If it exit is called differential coefficient of y w.r.t x and is denoted as f (x). 19 June 2009 Punjab EDUSAT Society (PES) 26.
27 Derivative Steps for derivative 1. Put y = given function. Change x to x and y to y Subtract (1) from (2) and obtain y and simplify. Divide both sides by x. Take limits both sides as x 0 keeping in mind 19 June 2009 Punjab EDUSAT Society (PES) 27.
28 Derivative Physical Meaning Imagine x is time and f(x) is distance travelled in time x. Let f(x+ x) be distance in x+ x time. Then distance f(x+ x) - f(x) is travelled in time x. Speed =Distance travelled / time. Speed in interval=(f(x+ x) - f(x) ) / x Speed at a point = 19 June 2009 Punjab EDUSAT Society (PES) 28.
29 Derivative Geometrical significance Let y=f(x) be function whose graph in xy plane is shown by curve PQ Q Y=f(x) P o 19 June 2009 Punjab EDUSAT Society (PES) 29
30 Derivative Geometrical significance Let y=f(x) be function whose graph in xy plane is shown by curve Let P(c,f(c)) and Q(c+h, f(c+h)) be ponits on curve Y=f(x) P Q o 19 June 2009 Punjab EDUSAT Society (PES) 30
31 Derivative Geometrical significance Let y=f(x) be function whose graph in xy plane is shown by curve Let P(c,f(c)) and Q(c+h, f(c+h)) be ponits on curve Y=f(x) P Q PT is tangent to curve at P T o 19 June 2009 Punjab EDUSAT Society (PES) 31
32 Derivative Geometrical significance Let y=f(x) be function whose graph in xy plane is shown by curve Let P(c,f(c)) and Q(c+h, f(c+h)) be ponits on curve PT is tangent to curve at P PQ chord meet OX at S T o Y=f(x) S P Q 19 June 2009 Punjab EDUSAT Society (PES) 32
33 Derivative Geometrical significance PQ chord meet OX at S Draw PM OX, QN OX, PR QN. XSQ= RPQ=θ XTP=α PR=MN=ON-OM =c+h-c = h RQ=QN-RN=QN-MP = f(c+h)-f(c) 19 June 2009 Punjab EDUSAT Society (PES) 33 o Y=f(x) θ T S M N P θ Q R
34 Geometrical significance PQ chord meet OX at S Draw PM OX, QN OX, PR QN. XSQ= RPQ=θ XTP=α PR=MN=ON-OM =c+h-c = h RQ=QN-RN=QN-MP = f(c+h)-f(c) 19 June 2009 Punjab EDUSAT Society (PES) 34 o Y=f(x) θ T S M N P θ Q R
35 Geometrical significance Y Q Now if Q approaches P along the curve the line PQ become the tangent to the curve at P in limiting case θ α as h 0 X S O α θ P M T R N X Y 19 June 2009 Punjab EDUSAT Society (PES) 35
36 Derivative Left hand derivative at a point. A function f is said to be derivable to the left of a point c D f iff exists finitely and is denoted by Lf (c ) and is called left hand derivative of f w.r.t. x at x=c. 19 June 2009 Punjab EDUSAT Society (PES) 36
37 Derivative Right hand derivative at a point A function f is said to be derivable to the right of a point c D f iff exists finitely and is denoted by Rf (c ) and is called right hand derivative of f w.r.t. x at x=c. 19 June 2009 Punjab EDUSAT Society (PES) 37
38 Derivative Derivative at a point A function f is said to be derivable at a point c D f iff exists finitely and is denoted by f (c ) and is called derivative of f w.r.t. x at x=c 19 June 2009 Punjab EDUSAT Society (PES) 38
39 Derivative Derivative and Continuity A function which is derivable at a point is a continuous at that point. But its converse may or may not be true. 19 June 2009 Punjab EDUSAT Society (PES) 39
40 Derivative Some Typical examples 1. Show that the function defined by f(x) = 3-2x if x<2 3x-7 if x 2 is continuous at x=2 but not derivable at x=2. 19 June 2009 Punjab EDUSAT Society (PES) 40
41 Derivative Another example Find left and right derivatives of f(x) = 2x+1 if x<1 6x+7 if x 1 at x=1. Is f is derivable at x=1? 19 June 2009 Punjab EDUSAT Society (PES) 41
42 Derivative Different names of first principle 1. By First Principle 2. From Definition 3. By Delta Method 4. By Ab-inito 19 June 2009 Punjab EDUSAT Society (PES) 42
43 Derivative Derivative by First Principle 1. Put y= given function. 2. Change x to x and y to y 3.Subtract (1) from (2) and obtain y and simplify. 4 Divide both sides by x. 5 Take limits both sides as x 0 keeping in mind 1 Y=f(x) 2 Y+ y =f(x+ x ) 3 Y+ y y= f(x+ x )-f(x) June 2009 Punjab EDUSAT Society (PES) 43
44 Derivative Some Standard Results. 19 June 2009 Punjab EDUSAT Society (PES) 44
45 Derivative Some Standard Results. 19 June 2009 Punjab EDUSAT Society (PES) 45
46 Derivative of Trigonometric Functions Derivative 1 D (sin x ) = cos x 2 D ( cos x) = - sin x 3 D ( tan x )= sec 2 x 4 D ( cot x ) = - cosec 2 x 5 D ( sec x ) = sec x. tan x 6 D ( cosec X ) = - cosec x. cot x 19 June 2009 Punjab EDUSAT Society (PES) 46
47 Derivative of Trigonometric Functions Derivative 1 D (sin ax+b ) = cos (ax+b).a 2 D ( cos ax+b) = - sin (ax+b).a 3 D ( tan ax+b )= sec 2 (ax+b).a 4 D ( cot ax+b ) = - cosec 2 (ax+b).a 5 D ( sec ax+b) = sec (ax+b). tan (ax+b).a 6 D( cosec ax+b )= - cosec (ax+b).cot (ax+b).a 19 June 2009 Punjab EDUSAT Society (PES) 47
48 Derivative Derivative of Composite Functions Let f(x) =(gοh)(x) =g(h(x)) Then f (x) =g (h(x)).h (x) 19 June 2009 Punjab EDUSAT Society (PES) 48
49 Chain Rule Derivative Let y= f(t) And t = g(x) be two functions. We want to find derivative of y w.r.t. x dy /dt =f (t) dt / dx = g (x) then dy /dx = dy /dt * dt /dx = f (t) * g (x) = f (g(x))* g (x) 19 June 2009 Punjab EDUSAT Society (PES) 49
50 Derivative Generalised Chain Rule Let y = f(t), t = g(u), u =h(x), And we want derivative of y w.r.t. x.then. 19 June 2009 Punjab EDUSAT Society (PES) 50
51 Q No. 1. Questions on chain rule Q No June 2009 Punjab EDUSAT Society (PES) 51
52 Q No. 1 Questions on Derivative Q No June 2009 Punjab EDUSAT Society (PES) 52
53 Important Results of Derivatives D( Constant) = 0 D( u±v) = D(u) ± D(v) D(u v) = u D(v) + v D(u) 19 June 2009 Punjab EDUSAT Society (PES) 53
54 Derivative of parametric equations Let x = f(t) and y = g(t) be two functions of t Then 19 June 2009 Punjab EDUSAT Society (PES) 54
55 Q No. 1. Questions on Parametric function Q. No. 2 =1 19 June 2009 Punjab EDUSAT Society (PES) 55
56 Derivative of one function w.r.t. other function Put one function of x is equal to y and put other function of x is equal to u Then 19 June 2009 Punjab EDUSAT Society (PES) 56
57 Example of Derivative of a function w.r.t. function Differentiate 7x 5-11x 2 w.r.t. 7x 2 15x Let u= 7x 5-11x 2, v = 7x 2 15x 19 June 2009 Punjab EDUSAT Society (PES) 57
58 Derivative w.r.t. another function Q No. 1. Q No. 2 x Differentiate log(xe x ) w.r.t. x log 19 June 2009 Punjab EDUSAT Society (PES) 58
59 Derivative of Implicit Functions Explicit functions -- When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f(x) Implicit functions - When a relationship between x and y is expressed in a way that it is not easy to solve for y and y is not expressed in terms of x. 19 June 2009 Punjab EDUSAT Society (PES) 59
60 Examples of Implicit functions x 6 +y x 2 y 2 =16 e x +e y =e x+y 19 June 2009 Punjab EDUSAT Society (PES) 60
61 Q No. 1 Logarithmic Derivative Q No June 2009 Punjab EDUSAT Society (PES) 61
62 Q No. 1 Logarithmic Derivative Q No June 2009 Punjab EDUSAT Society (PES) 62
63 Derivative of Inverse T Functions 19 June 2009 Punjab EDUSAT Society (PES) 63
64 Derivative of Inverse T Functions. 19 June 2009 Punjab EDUSAT Society (PES) 64
65 Derivative of Inverse T Functions. 19 June 2009 Punjab EDUSAT Society (PES) 65
66 Derivative of Inverse T Functions. 19 June 2009 Punjab EDUSAT Society (PES) 66
67 Some Important substitutions In case of ( 1 ) a 2 +x 2, put x = a tanθ ( 2 ) a 2 -x 2, put x = a sinθ ( 3 ) x 2 a 2,put x = a secθ 19 June 2009 Punjab EDUSAT Society (PES) 67
68 1 Properties of Inverse T- Functions Sin(sin -1 x)= x, x [-1,1] and sin -1 (sinx) = x, x [-π/2, π/2] Same result is true for other five trigonometric ratios 19 June 2009 Punjab EDUSAT Society (PES) 68
69 2 Properties of Inverse T- Functions cosec -1 x = sin -1 (1/x) x 1 or x -1 cos -1 x = sec -1 (1/x) x 1 or x -1 cot -1 x = tan -1 (1/x) x>0 19 June 2009 Punjab EDUSAT Society (PES) 69
70 3 Properties of Inverse T- Functions sin -1 (-x) = - sin -1 x, x [-1,1] cos -1 (-x) = π-cos -1 x x [-1,1] tan -1 (-x) = - tan -1 x x R cot -1 (-x) = - cot -1 x x R sec -1 (-x) = π - sec -1 x IxI 1 cosec -1 (-x) = -cesec -1 x IxI 1 19 June 2009 Punjab EDUSAT Society (PES) 70
71 4 Properties of Inverse T- Functions sin -1 x + cos -1 x = x [-1,1] tan -1 x + cot -1 x = x R sec -1 x + cosec -1 x = IxI 1 19 June 2009 Punjab EDUSAT Society (PES) 71
72 5 Properties of Inverse T- Functions 19 June 2009 Punjab EDUSAT Society (PES) 72
73 6 Properties of Inverse T- Functions 19 June 2009 Punjab EDUSAT Society (PES) 73
74 7 Properties of Inverse T- Functions 19 June 2009 Punjab EDUSAT Society (PES) 74
75 Q No. 1 Derivative of Inverse T- Function Q No June 2009 Punjab EDUSAT Society (PES) 75
76 Q No. 1 Derivative of Inverse T- Function Q No June 2009 Punjab EDUSAT Society (PES) 76
77 Multiple Angle Formulae 1 Sin 2A = 2 Sin A Cos A 2 Cos 2A = Cos 2 A Sin 2 A = 2 Cos 2 A 1 = 1 2 Sin 2 A 3 Tan 2A = 2 Tan A 1 Tan 2 A 19 June 2009 Punjab EDUSAT Society (PES) 77
78 Multiple Angle Formulae Sin 3A = 3 Sin A 4 Sin 3 A Cos 3A = 4 Cos 3 A 3 Cos A Tan 3A = 3 Tan A Tan 3 A 1-3 Tan 2 A 19 June 2009 Punjab EDUSAT Society (PES) 78
79 Multiple Angle Tan Form Sin 2A = 2 Tan A 1 + Tan 2 A Cos 2A = 1 Tan 2 A 1 + Tan 2 A Tan 2A = 2 Tan A 1 Tan 2 A 19 June 2009 Punjab EDUSAT Society (PES) 79
80 Another Form Sin A = ± Cos A =± Tan A = ± 19 June 2009 Punjab EDUSAT Society (PES) 80
81 Tan (π/4 ±A) Tan (π/4 + A) = 1 + Tan A 1 Tan A Tan (π/4 A ) = 1 Tan A 1 + Tan A 19 June 2009 Punjab EDUSAT Society (PES) 81
82 Q No. 1 Derivative of T-functions Q No June 2009 Punjab EDUSAT Society (PES) 82
83 Questions Q No. 1 Q No. 2 Q No June 2009 Punjab EDUSAT Society (PES) 83
84 Questions Q No. 1 Q No. 2 Q No June 2009 Punjab EDUSAT Society (PES) 84
85 Questions Q No. 1 Q No. 2 Q No June 2009 Punjab EDUSAT Society (PES) 85
86 Derivatives of Higher order Q No. 1 Q No June 2009 Punjab EDUSAT Society (PES) 86
87 Derivatives of Higher order Q No. 1 Q No June 2009 Punjab EDUSAT Society (PES) 87
88 Rolle s Theorem Statement: If a function f(x) defined on [a,b] is such that (i) f(x) is continuous in closed interval [a,b] (ii) f(x) is derivable in open interval(a,b) (iii) f(a)= f(b) then there exists at least one real number c (a,b) such that f ( c)= 0 19 June 2009 Punjab EDUSAT Society (PES) 88
89 Rolle s Theorem Geometrical Interpretation Let AB be the graph of function y=f(x) such that the point A and B of the graph correspond to the numbers a and b of the interval [a,b] Y P Y A B A B f(a) f ( c) f(b) f(a) P f ( c) f(b) O M N X O M N X 19 June 2009 Punjab EDUSAT Society (PES) 89
90 Rolle s Theorem Geometrical Interpretation f(x) is continuous in the interval [a,b] its graph is a continuous curve between A Y and B. Y A P B A P B f(a) f ( c) f(b) f(a) f ( c) f(b) O M N X O M N X 19 June 2009 Punjab EDUSAT Society (PES) 90
91 Rolle s Theorem Geometrical Interpretation 19 June 2009 Punjab EDUSAT Society (PES) 91
92 Varify Rolle s theorem in following Q No. 1 cases f(x)=x 3 +3x 2-24x-80 in interval [-4,5] Q No. 2 Q No. 3 f(x)= sin x-sin 2x in 0 x 2π Discuss the applicability of Rolle s theorem to the function 19 June 2009 Punjab EDUSAT Society (PES) 92
93 Lagrange s mean value Theorem Statement: If a function f(x) defined on [a,b] is such that (i) f(x) is continuous in closed interval [a,b] (ii) f(x) is derivable in open interval(a,b) then there exists at least one real number c (a,b) such that 19 June 2009 Punjab EDUSAT Society (PES) 93
94 L.M.V. Theorem Geometrical Interpretation Let AB be the graph of function y=f(x) such that the point A and B of the graph correspond to the numbers a and b of the interval [a,b] Y P Y A A f(a) f ( c) B f(b) f(a) P f ( c) B f(b) O M N X O M N X 19 June 2009 Punjab EDUSAT Society (PES) 94
95 L.M.V. Theorem Geometrical Interpretation f(x) is continuous in the interval [a,b] its graph is a continuous curve between A Y and B. Y A P B A P B f(a) f ( c) f(b) f(a) f ( c) f(b) O M N X O M N X 19 June 2009 Punjab EDUSAT Society (PES) 95
96 L.M.V. Theorem Geometrical Interpretation 19 June 2009 Punjab EDUSAT Society (PES) 96
97 Q No. 1 Verify L.M.V. in following cases f(x)= (x-3)(x-5)(x-9) in interval [3,5] Q No. 2 Q No. 3 Find point on the parabola y=(x-3) 2 where the tangent is parallel to the chord joining (3,0) and (4,1). 19 June 2009 Punjab EDUSAT Society (PES) 97
98 The End 19 June 2009 Punjab EDUSAT Society (PES) 98
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