Chapter 2.4 Differentiability PDF
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This document explores differentiability of functions at a point, covering the meaning of differentiability, right-hand and left-hand derivatives, and conditions for differentiability. It includes several illustrative examples and explanations in calculus concepts.
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2.4.1 Differentiability of a Function at a Point. (1) Meaning of differentiability at a point : Consider the function f (x ) defined on an open interval E3 Y (b, c) let P(a, f (a)) be a point on the curve y f (x ) and let Q (a h, f (a h)) and 60 Functions, Limits, Continuity and Differentiabil...
2.4.1 Differentiability of a Function at a Point. (1) Meaning of differentiability at a point : Consider the function f (x ) defined on an open interval E3 Y (b, c) let P(a, f (a)) be a point on the curve y f (x ) and let Q (a h, f (a h)) and 60 Functions, Limits, Continuity and Differentiability 113 [a + h, f R (a+h)] [a–h, f (a–h)] R(a h, f (a h)) be two neighbouring Q points on the left hand side and right hand side respectively of the point P. ID P [a, f (a)] f (a h) f (a) f (a h) f (a) = (a h) a h and, slope of chord PR f (a h) f (a) f (a h) f (a) =. aha h X O U Then slope of chord PQ Thus, lim h 0 D YG As h 0, point Q and R both tends to P from left hand and right hand respectively. Consequently, chords PQ and PR becomes tangent at point P. f (a h) f (a) lim (slope of chord PQ)= lim (slope of chord PQ) QP h 0 h Slope of the tangent at point P, which is limiting position of the chords drawn on the left f (a h) f (a) hand side of point P and lim = lim (slope of chord PR) = lim (slope of chord PR). h 0 R P h 0 h U Slope of the tangent at point P, which is the limiting position of the chords drawn on the right hand side of point P. Now, f (x ) is differentiable at x a lim h 0 f (a h) f (a) f (a h) f (a) lim h 0 h h ST There is a unique tangent at point P. Thus, f (x ) is differentiable at point P, iff there exists a unique tangent at point P. In other words, f (x ) is differentiable at a point P iff the curve does Y not have P as a corner point. i.e., "the function is not differentiable at those points on which function has jumps (or holes) and sharp edges.” Let us consider the function f (x ) | x 1 | , which can be graphically shown, f (x)=–x+1 f (x) = x – 1 f ' (x)= –1 f ' (x) = 1 Which show f (x ) is not differentiable at x 1. Since, f (x ) has sharp edge at x 1. O 1 2 3 X 114 Functions, Differentiability Limits, Continuity and Mathematically : The right hand derivative at x 1 is 1 and left-hand derivative at x 1 is –1. Thus, f (x ) is not differentiable at x 1. (2) Right hand derivative : Right hand derivative of f (x ) at x a , denoted by f ' (a 0 ) or f ' (a ) , is the lim h 0 f (a h) f (a). h f ' (a ) , is the lim h 0 60 (3) Left hand derivative : Left hand derivative of f (x ) at x a, denoted by f ' (a 0 ) or f (a h) f (a). h E3 (4) A function f (x ) is said to be differentiable (finitely) at x = a if f ' (a 0) f ' (a 0) = finite f (a h) f (a) f (a h) f (a) = finite and the common limit is called the derivative lim h 0 h 0 h h f (x ) f (a) of f (x ) at x a , denoted by f ' (a). Clearly, f ' (a) lim {x a from the left as well as from x a x a the right}. x2 Consider f (x ) | x | , x 0 0 , x 0 [EAMCET 1994] U Example: 1 ID i.e., lim (a) f (x ) is discontinuous everywhere D YG (b) f (x ) is continuous everywhere but not differentiable at x = 0 (c) f ' ( x ) exists in (–1, 1) (d) f ' ( x ) exists in (–2, 2) Solution: (b) x2 x, x 0 x2 x ,x 0 ,x 0 We have, f (x ) | x | = 0 0 , x 0 x2 x, x 0 x lim f (x ) lim x 0 , lim f (x ) lim x 0 and f (0 ) = 0. x 0 x 0 x 0 x 0 U So f (x ) is continuous at x 0. Also f (x ) is continuous for all other values of x. Hence, f (x ) is everywhere continuous. ST Also, Rf ' (0) f ' (0 0) = lim h 0 f (h) f (0 ) h0 1 = lim h 0 h h0 i.e. Rf ' (0 ) 1 and Lf ' (0) f ' (0 0) lim h 0 h f (h) f (0) 1 = lim h 0 h h i.e. Lf ' (0 ) 1 So, Lf ' (0 ) Rf ' (0) i.e., f (x ) is not differentiable at x 0. Example: 2 If the function f is defined by f (x ) (a) Everywhere Solution: (a) x , then at what points f is differentiable 1 | x | (b) Except at x 1 (c) Except at x 0 (d) Except at x 0 or 1 x h 1 x , x 0 0 f (h) f (0 ) x 1 We have, f (x ) = lim h =1 0 , x 0 ; Lf ' (0 ) lim h 0 h h 0 h 1 | x | x , x 0 1 x Functions, Limits, Continuity and Differentiability 115 f (h) f (0) = Rf ' (0) lim h 0 h So, Lf ' (0) Rf ' (0) 1 h 0 1 h 0 1 h = lim =1 h 0 1 h h lim So, f (x ) is differentiable at x 0 ; Also f (x ) is differentiable at all other points. Hence, f (x ) is everywhere differentiable. (a) –2 Solution: (b) (b) 0 60 The value of the derivative of | x 1 | | x 3 | at x 2 is (c) 2 (d) Not defined (x 1) (x 3) , x 1 2 x 4 , x 1 2 , 1x 3 Let f (x ) | x 1 | | x 3 | = (x 1) (x 3) , 1 x 3 = (x 1) (x 3) , x 3 2x 4 , x 3 E3 Example: 3 Since, f (x ) = 2 for 1 x 3. Therefore f ' ( x ) 0 for all x (1, 3). Hence, f ' ( x ) 0 at x 2. (a) Continuous and derivable at x 0 ID Example: 4 sin x 2 The function f defined by f (x ) x , x 0 0 , x 0 (b) Neither continuous nor derivable at x 0 (c) Continuous but not derivable at x 0 x 0 x 0 U We have, lim f (x ) lim sin x 2 sin x 2 lim x 0 x 2 x x 1 0 0 f (0) D YG Solution: (a) (d) None of these So, f (x ) is continuous at x 0 , f (x ) is also derivable at x 0, because lim x 0 f (x ) f (0) sin x 2 lim x 0 x 2 x 0 = 1 Y exists finitely. Example: 5 If f (x ) | log | x || , then (a) f (x ) is continuous and differentiable for all x in its domain X' (–1, 0) O (1, 0) f(x)=|log| x|| X U (b) f (x ) is continuous for all x in its domain but not Y' differentiable at x 1. (c) f (x ) is neither continuous nor differentiable at ST x 1 (d) None of these Solution: (b) It is evident from the graph of f (x ) = | log | x || that f (x ) is everywhere continuous but not differentiable at x 1.. Example: 6 The left hand derivative of f ( x ) [ x ] sin(x ) at x k (k is an integer), is (a) (1)k (k 1) Solution: (a) (b) (1)k 1 (k 1) (c) (1)k k f ( x ) [ x ] sin( x ) If x is just less than k, [x] = k – 1. f (x ) (k 1) sin(x ) , when x k k I Now L.H.D. at x k , (d) (1)k 1 k 116 Functions, Differentiability = = Example: 7 Continuity and lim (k 1) sin( x ) k sin( k ) (k 1) sin( x ) = lim [as sin( k ) 0, k integer] x k x k (x k ) lim (k 1) sin( (k h)) h x k h0 [Let x (k h) ] (k 1)(1)k 1 sin h = h0 h lim lim (k 1)(1)k 1 h0 sin h ( ) = (k 1)(1)k = (1)k (k 1). h The function f (x ) | x | | x 1| is (a) Continuous at x 1 , but not differentiable (b) Both continuous and differentiable at x 1 (c) Not continuous at x 1 (d) None of these E3 Solution: (a) 60 = Limits, x 0 2 x 1, 1 , 0 x 1 We have, f (x ) | x | | x 1 | 2 x 1, x 1 Since, lim f (x ) lim 1 1, lim f (x ) lim (2 x 1) 1 and f (1) 2 1 1 1 x 1 x 1 x 1 ID x 1 lim f (x ) lim f (x ) f (1). So, f (x ) is continuous at x = 1. x 1 x 1 Y and lim x 1 U 1 1 f (x ) f (1) f (1 h) f (1) lim lim 0 , Now, lim h 0 h 0 h h x 1 x 1 2(1 h) 1 1 f (x ) f (1) f (1 h) f (1) 2. lim = lim h 0 h 0 x 1 h h y=2x–1 y=1 y=– 2x+1 D YG (LHD at x 1 ) (RHD at x 1 ). So, f(x) is not differentiable at x = 1. X x =1/2 x =1 2 x 1 , x 0 1 , 0 x 1 is Trick : The graph of f (x ) | x | | x 1 | i.e. f (x ) 2 x 1 , x 1 By graph, it is clear that the function is not differentiable at x 0 , 1 as there it has sharp edges. Let f (x ) | x 1 | | x 1 | , then the function is (a) Continuous (b) Differentiable except x 1 (c) Both (a) and (b) (d) None of these U Example: 8 ST Solution: (c) 2 x , when x 1 Here f (x ) | x 1| | x 1| f (x ) 2 , when 1 x 1 2 x , when x 1 Graphical solution : The graph of the function is shown alongside, From the graph it is clear that the function is continuous at all real x, also differentiable at all real x except at x 1; Since sharp edges at x 1 and x 1. At x 1 we see that the slope from the right i.e., R.H.D. = 2, while slope from the left i.e., L.H.D.= 0 Similarly, at x 1 it is clear that R.H.D. = 0 while L.H.D. = – 2 Trick : In this method, first of all, we differentiate the function and on the derivative equality sign should be removed from Y doubtful points. 2 X’ –1 O 1 X 2 , x 1 Here, f ' (x ) 0 , 1 x 1 (No equality on –1 and +1) 2 , x 1 Functions, Limits, Continuity and Differentiability 117 Now, at x 1, f ' (1 ) 2 while f ' (1 ) 0 and 60 at x 1, f ' (1 ) 0 while f ' (1 ) 2 Thus, f (x ) is not differentiable at x 1. Note : , x 1 ax 2 b If the derivative of the function f (x ) 2 is everywhere continuous and differentiable bx ax 4 , x 1 E3 Example: 9 This method is not applicable when function is discontinuous. at x = 1 then (a) a = 2, b = 3 2 ax b , x 1 f (x ) 2 bx ax 4 , x 1 (d) a = – 3, b = – 2 x 1 x 1 U , 2ax f ' (x ) 2 bx a , (c) a = –2, b = – 3 ID Solution: (a) (b) a = 3, b = 2 To find a, b we must have two equations in a, b Since f (x ) is differentiable, it must be continuous at x 1. D YG R L V at x 1 for f(x) b a 4 a b 2a 4 i.e., a 2 Again f ' (x ) is continuous, it must be continuous at x 1. R L V at x 1 for f ' ( x ) 2b a 2a. Putting a 2, we get 2b 2 4 2b 6 or b 3. Example: 10 Let f be twice differentiable function such that f " (x ) f (x ) and f ' (x ) g(x ), h(x ) { f (x )}2 {g(x )}2. If U h(5 ) 11, then h(10 ) is equal to (a) 22 (c) 0 (d) None of these Differentiating the given relation h (x ) [ f (x )]2 [g(x )]2 w.r.t x, we get h' (x ) 2 f (x ) f ' (x ) 2 g(x )g' (x ).......(i) ST Solution: (b) (b) 11 But we are given f ' ' (x ) f (x ) and f ' (x ) g(x ) so that f ' ' (x ) g ' (x ). Then (1) may be re-written as h' (x ) 2 f ' ' (x ) f ' (x ) 2 f ' (x ) f ' ' (x ) 0. Thus h( x ) 0 Whence by intergrating, we get h (x ) constant = c (say). Hence h(x ) c, for all x. In particular, h(5 ) c. But we are given h(5 ) 11. It follows that c 11 and we have h(x ) 11 for all x. Therefore, h(10) = 11. Example: 11 | 2 x 3 | [ x ] , x 1 The function f (x ) x sin , x 1 2 (a) Is continuous at x 2 (c) Is continuous but not differentiable at x 1 Solution: (c) [2 h] 2, [2 h] 1, [1 h] 1, [1 h] 0 (b) (d) None of these Is differentiable at x 1 118 Functions, Differentiability Limits, Continuity and At x = 2, we will check R L V R lim | 4 2h 3 | [2 h] 2, V 1. 2 2 h 0 L lim | 4 2h 3 | [2 h] 1, R L , not continuous h 0 V | 1| 1 L lim sin h 0 2 (1 h) 1 Since R L V continuous at x 1. | 1 |. 1 1 | 2 2h 3 | [1 h] 1 1 1 lim lim 0 h 0 h 0 h 0 h h h L.H.D. lim h 0 E3 R.H.D. lim 60 At x 1, R lim| 2 2h 3 | [1 h] 1. 1 1, 1.0 1 1 | 2 2h 3 | [1 h] 1 lim lim h 0 h h 0 h h ID Since R.H.D. L.H.D. not differentiable. at x 1. 2.4.2 Differentiability in an Open Interval. U A function f (x ) defined in an open interval (a, b) is said to be differentiable or derivable in open interval (a, b) if it is differentiable at each point of (a, b). Differentiability in a closed interval : A function f : [a, b] R is said to be differentiable in [a, b] if D YG (1) f ' (x ) exists for every x such that a x b i.e. f is differentiable in (a, b). (2) Right hand derivative of f at x a exists. (3) Left hand derivative of f at x b exists. Everywhere differentiable function : If a function is differentiable at each x R , then it is said to be everywhere differentiable. e.g., A constant function, a polynomial function, sin x, cos x etc. are everywhere differentiable. U Some standard results on differentiability (1) Every polynomial function is differentiable at each x R. (2) The exponential function a x , a 0 is differentiable at each x R. ST (3) Every constant function is differentiable at each x R. (4) The logarithmic function is differentiable at each point in its domain. (5) Trigonometric and inverse trigonometric functions are differentiable in their domains. (6) The sum, difference, product and quotient of two differentiable functions is differentiable. (7) The composition of differentiable function is a differentiable function. Important Tips If f is derivable in the open interval (a, b) and also at the end points ‘a’ and ‘b’, then f is said to be derivable in the closed interval [a, b]. A function f is said to be a differentiable function if it is differentiable at every point of its domain. If a function is differentiable at a point, then it is continuous also at that point. Functions, Limits, Continuity and Differentiability 119 i.e. Differentiability Continuity, but the converse need not be true. a. If a function ‘f’ is not differentiable but is continuous at x = a, it geometrically implies a sharp corner or kink at x = If f(x) is differentiable at x = a and g(x) is not differentiable at x =a, then the product function f(x).g(x) can still be differentiable at x = a. 60 If f(x) and g(x) both are not differentiable at x = a then the product function f(x).g(x) can still be differentiable at x = a. If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x) + g(x) is also not differentiable at x = a The set of points where the function f (x ) 1 e x (b) (, 0) (0, ) (a) (, ) Solution: (b) 2 is differentiable (c) (1, ) (d) None of these Clearly, f (x ) is differentiable for all non-zero values of x, For x 0 , we have f ' (x ) ID Example: 12 E3 If f(x) and g(x) both are not differentiable at x = a, then the sum function may be a differentiable function. Now, (L.H.D. at x = 0) x 0 f (x ) f (0) f (0 h) f (0 ) 1 e h lim = lim h 0 x 0 h h0 h 2 lim h0 1 e h h 2 2 U = lim eh 1 = lim h0 h2 xe x 1 e x 1 eh 2 2 1 2 eh 1 1 f (x ) f (0) 1 e h 0 = lim 1. lim 2 h0 h0 x 0 h x 0 h2 eh So, f (x ) is not differentiable at x 0 , Hence, the points of differentiability of f (x ) are (,0 ) (0, ). 2 2 Example: 13 D YG and, (RHD at x = 0) = lim The function f ( x ) e | x| is (a) Continuous everywhere but not differentiable at x 0 (b) Continuous and differentiable everywhere (c) Not continuous at x 0 (d) None of these e x , x 0 We have, f (x ) x e ,x 0 U Solution: (a) Clearly, f (x ) is continuous and differentiable for all non-zero x. Now, lim f (x ) lim e x = 1 and lim f (x ) lim e x 1 x 0 ST x 0 Also, f (0 ) e 0 x 0 Y x 0 f(x)=e–|x| 1 (0, 1) So, f (x ) is continuous for all x. d [e x ] x 0 e 0 1 (LHD at x 0 ) = (e x ) dx x 0 d [e x ] x 0 1 (RHD at x 0 ) = (e x ) dx x 0 X' O y X Y' So, f (x ) is not differentiable at x 0. Hence, f ( x ) e | x| is everywhere continuous but not differentiable at x 0. This fact is also evident from the graph of the function. Example: 14 If f (x ) 1 1 x 2 , then f (x ) is 120 Functions, Differentiability Limits, Continuity and (a) Continuous on [–1, 1] and differentiable on (–1, 1) differentiable on (–1, 0) (0, 1) (c) Continuous and differentiable on [–1, 1] (d) None of these We have, f (x ) 1 1 x 2. The domain of definition of f (x ) is [–1, 1]. 1 For x 0, x 1, x 1 we have f ' (x ) x 1 x2 1 1 x2 60 Solution: (b) (b) Continuous on [–1,1] and Since f (x ) is not defined on the right side of x 1 and on the left side of x 1. Also, f ' ( x ) when x 1 or x 1 . So, we check the differentiability at x = 0. x 0 h 0 1 {1 (1 / 2)h 2 (3 / 8 )h 4 ....} 1 3 2 1 1 1 h2 0 = lim h ..... lim h0 2 h 0 8 h h 2 1 Similarly, (RHD at x = 0) = 2 ID = lim f (x ) f (0) f (0 h) f (0) lim h 0 x 0 h E3 Now, (LHD at x 0 ) = lim Hence, f (x ) is not differentiable at x = 0. Example: 15 Let f (x ) be a function differentiable at x c. Then lim f ( x ) equals x c (b) f "(c) (c) U (a) f ' (c ) 1 f (c) (d) None of these Since f (x ) is differentiable at x c, therefore it is continuous at x c. Hence, lim f (x ) f (c). Example: 16 The function f (x ) (x 1)| x 3 x 2 | cos(| x |) D YG Solution: (d) 2 2 (a) – 1 Solution: (d) (x 2 (b) 0 x c is not differentiable at (c) 1 [IIT Screening 1999] (d) 2 3 x 2) (x 1)(x 2) ive When x 1 or 2, –ive when 1 x 2 Also cos | x | cos x (since cos( x ) cos x ) f (x ) (x 2 1)(x 2 3 x 2) cos x, 1 x 2 f (x ) (x 2 1)(x 2 3 x 2) cos x, x 2.........(i) U Evidently f (x ) is not differentiable at x 2 as L R Note : 1 1 If f (x ) xe | x | x , x 0 , then f (x ) is [AIEEE 2003] 0 , x 0 (a) Continuous as well as differentiable for all x (b) Continuous for all x but not differentiable at ST Example: 17 For all other values like x 0, 0 x 1, f (x ) is same as given by (i). x 0 (c) Neither differentiable nor continuous at x 0 Solution: (b) f (0 ) 0 and f (x ) 1 1 | x| x xe R.H.L. = lim (0 h)e 2 / h lim h0 L.H.L. = lim (0 h)e h0 1 1 h h h0 f (x ) is continuous. 0 h e 2/h 0 (d) Discontinuous every where R f (x ) at (x 0) lim h 0 Functions, Limits, Continuity and Differentiability 121 f (0 h) f (0) he 2 / h = lim e 0 h 0 h h 1 1 h f (0 h) f (0) he h = lim L f (x ) at (x 0) lim h 0 h 0 h h = +1 L f (x ) R f (x ) Solution: (d) 1 , x 0, f (0) 0 at x 0 x (a) Is continuous but not differentiable (b) Is discontinuous (c) Is having continuous derivative (d) Is continuous and differentiable 1 1 lim f (x ) x 2 sin but 1 sin 1 and x 0 x 0 x x The function f (x ) x 2 sin lim f (x ) 0 lim f (x ) f (0) x 0 x 0 Therefore f (x ) is continuous at x 0. Also, the function f (x ) x 2 sin h 2 sin h 0 Which of the following is not true (a) A polynomial function is always continuous (b) A continuous function is always differentiable U Example: 19 1 is differentiable because x ID R f (x ) lim 1 1 h 2 sin 0 h 0. h 0 , L f (x ) lim h 0 h h [MP PET 2003] E3 Example: 18 60 f (x ) is not differentiable at x 0. Example: 20 If f (x ) sgn (x 3 ), then [DCE 2001] (a) f is continuous but not derivable at x 0 (b) f ' (0 ) 2 (c) f ' (0 ) 1 (d) f is not derivable at x 0 x 3 | x 3 | 0 x Here, f (x ) sgn x 3 | x | 0 1, 0 , 1, ST U Solution: (d) D YG Solution: (b) (c) A differentiable function is always continuous (d) e x is continuous for all x A continuous function may or may not be differentiable. So (b) is not true. for x3 0 for x3 0 for x 0 for x 0. Thus, f (x ) sgn x 3 sgn x , which is neither continuous nor x 0 x 0 x 0 derivable at 0. Note : f (0 ) lim h 0 1 0 10 f (0 h) f (0) f (0 h) f (0) lim . lim and f (0 ) lim h h h h h 0 h 0 h 0 f (0 ) f (0 ) , f is not derivable at x 0. Example: 21 1 x , x 2 A function f (x ) is 5 x , x 2 (a) Not continuous at x 2 (c) Continuous but not differentiable at x 2 [AMU 2001] (b) (d) None of the above Differentiable at x 2 122 Functions, Differentiability Solution: (c) Limits, Continuity and lim 1 (2 h) 3 , lim 5 (2 h) 3 , f (2) 3 h 0 h 0 Hence, f is continuous at x 2 Now R f (x ) lim h 0 h 0 1 (2 h) 3 1 h 60 L f (x ) lim 5 (2 h) 3 1 h R f (x ) L f (x ) Example: 22 E3 f is not differentiable at x 2. Let f : R R be a function. Define g : R R by g(x ) | f (x )| for all x. Then g is (a) Onto if f is onto (b) One-one if f is one-one (c) Continuous if f is continuous (d) Solution: (c) Differentiable ID differentiable if f is g(x ) | f (x )| 0. So g(x ) cannot be onto. If f (x ) is one-one and f (x 1 ) f (x 2 ) then Y ensure that g(x ) is one-one. U g(x 1 ) g(x 2 ). So, ‘ f (x ) is one-one’ does not If f (x ) is continuous for x R,| f (x )| is also x R. This is obvious from following D YG continuous for the P y = f (x) O X Y y = | f (x)| y = | f (x)| P O X y = f (x) graphical consideration. So the answer (c) is correct. The fourth answer (d) is not correct from the above graphs y f (x ) is ST U differentiable at P while y | f (x )| has two tangents at P, i.e. not differentiable at P.
