CONTINUITY AND DIFFERENTIABILITY

Transcription

1 5. Introduction The whole of science is nothing more than a refinement of everyday thinking. ALBERT EINSTEIN This chapter is essentially a continuation of our stu of differentiation of functions in Class XI. We had learnt to differentiate certain functions like polynomial functions and trigonometric functions. In this chapter, we introduce the very important concepts of continuity, differentiability and relations between them. We will also learn differentiation of inverse trigonometric functions. Further, we introduce a new class of functions called eponential and logarithmic functions. These functions lead to powerful techniques of differentiation. We illustrate certain geometrically obvious conditions through differential calculus. In the process, we will learn some fundamental theorems in this area. 5. Continuity CONTINUITY AND DIFFERENTIABILITY We start the section with two informal eamples to get a feel of continuity. Consider the function, if 0 f ( ), if > 0 This function is of course defined at every point of the real line. Graph of this function is given in the Fig 5.. One can deduce from the graph that the value of the function at nearby points on -ais remain close to each other ecept at 0. At the points near and to the left of 0, i.e., at points like 0., 0.0, 0.00, the value of the function is. At the points near and to the right of 0, i.e., at points like 0., 0.0, Chapter 5 Sir Issac Newton (64-77) Fig 5.

2 48 MATHEMATICS 0.00, the value of the function is. Using the language of left and right hand limits, we may say that the left (respectively right) hand limit of f at 0 is (respectively ). In particular the left and right hand limits do not coincide. We also observe that the value of the function at 0 concides with the left hand limit. Note that when we try to draw the graph, we cannot draw it in one stroke, i.e., without lifting pen from the plane of the paper, we can not draw the graph of this function. In fact, we need to lift the pen when we come to 0 from left. This is one instance of function being not continuous at 0. Now, consider the function defined as, if 0 f( ), if 0 This function is also defined at every point. Left and the right hand limits at 0 are both equal to. But the value of the function at 0 equals which does not coincide with the common value of the left and right hand limits. Again, we note that we cannot draw the graph of the function without lifting the pen. This is yet another instance of a function being not continuous at 0. Naively, we may say that a function is continuous at a fied point if we can draw the graph of the function around that point without Fig 5. lifting the pen from the plane of the paper. Mathematically, it may be phrased precisely as follows: Definition Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if lim f ( ) f ( c ) c More elaborately, if the left hand limit, right hand limit and the value of the function at c eist and equal to each other, then f is said to be continuous at c. Recall that if the right hand and left hand limits at c coincide, then we say that the common value is the limit of the function at c. Hence we may also rephrase the definition of continuity as follows: a function is continuous at c if the function is defined at c and if the value of the function at c equals the limit of the function at c. If f is not continuous at c, we say f is discontinuous at c and c is called a point of discontinuity of f.

3 CONTINUITY AND DIFFERENTIABILITY 49 Eample Check the continuity of the function f given by f () + 3 at. Solution First note that the function is defined at the given point and its value is 5. Then find the limit of the function at. Clearly lim f( ) lim (+ 3) () Thus lim f( ) 5 f() Hence, f is continuous at. Eample Eamine whether the function f given by f () is continuous at 0. Solution First note that the function is defined at the given point 0 and its value is 0. Then find the limit of the function at 0. Clearly 0 0 lim f( ) lim 0 0 Thus lim f( ) 0 f(0) 0 Hence, f is continuous at 0. Eample 3 Discuss the continuity of the function f given by f() at 0. Solution By definition,if < 0 f (), if 0 Clearly the function is defined at 0 and f (0) 0. Left hand limit of f at 0 is lim f( ) lim ( ) Similarly, the right hand limit of f at 0 is lim f( ) lim Thus, the left hand limit, right hand limit and the value of the function coincide at 0. Hence, f is continuous at 0. Eample 4 Show that the function f given by f () 3 + 3, if 0, if 0 is not continuous at 0.

4 50 MATHEMATICS Solution The function is defined at 0 and its value at 0 is. When 0, the function is given by a polynomial. Hence, 3 3 lim f ( ) lim ( + 3) Since the limit of f at 0 does not coincide with f (0), the function is not continuous at 0. It may be noted that 0 is the only point of discontinuity for this function. Eample 5 Check the points where the constant function f () k is continuous. Solution The function is defined at all real numbers and by definition, its value at any real number equals k. Let c be any real number. Then lim f ( ) lim k k c c Since f (c) k lim c f () for any real number c, the function f is continuous at every real number. Eample 6 Prove that the identity function on real numbers given by f () is continuous at every real number. Solution The function is clearly defined at every point and f (c) c for every real number c. Also, lim f ( ) lim c c c Thus, lim f () c f (c) and hence the function is continuous at every real number. c Having defined continuity of a function at a given point, now we make a natural etension of this definition to discuss continuity of a function. Definition A real function f is said to be continuous if it is continuous at every point in the domain of f. This definition requires a bit of elaboration. Suppose f is a function defined on a closed interval [a, b], then for f to be continuous, it needs to be continuous at every point in [a, b] including the end points a and b. Continuity of f at a means lim f ( ) f (a) + a and continuity of f at b means a lim f ( ) f(b) b Observe that lim f ( ) and lim f ( ) do not make sense. As a consequence + b of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of f is a singleton, f is a continuous function.

5 CONTINUITY AND DIFFERENTIABILITY 5 Eample 7 Is the function defined by f (), a continuous function? Solution We may rewrite f as f (),if < 0, if 0 By Eample 3, we know that f is continuous at 0. Let c be a real number such that c < 0. Then f (c) c. Also lim f ( ) lim ( ) c (Why?) c c Since lim f ( ) f ( c ), f is continuous at all negative real numbers. c Now, let c be a real number such that c > 0. Then f (c) c. Also lim f ( ) lim c (Why?) c c Since lim f ( ) f ( c ), f is continuous at all positive real numbers. Hence, f c is continuous at all points. Eample 8 Discuss the continuity of the function f given by f () 3 +. Solution Clearly f is defined at every real number c and its value at c is c 3 + c. We also know that 3 3 lim f ( ) lim ( + ) c + c c c Thus lim f ( ) f ( c ), and hence f is continuous at every real number. This means c f is a continuous function. Eample 9 Discuss the continuity of the function f defined by f (), 0. Solution Fi any non zero real number c, we have lim f( ) lim c c c Also, since for c 0, f() c, we have lim ( ) ( ) c f f c and hence, f is continuous c at every point in the domain of f. Thus f is a continuous function.

6 5 MATHEMATICS We take this opportunity to eplain the concept of infinity. This we do by analysing the function f () near 0. To carry out this analysis we follow the usual trick of finding the value of the function at real numbers close to 0. Essentially we are trying to find the right hand limit of f at 0. We tabulate this in the following (Table 5.). Table n f () n We observe that as gets closer to 0 from the right, the value of f () shoots up higher. This may be rephrased as: the value of f () may be made larger than any given number by choosing a positive real number very close to 0. In symbols, we write lim f( ) (to be read as: the right hand limit of f () at 0 is plus infinity). We wish to emphasise that + is NOT a real number and hence the right hand limit of f at 0 does not eist (as a real number). Similarly, the left hand limit of f at 0 may be found. The following table is self eplanatory. Table n f () n From the Table 5., we deduce that the value of f () may be made smaller than any given number by choosing a negative real number very close to 0. In symbols, we write lim f( ) 0 (to be read as: the left hand limit of f () at 0 is minus infinity). Again, we wish to emphasise that is NOT a real number and hence the left hand limit of f at 0 does not eist (as a real number). The graph of the reciprocal function given in Fig 5.3 is a geometric representation of the above mentioned facts. Fig 5.3

7 CONTINUITY AND DIFFERENTIABILITY 53 Eample 0 Discuss the continuity of the function f defined by f () +, if, if > Solution The function f is defined at all points of the real line. Case If c <, then f (c) c +. Therefore, lim f( ) lim f( + ) c+ Thus, f is continuous at all real numbers less than. Case If c >, then f (c) c. Therefore, lim f( ) lim ( ) c f (c) c c Thus, f is continuous at all points >. Case 3 If c, then the left hand limit of f at is lim f ( ) lim ( + ) + 3 The right hand limit of f at is lim f( ) lim ( ) + + c c Since the left and right hand limits of f at Fig 5.4 do not coincide, f is not continuous at. Hence is the only point of discontinuity of f. The graph of the function is given in Fig 5.4. Eample Find all the points of discontinuity of the function f defined by +, if < f () 0, if, if > Solution As in the previous eample we find that f is continuous at all real numbers. The left hand limit of f at is lim f ( ) lim ( + ) + 3 The right hand limit of f at is lim f( ) lim ( ) + + Since, the left and right hand limits of f at do not coincide, f is not continuous at. Hence is the only point of discontinuity of f. The graph of the function is given in the Fig 5.5. Fig 5.5

8 54 MATHEMATICS Eample Discuss the continuity of the function defined by f () +, if < 0 +, if > 0 Solution Observe that the function is defined at all real numbers ecept at 0. Domain of definition of this function is D D where D { R : < 0} and D { R : > 0} Case If c D, then lim f( ) lim ( + ) c c c + f (c) and hence f is continuous in D. Case If c D, then lim f( ) lim ( + ) c c c + f (c) and hence f is continuous in D. Since f is continuous at all points in the domain of f, we deduce that f is continuous. Graph of this function is given in the Fig 5.6. Note that to graph Fig 5.6 this function we need to lift the pen from the plane of the paper, but we need to do that only for those points where the function is not defined. Eample 3 Discuss the continuity of the function f given by, if 0 f (),if< 0 Solution Clearly the function is defined at every real number. Graph of the function is given in Fig 5.7. By inspection, it seems prudent to partition the domain of definition of f into three disjoint subsets of the real line. Let D { R : < 0}, D {0} and D 3 { R : > 0} Fig 5.7 Case At any point in D, we have f () and it is easy to see that it is continuous there (see Eample ). Case At any point in D 3, we have f () and it is easy to see that it is continuous there (see Eample 6).

9 CONTINUITY AND DIFFERENTIABILITY 55 Case 3 Now we analyse the function at 0. The value of the function at 0 is f (0) 0. The left hand limit of f at 0 is The right hand limit of f at 0 is Thus lim f( ) lim 0 0 lim f( ) lim lim f( ) 0 f (0) and hence f is continuous at 0. This means that f is continuous at every point in its domain and hence, f is a continuous function. Eample 4 Show that every polynomial function is continuous. Solution Recall that a function p is a polynomial function if it is defined by p() a 0 + a a n n for some natural number n, a n 0 and a i R. Clearly this function is defined for every real number. For a fied real number c, we have lim p ( ) p ( c ) c By definition, p is continuous at c. Since c is any real number, p is continuous at every real number and hence p is a continuous function. Eample 5 Find all the points of discontinuity of the greatest integer function defined by f () [], where [] denotes the greatest integer less than or equal to. Solution First observe that f is defined for all real numbers. Graph of the function is given in Fig 5.8. From the graph it looks like that f is discontinuous at every integral point. Below we eplore, if this is true. Fig 5.8

10 56 MATHEMATICS Case Let c be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to c the value of the function is equal to [c]; i.e., lim f( ) lim [ ] [ c]. Also f (c) [c] and hence the function is continuous at all real c c numbers not equal to integers. Case Let c be an integer. Then we can find a sufficiently small real number r > 0 such that [c r] c whereas [c + r] c. This, in terms of limits mean that lim f () c, lim f () c c c + Since these limits cannot be equal to each other for any c, the function is discontinuous at every integral point. 5.. Algebra of continuous functions In the previous class, after having understood the concept of limits, we learnt some algebra of limits. Analogously, now we will stu some algebra of continuous functions. Since continuity of a function at a point is entirely dictated by the limit of the function at that point, it is reasonable to epect results analogous to the case of limits. Theorem Suppose f and g be two real functions continuous at a real number c. Then () f + g is continuous at c. () f g is continuous at c. (3) f. g is continuous at c. f (4) g is continuous at c, (provided g (c) 0). Proof We are investigating continuity of (f + g) at c. Clearly it is defined at c. We have lim( f + g)( ) lim[ f( ) + g( )] (by definition of f + g) c c lim f( ) + lim g( ) (by the theorem on limits) c c f (c) + g(c) (as f and g are continuous) (f + g) (c) (by definition of f + g) Hence, f + g is continuous at c. Proofs for the remaining parts are similar and left as an eercise to the reader.

11 CONTINUITY AND DIFFERENTIABILITY 57 Remarks (i) As a special case of (3) above, if f is a constant function, i.e., f () λ for some real number λ, then the function (λ. g) defined by (λ. g) () λ. g() is also continuous. In particular if λ, the continuity of f implies continuity of f. (ii) As a special case of (4) above, if f is the constant function f () λ, then the λ λ λ function defined by ( ) is also continuous wherever g() 0. In g g g( ) particular, the continuity of g implies continuity of g. The above theorem can be eploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. The following eamples illustrate this: Eample 6 Prove that every rational function is continuous. Solution Recall that every rational function f is given by p ( ) f( ), q( ) 0 q ( ) where p and q are polynomial functions. The domain of f is all real numbers ecept points at which q is zero. Since polynomial functions are continuous (Eample 4), f is continuous by (4) of Theorem. Eample 7 Discuss the continuity of sine function. Solution To see this we use the following facts lim sin 0 0 We have not proved it, but is intuitively clear from the graph of sin near 0. Now, observe that f () sin is defined for every real number. Let c be a real number. Put c + h. If c we know that h 0. Therefore lim f ( ) c c lim sin lim sin( c+ h) h 0 lim [sin ccos h+ coscsin h] h 0 lim [sin ccos h] + lim [coscsin h] h 0 h 0 sin c + 0 sin c f (c) Thus lim f () f (c) and hence f is a continuous function. c

12 58 MATHEMATICS Remark A similar proof may be given for the continuity of cosine function. Eample 8 Prove that the function defined by f () tan is a continuous function. Solution The function f () tan sin. This is defined for all real numbers such cos that cos 0, i.e., (n +) π. We have just proved that both sine and cosine functions are continuous. Thus tan being a quotient of two continuous functions is continuous wherever it is defined. An interesting fact is the behaviour of continuous functions with respect to composition of functions. Recall that if f and g are two real functions, then (f o g) () f (g ()) is defined whenever the range of g is a subset of domain of f. The following theorem (stated without proof) captures the continuity of composite functions. Theorem Suppose f and g are real valued functions such that (f o g) is defined at c. If g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c. The following eamples illustrate this theorem. Eample 9 Show that the function defined by f () sin ( ) is a continuous function. Solution Observe that the function is defined for every real number. The function f may be thought of as a composition g o h of the two functions g and h, where g () sin and h (). Since both g and h are continuous functions, by Theorem, it can be deduced that f is a continuous function. Eample 0 Show that the function f defined by f () +, where is any real number, is a continuous function. Solution Define g by g () + and h by h () for all real. Then (h o g) () h (g ()) h ( + ) + f () In Eample 7, we have seen that h is a continuous function. Hence g being a sum of a polynomial function and the modulus function is continuous. But then f being a composite of two continuous functions is continuous.

13 CONTINUITY AND DIFFERENTIABILITY 59 EXERCISE 5.. Prove that the function f () 5 3 is continuous at 0, at 3 and at 5.. Eamine the continuity of the function f () at Eamine the following functions for continuity. (a) f () 5 (b) f () 5 5 (c) f () (d) f () Prove that the function f () n is continuous at n, where n is a positive integer. 5. Is the function f defined by,if f( ) 5, if > continuous at 0? At? At? Find all points of discontinuity of f, where f is defined by , if f( ) 3, if > , if 3 f( ), if 3 < <3 6+, if 3 8.,if 0 f( ) 0, if 0 9.,if< 0 f( ), if 0 +, if 0. f( ) +, if <. f( ) 3 3, if +, if >. 0, if f( ), if > 3. Is the function defined by a continuous function? + 5, if f ( ) 5, if >

14 60 MATHEMATICS Discuss the continuity of the function f, where f is defined by 4. 3, if 0 f( ) 4, if < < 3 5, if , if < 0 f ( ) 0, if 0 4, if > 6., if f( ), if <, if > 7. Find the relationship between a and b so that the function f defined by a +, if 3 f ( ) b + 3, if > 3 is continuous at For what value of λ is the function defined by λ ( ), if 0 f( ) 4+, if > 0 continuous at 0? What about continuity at? 9. Show that the function defined by g () [] is discontinuous at all integral points. Here [] denotes the greatest integer less than or equal to. 0. Is the function defined by f () sin + 5 continuous at π?. Discuss the continuity of the following functions: (a) f () sin + cos (b) f () sin cos (c) f () sin. cos. Discuss the continuity of the cosine, cosecant, secant and cotangent functions. 3. Find all points of discontinuity of f, where sin, if < 0 f( ) +, if 0 4. Determine if f defined by sin, if 0 f( ) 0, if 0 is a continuous function?

15 5. Eamine the continuity of f, where f is defined by CONTINUITY AND DIFFERENTIABILITY 6 sin cos, if 0 f( ), if 0 Find the values of k so that the function f is continuous at the indicated point in Eercises 6 to kcos π, if π f( ) π 3, if at π 7. k, if f( ) 3, if > at 8. k +, if π f ( ) cos, if >π at π k +, if 5 9. f( ) at 5 3 5, if > Find the values of a and b such that the function defined by 5, if f( ) a+ b, if < < 0, if 0 is a continuous function. 3. Show that the function defined by f () cos ( ) is a continuous function. 3. Show that the function defined by f () cos is a continuous function. 33. Eamine that sin is a continuous function. 34. Find all the points of discontinuity of f defined by f () Differentiability Recall the following facts from previous class. We had defined the derivative of a real function as follows: Suppose f is a real function and c is a point in its domain. The derivative of f at c is defined by f ( c+ h) f( c) lim h 0 h

16 6 MATHEMATICS d provided this limit eists. Derivative of f at c is denoted by f (c) or ( f( )) c. The function defined by f ( + h) f( ) f ( ) lim h 0 h wherever the limit eists is defined to be the derivative of f. The derivative of f is d denoted by f () or ( f ( )) or if y f () by or y. The process of finding derivative of a function is called differentiation. We also use the phrase differentiate f () with respect to to mean find f (). The following rules were established as a part of algebra of derivatives: () (u ± v) u ± v () (uv) u v + uv (Leibnitz or product rule) (3) u u v uv, wherever v 0 (Quotient rule). v v The following table gives a list of derivatives of certain standard functions: Table 5.3 f () n sin cos tan f () n n cos sin sec Whenever we defined derivative, we had put a caution provided the limit eists. Now the natural question is; what if it doesn t? The question is quite pertinent and so is f ( c+ h) f( c) its answer. If lim does not eist, we say that f is not differentiable at c. h 0 h In other words, we say that a function f is differentiable at a point c in its domain if both f ( c+ h) f( c) f ( c+ h) f( c) lim and lim are finite and equal. A function is said + h 0 h h 0 h to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. As in case of continuity, at the end points a and b, we take the right hand limit and left hand limit, which are nothing but left hand derivative and right hand derivative of the function at a and b respectively. Similarly, a function is said to be differentiable in an interval (a, b) if it is differentiable at every point of (a, b).

17 CONTINUITY AND DIFFERENTIABILITY 63 Theorem 3 If a function f is differentiable at a point c, then it is also continuous at that point. Proof Since f is differentiable at c, we have But for c, we have f( ) f( c) lim f ( c) c c f () f (c) Therefore lim [ f( ) f( c)] c f( ) f( c).( c ) c f( ) f( c) lim. ( c) c c or lim [ f( )] lim [ f( c)] f( ) f ( c) lim. lim [( c)] c c c c c f (c). 0 0 or lim f ( ) f (c) c Hence f is continuous at c. Corollary Every differentiable function is continuous. We remark that the converse of the above statement is not true. Indeed we have seen that the function defined by f () is a continuous function. Consider the left hand limit f(0 + h) f(0) h lim h 0 h h The right hand limit f(0 + h) f(0) h lim + h 0 h h f (0 + h) f(0) Since the above left and right hand limits at 0 are not equal, lim h 0 h does not eist and hence f is not differentiable at 0. Thus f is not a differentiable function Derivatives of composite functions To stu derivative of composite functions, we start with an illustrative eample. Say, we want to find the derivative of f, where f () ( + ) 3

18 64 MATHEMATICS One way is to epand ( + ) 3 using binomial theorem and find the derivative as a polynomial function as illustrated below. d f ( ) d 3 ( ) + d 3 ( ) ( + ) Now, observe that f () (h o g) () where g() + and h() 3. Put t g() +. Then f() h(t) t 3. Thus df 6 ( + ) 3( + ). 3t. dh dt dt The advantage with such observation is that it simplifies the calculation in finding the derivative of, say, ( + ) 00. We may formalise this observation in the following theorem called the chain rule. Theorem 4 (Chain Rule) Let f be a real valued function which is a composite of two functions u and v; i.e., f v o u. Suppose t u() and if both dt and dv dt eist, we have df dv dt dt We skip the proof of this theorem. Chain rule may be etended as follows. Suppose f is a real valued function which is a composite of three functions u, v and w; i.e., f (w o u) o v. If t v () and s u (t), then df d( wo u) dt dw ds dt dt ds dt provided all the derivatives in the statement eist. Reader is invited to formulate chain rule for composite of more functions. Eample Find the derivative of the function given by f () sin ( ). Solution Observe that the given function is a composite of two functions. Indeed, if t u() and v(t) sin t, then f () (v o u) () v(u()) v( ) sin

19 CONTINUITY AND DIFFERENTIABILITY 65 Put t u(). Observe that dv cost dt and dt eist. Hence, by chain rule df dv dt cost dt It is normal practice to epress the final result only in terms of. Thus df cost cos Alternatively, We can also directly proceed as follows: y sin ( ) d (sin ) Eample Find the derivative of tan ( + 3). cos d ( ) cos Solution Let f () tan ( + 3), u () + 3 and v(t) tan t. Then (v o u) () v(u()) v( + 3) tan ( + 3) f () Thus f is a composite of two functions. Put t u() + 3. Then dt eist. Hence, by chain rule df dv dt sec ( + 3) dt dv sec t dt and Eample 3 Differentiate sin (cos ( )) with respect to. Solution The function f () sin (cos ( )) is a composition f () (w o v o u) () of the three functions u, v and w, where u(), v(t) cos t and w(s) sin s. Put dw ds dt t u() and s v(t) cos t. Observe that cos s, sin t and ds dt eist for all real. Hence by a generalisation of chain rule, we have df dw ds dt (cos s). ( sin t). () sin ds dt. cos (cos )

20 66 MATHEMATICS Alternatively, we can proceed as follows: y sin (cos ) d Therefore sin (cos ) cos (cos ) d (cos ) cos (cos ) ( sin ) d ( ) sin cos (cos ) () sin cos (cos ) EXERCISE 5. Differentiate the functions with respect to in Eercises to 8.. sin ( + 5). cos (sin ) 3. sin (a + b) 4. sec (tan ( )) 5. sin ( a + b) cos ( c + d) 7. ( cot ) 8. cos( ) 9. Prove that the function f given by f (), R is not differentiable at. 0. Prove that the greatest integer function defined by f () [], 0 < < 3 is not differentiable at and. 6. cos 3. sin ( 5 ) 5.3. Derivatives of implicit functions Until now we have been differentiating various functions given in the form y f (). But it is not necessary that functions are always epressed in this form. For eample, consider one of the following relationships between and y: y π 0 + sin y y 0 In the first case, we can solve for y and rewrite the relationship as y π. In the second case, it does not seem that there is an easy way to solve for y. Nevertheless, there is no doubt about the dependence of y on in either of the cases. When a relationship between and y is epressed in a way that it is easy to solve for y and write y f (), we say that y is given as an eplicit function of. In the latter case it

21 CONTINUITY AND DIFFERENTIABILITY 67 is implicit that y is a function of and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to differentiate implicit functions. Eample 4 Find if y π. Solution One way is to solve for y and rewrite the above as y π But then Alternatively, directly differentiating the relationship w.r.t.,, we have d ( y ) dπ Recall that d π means to differentiate the constant function taking value π everywhere w.r.t.,. Thus which implies that d d ( ) ( y) 0 Eample 5 Find, if y + sin y cos. Solution We differentiate the relationship directly with respect to, i.e., d d + (sin y) (cos ) which implies using chain rule This gives where + cos y sin sin + cosy y (n + ) π

22 68 MATHEMATICS Derivatives of inverse trigonometric functions We remark that inverse trigonometric functions are continuous functions, but we will not prove this. Now we use chain rule to find derivatives of these functions. Eample 6 Find the derivative of f given by f () sin assuming it eists. Solution Let y sin. Then, sin y. Differentiating both sides w.r.t., we get cos y which implies that cos y cos(sin ) π π Observe that this is defined only for cos y 0, i.e., sin,, i.e.,,, i.e., (, ). To make this result a bit more attractive, we carry out the following manipulation. Recall that for (, ), sin (sin ) and hence cos y (sin y) (sin (sin )) π π Also, since y,, cos y is positive and hence cos y Thus, for (, ), cos y Eample 7 Find the derivative of f given by f () tan assuming it eists. Solution Let y tan. Then, tan y. Differentiating both sides w.r.t., we get which implies that sec y sec y + tan y + (tan (tan )) +

23 CONTINUITY AND DIFFERENTIABILITY 69 Finding of the derivatives of other inverse trigonometric functions is left as eercise. The following table gives the derivatives of the remaining inverse trigonometric functions (Table 5.4): Table 5.4 f () cos cot sec cosec f () + Domain of f (, ) R (, ) (, ) (, ) (, ) EXERCISE 5.3 Find in the following:. + 3y sin. + 3y sin y 3. a + by cos y 4. y + y tan + y 5. + y + y y + y + y sin y + cos y π 8. sin + cos y 9. y sin y tan, 3 < < y cos, 0 < < + y sin, 0 < < + 3. y cos, < < + y, < < 4. sin ( ) 5., y sec 0 < <

24 70 MATHEMATICS 5.4 Eponential and Logarithmic Functions Till now we have learnt some aspects of different classes of functions like polynomial functions, rational functions and trigonometric functions. In this section, we shall learn about a new class of (related) functions called eponential functions and logarithmic functions. It needs to be emphasized that many statements made in this section are motivational and precise proofs of these are well beyond the scope of this tet. The Fig 5.9 gives a sketch of y f (), y f (), y f 3 () 3 and y f 4 () 4. Observe that the curves get steeper as the power of increases. Steeper the curve, faster is the rate of growth. What this means is that for a fied increment in the value of (> ), the Fig 5.9 increment in the value of y f n () increases as n increases for n,, 3, 4. It is conceivable that such a statement is true for all positive values of n, where f n () n. Essentially, this means that the graph of y f n () leans more towards the y-ais as n increases. For eample, consider f 0 () 0 and f 5 () 5. If increases from to, f 0 increases from to 0 whereas f 5 increases from to 5. Thus, for the same increment in, f 5 grow faster than f 0. Upshot of the above discussion is that the growth of polynomial functions is dependent on the degree of the polynomial function higher the degree, greater is the growth. The net natural question is: Is there a function which grows faster than any polynomial function. The answer is in affirmative and an eample of such a function is y f () 0. Our claim is that this function f grows faster than f n () n for any positive integer n. For eample, we can prove that 0 grows faster than f 00 () 00. For large values of like 0 3, note that f 00 () (0 3 ) whereas f (0 3 3 ) Clearly f () is much greater than f 00 (). It is not difficult to prove that for all > 0 3, f () > f 00 (). But we will not attempt to give a proof of this here. Similarly, by choosing large values of, one can verify that f () grows faster than f n () for any positive integer n.

25 CONTINUITY AND DIFFERENTIABILITY 7 Definition 3 The eponential function with positive base b > is the function y f () b The graph of y 0 is given in the Fig 5.9. It is advised that the reader plots this graph for particular values of b like, 3 and 4. Following are some of the salient features of the eponential functions: () Domain of the eponential function is R, the set of all real numbers. () Range of the eponential function is the set of all positive real numbers. (3) The point (0, ) is always on the graph of the eponential function (this is a restatement of the fact that b 0 for any real b > ). (4) Eponential function is ever increasing; i.e., as we move from left to right, the graph rises above. (5) For very large negative values of, the eponential function is very close to 0. In other words, in the second quadrant, the graph approaches -ais (but never meets it). Eponential function with base 0 is called the common eponential function. In the Appendi A..4 of Class XI, it was observed that the sum of the series !! is a number between and 3 and is denoted by e. Using this e as the base we obtain an etremely important eponential function y e. This is called natural eponential function. It would be interesting to know if the inverse of the eponential function eists and has nice interpretation. This search motivates the following definition. Definition 4 Let b > be a real number. Then we say logarithm of a to base b is if b a. Logarithm of a to base b is denoted by log b a. Thus log b a if b a. Let us work with a few eplicit eamples to get a feel for this. We know 3 8. In terms of logarithms, we may rewrite this as log 8 3. Similarly, is equivalent to saying log Also, is equivalent to saying log or log On a slightly more mature note, fiing a base b >, we may look at logarithm as a function from positive real numbers to all real numbers. This function, called the logarithmic function, is defined by log b : R + R log b y if b y

26 7 MATHEMATICS As before if the base b 0, we say it is common logarithms and if b e, then we say it is natural logarithms. Often natural logarithm is denoted by ln. In this chapter, log denotes the logarithm function to base e, i.e., ln will be written as simply log. The Fig 5.0 gives the plots of logarithm function to base, e and 0. Some of the important observations about the logarithm function to any base b > are listed below: Fig 5.0 () We cannot make a meaningful definition of logarithm of non-positive numbers and hence the domain of log function is R +. () The range of log function is the set of all real numbers. (3) The point (, 0) is always on the graph of the log function. (4) The log function is ever increasing, i.e., as we move from left to right the graph rises above. (5) For very near to zero, the value of log can be made lesser than any given real number. In other words in the fourth quadrant the graph approaches y-ais (but never meets it). (6) Fig 5. gives the plot of y e and y ln. It is of interest to observe that the two curves are the mirror Fig 5. images of each other reflected in the line y. Two properties of log functions are proved below: () There is a standard change of base rule to obtain log a p in terms of log b p. Let log a p α, log b p β and log b a γ. This means a α p, b β p and b γ a. Substituting the third equation in the first one, we have (b γ ) α b γα p Using this in the second equation, we get b β p b γα

27 CONTINUITY AND DIFFERENTIABILITY 73 which implies β αγ or α β. But then γ log a p log b p logba () Another interesting property of the log function is its effect on products. Let log b pq α. Then b α pq. If log b p β and log b q γ, then b β p and b γ q. But then b α pq b β b γ b β + γ which implies α β + γ, i.e., log b pq log b p + log b q A particularly interesting and important consequence of this is when p q. In this case the above may be rewritten as log b p log b p + log b p log p An easy generalisation of this (left as an eercise!) is log b p n n log p for any positive integer n. In fact this is true for any real number n, but we will not attempt to prove this. On the similar lines the reader is invited to verify log b y log b log b y Eample 8 Is it true that e log for all real? Solution First, observe that the domain of log function is set of all positive real numbers. So the above equation is not true for non-positive real numbers. Now, let y e log. If y > 0, we may take logarithm which gives us log y log (e log ) log. log e log. Thus y. Hence e log is true only for positive values of. One of the striking properties of the natural eponential function in differential calculus is that it doesn t change during the process of differentiation. This is captured in the following theorem whose proof we skip. Theorem 5 () The derivative of e w.r.t., is e ; i.e., d (e ) e. () The derivative of log w.r.t., is ; i.e., d (log ).

28 74 MATHEMATICS Eample 9 Differentiate the following w.r.t. : (i) e (ii) sin (log ), > 0 (iii) cos (e ) (iv) e cos Solution (i) (ii) (iii) (iv) Let y e. Using chain rule, we have d e ( ) e Let y sin (log ). Using chain rule, we have d cos (log ) cos (log ) (log ) Let y cos (e ). Using chain rule, we have d e ( e ) ( e ) e Let y e cos. Using chain rule, we have Differentiate the following w.r.t. : cos cos e ( sin ) (sin ) e EXERCISE 5.4. e sin. sin e 3. 3 e 4. sin (tan e ) 5. log (cos e ) 6. 5 e + e e 7. e, > 0 8. log (log ), > cos (log + e ), > Logarithmic Differentiation cos, > 0 log In this section, we will learn to differentiate certain special class of functions given in the form y f () [u()] v () By taking logarithm (to base e) the above may be rewritten as log y v() log [u()]

29 CONTINUITY AND DIFFERENTIABILITY 75 Using chain rule we may differentiate this to get v ( ). u () + v (). log [u()] y u( ) which implies that v( ) y u ( ) v( ) log [ u( ) ] + u( ) The main point to be noted in this method is that f () and u() must always be positive as otherwise their logarithms are not defined. This process of differentiation is known as logarithms differentiation and is illustrated by the following eamples: Eample 30 Differentiate ( 3)( + 4) w.r.t.. Solution Let y ( 3)( + 4) ( ) Taking logarithm on both sides, we have log y [log ( 3) + log ( + 4) log ( )] Now, differentiating both sides w.r.t., we get or 6+ 4 y + ( 3) y ( 3) ( 3)( + 4) ( 3) Eample 3 Differentiate a w.r.t., where a is a positive constant. Solution Let y a. Then log y log a Differentiating both sides w.r.t., we have y log a

30 76 MATHEMATICS or y log a d Thus ( a ) d Alternatively ( a ) a log a Eample 3 Differentiate sin, > 0 w.r.t.. d loga log a d ( e ) e ( log a) e log a. log a a log a. Solution Let y sin. Taking logarithm on both sides, we have log y sin log Therefore. y sin d d (log ) + log (sin ) or y (sin ) log cos + or sin y + cos log Eample 33 Find, if y + y + a b. sin sin + cos log sin sin sin + cos log Solution Given that y + y + a b. Putting u y, v y and w, we get u + v + w a b du dv dw Therefore () Now, u y. Taking logarithm on both sides, we have log u log y Differentiating both sides w.r.t., we have

31 CONTINUITY AND DIFFERENTIABILITY 77 du d d (log y) + log y ( ) u + log y y So Also v y Taking logarithm on both sides, we have log v y log Differentiating both sides w.r.t., we have So Again du u + log y y + log y y y dv d y (log ) + log v dv y + log y v + log y y log + w Taking logarithm on both sides, we have log w log. Differentiating both sides w.r.t., we have... ()... (3) dw d d (log ) + log ( ) w + log i.e. dw w ( + log ) ( + log )... (4)

32 78 MATHEMATICS From (), (), (3), (4), we have or Therefore y y y + log y log y ( + log ) 0 (. y + y. log ) ( + log ) y. y y log y y [ y log y+ y. + (+ log )] y. y + log EXERCISE 5.5 Differentiate the functions given in Eercises to w.r.t... cos. cos. cos (log ) cos 4. sin 5. ( + 3). ( + 4) 3. ( + 5) 4 6. ( )( ) ( 3)( 4)( 5) (log ) + log 8. (sin ) + sin 9. sin + (sin ) cos 0.. ( cos ) + ( sin ) cos + Find of the functions given in Eercises to 5.. y + y 3. y y 4. (cos ) y (cos y) 5. ( y) y e 6. Find the derivative of the function given by f () ( + ) ( + ) ( + 4 ) ( + 8 ) and hence find f (). 7. Differentiate ( 5 + 8) ( ) in three ways mentioned below: (i) by using product rule (ii) by epanding the product to obtain a single polynomial. (iii) by logarithmic differentiation. Do they all give the same answer? +

33 CONTINUITY AND DIFFERENTIABILITY If u, v and w are functions of, then show that d du dv (u. v. w) v. w + u.. w + u. v dw in two ways - first by repeated application of product rule, second by logarithmic differentiation. 5.6 Derivatives of Functions in Parametric Forms Sometimes the relation between two variables is neither eplicit nor implicit, but some link of a third variable with each of the two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is epressed via a third variable. The third variable is called the parameter. More precisely, a relation epressed between two variables and y in the form f (t), y g (t) is said to be parametric form with t as a parameter. In order to find derivative of function in such form, we have by chain rule. dt dt or Thus dt whenever dt 0 dt g () t as g ( t) and f ( t) f () t dt dt [provided f (t) 0] Eample 34 Find, if a cos θ, y a sin θ. Solution Given that Therefore dθ a cos θ, y a sin θ a sin θ, dθ a cos θ Hence dθ acosθ cot θ asin θ dθ

34 80 MATHEMATICS Eample 35 Find, if at, y at. Solution Given that at, y at So at and dt dt a Therefore dt a at t dt Eample 36 Find, if a (θ + sin θ), y a ( cos θ). Solution We have a( + cos θ), dθ dθ a (sin θ) Therefore dθ asin θ θ tan a ( + cos θ) dθ Note It may be noted here that is epressed in terms of parameter only without directly involving the main variables and y. Eample 37 Find,if y a. Solution Let a cos 3 θ, y a sin 3 θ. Then y ( acos θ ) + ( asin θ) 3 3 a (cos θ+ (sin θ ) a Hence, a cos 3 θ, y a sin 3 θ is parametric equation of Now y a dθ 3a cos θ sin θ and dθ 3a sin θ cos θ

35 CONTINUITY AND DIFFERENTIABILITY 8 Therefore dθ 3asin θcosθ y tan 3 θ 3acos θsinθ dθ Note Had we proceeded in implicit way, it would have been quite tedious. EXERCISE 5.6 If and y are connected parametrically by the equations given in Eercises to 0, without eliminating the parameter, Find.. at, y at 4. a cos θ, y b cos θ 3. sin t, y cos t 4. 4t, y 4 t 5. cos θ cos θ, y sin θ sin θ 6. a (θ sin θ), y a ( + cos θ) sin t cos t, cos t y cost t 8. a cost+ log tan y a sin t 9. a sec θ, y b tan θ 0. a (cos θ + θ sin θ), y a (sin θ θ cos θ). If sin t cos t y a, y a, show that 5.7 Second Order Derivative Let y f (). Then f ()... () If f () is differentiable, we may differentiate () again w.r.t.. Then, the left hand side becomes d which is called the second order derivative of y w.r.t. and is denoted by d y. The second order derivative of f () is denoted by f (). It is also

36 8 MATHEMATICS denoted by D y or y or y if y f (). We remark that higher order derivatives may be defined similarly. Eample 38 Find d y, if y 3 + tan. Solution Given that y 3 + tan. Then Therefore 3 + sec d y d 6 + sec. sec tan 6 + sec tan ( 3 + sec ) Eample 39 If y A sin + B cos, then prove that Solution We have A cos B sin y 0 d y +. and d y d (A cos B sin ) A sin B cos y Hence d y + y 0 d y Eample 40 If y 3e + e 3, prove that 5 6y 0 Solution Given that y 3e + e 3. Then 6e + 6e 3 6 (e + e 3 ) Therefore d y e + 8e 3 6 (e + 3e 3 ) Hence d y 5 + 6y 6 (e + 3e 3 ) 30 (e + e 3 ) + 6 (3e + e 3 ) 0

37 CONTINUITY AND DIFFERENTIABILITY 83 Eample 4 If y sin, show that ( ) Solution We have y sin. Then ( ) d y 0. or ( ) d So ( ). 0 or ( ) d y d ( ) ( ) 0 + or d y ( ) 0 d y Hence ( ) 0 Alternatively, Given that y sin, we have y, i.e., ( ) So y ( ). y y + y (0 ) 0 Hence ( ) y y 0 EXERCISE 5.7 Find the second order derivatives of the functions given in Eercises to cos 4. log 5. 3 log 6. e sin 5 7. e 6 cos 3 8. tan 9. log (log ) 0. sin (log ) d y. If y 5 cos 3 sin, prove that y 0 +

38 84 MATHEMATICS. If y cos, Find d y in terms of y alone. 3. If y 3 cos (log ) + 4 sin (log ), show that y + y + y 0 4. If y Ae m + Be n, show that d y ( m n) mny If y 500e e 7, show that d y 49y 6. If e y ( + ), show that d y 7. If y (tan ), show that ( + ) y + ( + ) y 5.8 Mean Value Theorem In this section, we will state two fundamental results in Calculus without proof. We shall also learn the geometric interpretation of these theorems. Theorem 6 (Rolle s Theorem) Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b), such that f(a) f(b), where a and b are some real numbers. Then there eists some c in (a, b) such that f (c) 0. In Fig 5. and 5.3, graphs of a few typical differentiable functions satisfying the hypothesis of Rolle s theorem are given. Fig 5. Fig 5.3 Observe what happens to the slope of the tangent to the curve at various points between a and b. In each of the graphs, the slope becomes zero at least at one point. That is precisely the claim of the Rolle s theorem as the slope of the tangent at any point on the graph of y f () is nothing but the derivative of f () at that point.

39 CONTINUITY AND DIFFERENTIABILITY 85 Theorem 7 (Mean Value Theorem) Let f : [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Then there eists some c in (a, b) such that f () c f () b f() a b a Observe that the Mean Value Theorem (MVT) is an etension of Rolle s theorem. Let us now understand a geometric interpretation of the MVT. The graph of a function y f() is given in the Fig 5.4. We have alrea interpreted f (c) as the slope of the f ( b) f( a) tangent to the curve y f() at (c, f(c)). From the Fig 5.4 it is clear that b a is the slope of the secant drawn between (a, f(a)) and (b, f(b)). The MVT states that there is a point c in (a, b) such that the slope of the tangent at (c, f(c)) is same as the slope of the secant between (a, f (a)) and (b, f(b)). In other words, there is a point c in (a, b) such that the tangent at (c, f(c)) is parallel to the secant between (a, f(a)) and (b, f(b)). Fig 5.4 Eample 4 Verify Rolle s theorem for the function y +, a and b. Solution The function y + is continuous in [, ] and differentiable in (, ). Also f( ) f( ) 6 and hence the value of f() at and coincide. Rolle s theorem states that there is a point c (, ), where f (c) 0. Since f (), we get c 0. Thus at c 0, we have f (c) 0 and c 0 (, ). Eample 43 Verify Mean Value Theorem for the function f() in the interval [, 4]. Solution The function f () is continuous in [, 4] and differentiable in (, 4) as its derivative f () is defined in (, 4).

40 86 Now, MATHEMATICS f() 4 and f (4) 6. Hence f( b) f( a) b a 4 MVT states that there is a point c (, 4) such that f (c) 6. But f () which implies c 3. Thus at c 3 (, 4), we have f (c) 6. EXERCISE 5.8. Verify Rolle s theorem for the function f() + 8, [ 4, ].. Eamine if Rolle s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle s theorem from these eample? (i) f() [] for [5, 9] (ii) f() [] for [, ] (iii) f() for [, ] 3. If f : [ 5, 5] R is a differentiable function and if f () does not vanish anywhere, then prove that f( 5) f(5). 4. Verify Mean Value Theorem, if f() 4 3 in the interval [a, b], where a and b Verify Mean Value Theorem, if f () in the interval [a, b], where a and b 3. Find all c (, 3) for which f (c) Eamine the applicability of Mean Value Theorem for all three functions given in the above eercise. Miscellaneous Eamples Eample 44 Differentiate w.r.t., the following function: sec (i) (ii) e + 3cos (iii) log 7 (log ) + 4 Solution (i) Let y (3+ ) + ( + 4) Note that this function is defined at all real numbers >. Therefore 3 d d (3+ ) (3+ ) + ( + 4) ( + 4)

41 CONTINUITY AND DIFFERENTIABILITY 87 (ii) 3 (3 + ) (3) ( + 4) ( ) 3 This is defined for all real numbers Let sec y e + 3cos >. 3 This is defined at every real number in [,] {} 0. Therefore e d (sec ) + 3 sec sec d e sec (sec ) + 3 sec sec (sec tan ) e + 3 sec sec tan e + 3 Observe that the derivative of the given function is valid only in [,] {} 0 as the derivative of cos eists only in (, ) and the function itself is not defined at 0. log (log ) (iii) Let y log 7 (log ) (by change of base formula). log7 The function is defined for all real numbers >. Therefore d (log (log )) log 7 d (log ) log7 log log 7 log

42 88 MATHEMATICS Eample 45 Differentiate the following w.r.t.. + (i) cos sin (sin ) (ii) tan (iii) sin + cos + 4 Solution (i) Let f () cos (sin ). Observe that this function is defined for all real numbers. We may rewrite this function as f() cos (sin ) π cos cos π Thus f (). sin (ii) Let f() tan. Observe that this function is defined for all real + cos numbers, where cos ; i.e., at all odd multiplies of π. We may rewrite this function as sin f() tan + cos sin cos tan cos tan tan Observe that we could cancel cos in both numerator and denominator as it is not equal to zero. Thus f (). + (iii) Let f() sin. To find the domain of this function we need to find all + 4 such that +. Since the quantity in the middle is always positive, + 4

43 CONTINUITY AND DIFFERENTIABILITY 89 + we need to find all such that, i.e., all such that We + 4 may rewrite this as + which is true for all. Hence the function is defined at every real number. By putting tan θ, this function may be rewritten as + f() sin + 4 sin ( + ) sin tanθ + tan θ sin [sin θ] θ tan ( ) d Thus f () ( ) ( + ) log + 4 ( )log Eample 46 Find f () if f() (sin ) sin for all 0 < < π. Solution The function y (sin ) sin is defined for all positive real numbers. Taking logarithms, we have log y log (sin ) sin sin log (sin ) Then y d (sin log (sin )) cos log (sin ) + sin. cos log (sin ) + cos ( + log (sin )) cos d (sin ) sin

44 90 MATHEMATICS Thus y(( + log (sin )) cos ) ( + log (sin )) ( sin )sin cos Eample 47 For a positive constant a find, where t t + y a,an t+ t Solution Observe that both y and are defined for all real t 0. Clearly Similarly dt dt d t+ ( t ) dt a dt 0 only if t ±. Thus for t ±, t t + d a t+ log a dt t t t + a loga t a d a t+ t+ t dt t a a t+ t t a dt dt t+ t a loga t a a t+ t t t+ t a log a a t+ t Eample 48 Differentiate sin w.r.t. e cos. Solution Let u () sin and v () e cos. We want to find du du /. Clearly dv dv / du dv sin cos and e cos ( sin ) (sin ) e cos a

45 CONTINUITY AND DIFFERENTIABILITY 9 Thus du dv sin cos cos cos cos sin e e Miscellaneous Eercise on Chapter 5 Differentiate w.r.t. the function in Eercises to.. ( ) 9. sin 3 + cos 6 3. (5) 3cos 4. sin ( ), 0 5. cos + 7, < < 6. sin sin cot sin sin, 0 < < π 7. (log ) log, > 8. cos (a cos + b sin ), for some constant a and b. 9. π 3π (sin cos ) (sin cos ), < < a + a + a a, for some fied a > 0 and > ( 3), for > 3. Find, if y ( cos t), 0 (t sin t), π π < t < 3. Find, if y sin + sin, 4. If + y + y + 0, for, < <, prove that ( + ) 5. If ( a) + (y b) c, for some c > 0, prove that 3 + d y is a constant independent of a and b.