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- 1. Limits and ContinuityWe discuss a number of functions. Our aim is to isolate animportant property of a function called continuity. 1. Let f (x) = sin(x). This is defined for all x . [Recall we use radians automatically in order to have the derivative of sin x being cos x.] 2. Let f (x) = log(x). This is defined for x > 0, and so naturally has a restricted domain. Note also that the domain is an open set. 3. Let f (x) = when x a, and suppose f (a) = 2a. 4. Let f (x) = 5. Let f (x) = 0 if x < 0, and let f (x) = 1 for x 0. 6. Let f (x) = sin when x 0 and let f (0) = 0. 7. In each case we are trying to study the behaviour of the function near a particular point. In example 1, the function is well behaved everywhere, there are no problems, and so there is no need to pick out particular points for special care. In example 2, the function is still well behaved wherever it is defined, but we had to restrict the domain. In all of what
- 2. follows, we will assume the domain of all of our functions is suitably restricted.8. We wont spend time in this course discussing standard functions. It is assumed that you know about functions such as sin x, cos x, tan x, log x, exp x, tan-1x and sin-1x, as well as the ``obvious ones like polynomials and rational functions -- those functions of the form p(x)/q(x), where p and q are polynomials. In particular, it is assumed that you know these are differentiable everywhere they are defined. We shall see later that this is quite a strong piece of information. In particular, it means they are examples of continuous functions. Note also that even a function like f (x) = 1/x is continuous, because, wherever it is defined (ie on - {0}), it is continuous.9. In example 3, the function is not defined at a, but rewriting the function 10. = x + a if x a,11. we see that as x approaches a, where the function is not defined, the value of the function approaches 2a. It thus seems very reasonable to extend the definition of f by defining f (a) = 2a. In fact, what we have observed is that 12. = (x + a) = 2a. Definition 4.3 Say that f (x) tends to l as x a iff given > 0, there is some > 0 such that whenever 0 < | x - a| < , then | f (x) - l| < . Example : Let f (x) = for x 2. Show how to define f (2) in order to make f a continuous function at 2.13. Solution. We have
- 3. = = (x2 + 2x + 4)14. Thus f (x) (22 + 2.2 + 4) = 12 as x 2. So defining f (2) = 12 makes f continuous at 2, (and hence for all values of x).15. [Can you work out why this has something to do with the derivative of f (x) = x3 at the point x = 2?] One sided limits Definition Say that f (x) = l, or that f has a limit from the left iff given > 0, there is some > 0 such that whenever a - < x < a, then | f (x) - f (a)| < . There is a similar definition of ``limit from the right, writen as f (x) = l Example Define f (x) as follows:- f (x) = Calculate the left and right hand limits of f (x) at 2. Solution. As x 2 -, f (x) = 3 - x 1 +, so the left hand limit is 1. As x 2 +, f (x) = x/2 1 +, so the right hand limit is 1. Thus the left and right hand limits agree (and disagree with f (2), so f is not continuous at 2). Note our convention: if f (x) 1 and always f (x) 1 as x 2 -, we say that f (x) tends to 1 from above, and write f (x) 1 + etc.
- 4. Proposition If f (x) exists, then both one sided limtsexist and are equal. Conversely, if both one sided limits exitsand are equal, then f (x) exists.Proposition (Continuity Test) The function f is continuousat a iff both one sided limits exits and are equal to f (a).Example : Let f (x) = Show that f iscontinuous at 1. [In fact f is continuous everywhere].Solution. We use the above criterion. Note that f (1) = 1. Also f (x) = x2 = 1 while f (x) = x = 1 = f (1).so f is continuous at 1.Exercise Let f (x) = Show that f iscontinuous at 0. [In fact f is continuous everywhere]Example : Let f (x) = | x|. Then f is continuous in .Solution. Note that if x < 0 then | x| = - x and so is continuous,while if x > 0, then | x| = x and so also is continuous. It remainsto examine the function at 0. From these identifications, we seethat | x| = 0 +, while | x| = 0 +. Since 0 + = 0 - = 0= | 0|, by the 4.12, | x| is continuous at 0Results giving Coninuity
- 5. Just as for sequences, building continuity directly by calculating limits soon becomes hard work. Instead we give a result that enables us to build new continuous functions from old ones, just as we did for sequences. Note that if f and g are functions andk is a constant, then k.f, f + g, fg and (often) f /g are also functions. Proposition Let f and g be continuous at a, and let k be a constant. Then k.f, f + g and fg are continuous at f. Also, if g(a) 0, then f /g is continuous at a. Proof. We show that f + g is continuous at a. Since, by definition, we have (f + g)(a) = f (a) + g(a), it is enough to show that (f (x) + g(x)) = f (a) + g(a). Pick > 0; then there is some such that if | x - a| < , then | f (x) - f (a)| < /2. Similarly there is some such that if | x- a| < , then | g(x) - g(a)| < /2. Let = min( , ), and pick x with | x - a| < . Then | f (x) + g(x) - (f (a) + g(a))| | f (x) - f (a)| + | g(x) - g(a)| < /2 + /2 = . This gives the result Note: Just as when dealing with sequences, we need to know that f /g is defined in some neighbourhood of a. This can be shown using a very similar proof to the corresponding result for sequences.**Proposition Let f be continuous at a, and let g be continuousat f (a). Then gof is continuous at a
- 6. Proof. Pick > 0. We must find > 0 such that if | x - a|< , then g(f (x)) - g(f (a))| < . We find using the givenproperties of f and g. Since g is continuous at f (a), there issome > 0 such that if | y - f (a)| < , then | g(y) - g(f (a))|< . Now use the fact that f is continuous at a, so there issome > 0 such that if | x - a| < , then | f (x) - f (a)| < .Combining these results gives the required inequality.Example : The function f : x sin3x is continuous.Solution. Write g(x) = sin(x) and h(x) = x3. Note that eachof g and h are continuous, and that f = goh. Thus f iscontinuous.Example: Let f (x) = tan . Show that f iscontinuous at every point of its domain.Solution. Let g(x) = . Since -1 < g(x) < 1, the functionis properly defined for all values of x (whilst tan x is undefinedwhen x = (2k + 1) /2 ), and the quotient is continuous, sinceeach term is, and since x2 + a2 0 for any x. Thus f iscontinuous, since f = tanog.
- 7. **Example : Suppose that sin(1/x) = l; in other words,assume, to get a contradiction, that the limit exists. Let xn= 1/( n);then xn 0 as n , and so by assumption, sin(1/xn) = sin(n )=0 l as n . Thus, just by looking at a single sequence, we see thatthe limit (if it exists) can only be l. But instead, consider thesequence xn = 2/(4n + 1) , so again xn 0 as n . In thiscase, sin(1/xn) = sin((4n + 1) /2) = 1, and we must also have l = 1.Thus l does not exist. Note: Sequences often provide a quick way of demonstrating that a function is not continuous, while, if f is well behaved on each sequence which converges to a, then in fact f is continuous at a. The proof is a little harder than the one we have just given, and is left until next year.**Infinite limits There are many more definitions and results about limits. First one that is close to the sequence definition: Definition Say that f (x) = l iff given > 0, there is some K such that whenever x > K, then | f (x) - l| < . Example : Evaluate . Solution. The idea here should be quite familiar from our sequence work. We use the fact that 1/x 0 as x . Thus
- 8. = as x .LIMITSLet f and g be two real valued functions with the same domainsuch that f(x)≤g(x)For all x in domain of definition, for some a, if both ≤Identities involving calculus [ Angle B= , A→C as →0 ⇨ CA→0 and
- 9. BA→BC as →0] These can be seen from looking at the diagrams.Sine and angle ratio identityProof: Area of OCD < Area of sector OAC < Area of OAB ½ r2sinx < ½ r2x < ½ r2tanx [In OAP, AB = OA tanx] , so , so , or
- 10. , so , but , so (By sandwich theorem)Cosine and angle ratio identityProof:
- 11. The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero. Cosine and square of angle ratio identity Proof: As in the preceding proof, The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2. INFORMAL APPROACH TO LIMIT Consider a function f(x) = . It is 0/0 form (known as indeterminate form) at x =2, this functionis defined ∀ x except x = 2.
- 12. If x ≠2, then f(x) = = x+2. The following table exhibits the values of f(x) at points which areclose to 2 on its two sides ( left & right on the real line). x 1. 1. 1. 1. 1. 1. 1.9 2 2.0 2. 2. 2. 2. 2. 2. 4 5 6 7 8 9 9 1 1 2 3 4 5 6 f( 3. 3. 3. 3. 3. 3. 3.9 0/ 4.0 4. 4. 4. 4. 4. 4. x) 4 5 6 7 8 9 9 0 1 1 2 3 4 5 6 Graph of f(x) 4 2 -2 0 2 It is evident from the above table and the graph of f(x) that as x increases and comes closer to 2 from left hand side of 2, the values of f(x) increase and come closer to 4. x→ 2- , f(x)→4 or , =4
- 13. From right hand side of 2 , using notation x→ 2+ , f(x)→4 or , =4Question If f(x) = x sin(1/x) , x≠ = 0, x=0 , then testthe continuity of f(x) at x=0. Answer L.H.L. = 0 [∵ -1≤ sin(1/x)≤1 and 0. Finitevalue =0 ] R.H.L. =0 ⇨ f(x) is cts.PROPERTIES OF INFINITY (i) c.∞ → ∞ , if c > 0 (ii) c.∞ = 0 , c = 0 (iii) c.∞ →-∞ , c < 0. (iv) c∞ = ∞ if c > 1 (v) =0,0≤c≤1 (vi) = 1 , c = 1. (vii) = -∞ , = ∞. INTERDETERMINATE FORMSEXAMPLES: ( ) ( ) ( ) ( - )1∞ ( )00 ( )∞0 (
- 14. o PROBLEM 3 : Determine if the following function is continuous at x=0 .SOLUTION : Function f is defined at x=0 since i.) f(0) = 2 .The left-hand limit =2.The right-hand limit =2.Thus, exists with
- 15. ii.) .Since iii.) ,all three conditions are satisfied, and f is continuous at x=0 .SOLUTION : Function h is not defined at x=-1 since it leads to divisionby zero. Thus, i.) h(-1)does not exist, condition i.) is violated, and function h is NOTcontinuous at x = -1 . o PROBLEM 5 : Check the following function for continuity at x=3 and x=-3 .SOLUTION 5 : First, check for continuity at x=3 . Function f is definedat x=3 since i.) .The limit
- 16. (Circumvent this indeterminate form by factoring the numerator andthe denominator.)(Recall that A2 - B2 = (A-B)(A+B) and A3 - B3 = (A-B)(A2+AB+B2 ) . )(Divide out a factor of (x-3) . ) = ,i.e., ii.) .Since,
- 17. iii.) ,all three conditions are satisfied, and f is continuous at x=3 . Now,check for continuity at x=-3 . Function f is not defined at x = -3 becauseof division by zero. Thus, i.) f(-3)does not exist, condition i.) is violated, and f is NOT continuous at x=-3. o PROBLEM 11 : For what values of x is the following function continuous ?SOLUTION 11 : Consider separately the three component functionswhich determine f . Function is continuous for x > 1 since itis the quotient of continuous functions and the denominator is neverzero. Function y = 5 -3x is continuous for since it is apolynomial. Function is continuous for x < -2 since it is thequotient of continuous functions and the denominator is never zero.Now check for continuity of f where the three components are joinedtogether, i.e., check for continuity at x=1 and x=-2 . For x = 1 function fis defined since i.) f(1) = 5 - 3(1) = 2 .
- 18. The right-hand limit =(Circumvent this indeterminate form one of two ways. Either factorthe numerator as the difference of squares, or multiply by theconjugate of the denominator over itself.) =2.The left-hand limit = = 5 - 3(1) =2.Thus, ii.) .Since
- 19. iii.) ,all three conditions are satisfied, and function f is continuous at x=1 .Now check for continuity at x=-2 . Function f is defined at x=-2 since i.) f(-2) = 5 - 3(-2) = 11 .The right-hand limit = = 5 - 3( -2) = 11 .The left-hand limit = = -1 .Since the left- and right-hand limits are different, ii.) does NOT exist,condition ii.) is violated, and function f is NOT continuous at x=-2 .Summarizing, function f is continuous for all values of x EXCEPT x=-2 .
- 20. o PROBLEM 13 : Determine all values of the constants A and B so that the following function is continuous for all values of x .SOLUTION 13 : First, consider separately the three components whichdetermine function f . Function y = Ax - B is continuous for for 2any values of A and B since it is a polynomial. Function y = 2x + 3Ax +B is continuous for for any values of A and B since it is apolynomial. Function y = 4 is continuous for x > 1 since it is apolynomial. Now determine A and B so that function f is continuous atx= -1 and x= 1 . First, consider continuity at x= -1 . Function f must bedefined at x= -1 , so i.) f(-1)= A(-1) - B = - A - B .The left-hand limit = = A (-1) - B =-A-B.The right-hand limit = = 2(-1)2 + 3A(-1) + B
- 21. = 2 - 3A + B .For the limit to exist, the right- and left-hand limits must exist and beequal. Thus, ii.) ,so that 2A - 2B = 2 ,or(Equation 1) A-B=1.Now consider continuity at x=1 . Function f must be defined at x=1 , so i.) f(1)= 2(1)2 + 3A(1) + B = 2 + 3A + B .The left-hand limit = = 2(1)2 + 3A(1) + B = 2 + 3A + B .The right-hand limit = =4.
- 22. For the limit to exist, the right- and left-hand limits must exist and beequal. Thus, ii.) ,or(Equation 2) 3A + B = 2 .Now solve Equations 1 and 2 simultaneously. Thus, A - B = 1 and 3A + B = 2are equivalent to A = B + 1 and 3A + B = 2 .Use the first equation to substitute into the second, getting 3 (B + 1 ) + B = 2 , 3B+3+B=2,and 4 B = -1 .Thus,and
- 23. .For this choice of A and B it can easily be shown that iii.)and iii.) ,so that all three conditions are satisfied at both x=1 and x=-1 , andfunction f is continuous at both x=1 and x=-1 . Therefore, function f iscontinuous for all values of x if and . o PROBLEM 14 : Show that the following function is continuous for all values of x .SOLUTION 14 : First describe f using functional composition. Let g(x) =-1/x2 and h(x) = ex . Function h is well-known to be continuous for allvalues of x . Function g is the quotient of functions continuous for allvalues of x , and is therefore continuous for all values of x except x=0 ,that x which makes the denominator zero. Thus, for all values of xexcept x=0 ,
- 24. f(x) = h ( g(x) ) = e g(x) = e -1/x2is a continuous function (the functional composition of continuousfunctions). Now check for continuity of f at x=0 . Function f is definedat x=0 since i.) f(0) = 0 .The limit(The numerator approaches -1 and the denominator is a positivenumber approaching zero.) ,so that =0,i.e.,
- 25. ii.) .Since iii.) ,all three conditions are satisfied, and f is continuous at x=0 . Thus, f iscontinuous for all values of x . o PROBLEM 15 : LetShow that f is continuous for all values of x .SOLUTION 15 : First show that f is continuous for all values of x .Describe f using functional composition. Let , , and 2k(x) = x . Function h is well-known to be continuous for all values of x. Function k is a polynomial and is therefore continuous for all valuesof x . Function g is the quotient of functions continuous for all valuesof x , and is therefore continuous for all values of x except x=0 , that xwhich makes the denominator zero. Thus, for all values of x exceptx=0 ,is a continuous function (the product and functional composition ofcontinuous functions). Now check for continuity of f at x=0 . Function fis defined at x=0 since i.) f(0) = 0 .
- 26. The limit does not exist since the values of oscillatebetween -1 and +1 as x approaches zero. However, forso that .Since ,it follows from the Squeeze Principle that ii.) .Since iii.) ,all three conditions are satisfied, and f is continuous at x=0 . Thus, f iscontinuous for all values of x . o PROBLEM 1 : Compute .SOLUTION1 : Note that DOES NOT EXIST since values of
- 27. oscillate between -1 and +1 as x approaches 0 from the left.However, this does NOT necessarily mean that does notexist ! ? #. Indeed, x3 < 0 andfor x < 0. Multiply each component by x3, reversing the inequalitiesand gettingor .Since ,it follows from the Squeeze Principle that . o PROBLEM 2 : Compute .
- 28. SOLUTION2 : First note that ,so thatand .Since we are computing the limit as x goes to infinity, it is reasonableto assume that x+100 > 0. Thus, dividing by x+100 and multiplying byx2, we getand .Then =
- 29. = = = .Similarly, = .Thus, it follows from the Squeeze Principle that = (does not exist).DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONSNone of the six basic trigonometry functions is a one-to-one function.However, in the following list, each trigonometry function is listedwith an appropriately restricted domain, which makes it one-to-one. 1. for 2. for 3. for 4. for , except
- 30. 5. for , except x = 0 6. forBecause each of the above-listed functions is one-to-one, each has aninverse function. The corresponding inverse functions are 1. for 2. for 3. for 4. arc for , except 5. arc for , except y = 0 6. arc forIn the following discussion and solutions the derivative of a functionh(x) will be denoted by or h(x) . The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometryidentities, implicit differentiation, and the chain rule. They are asfollows. 1. 2. 3. 4. arc 5. arc
- 31. 6. arcImportant points: (1)when exists but either f© does notexist or ≠f(c), we say that f has a reomovable discts.;otherwise f has non- reomovable discts.(2) composition of real valued fns. of f and g (fog) is defined at c. if gis cts. At c and if f is cts. At g© then fog is cts. At c, fog(x)=f(g(x)) isdefined whenever the range of g is subset of domain of f. [Ex. sin(x²)]Some questions on differentiability:1. Show that f(x)=x2 is diff. at x=1, find f’(1)2. Prove that f(x) = [x], 0<x<3 is not diff. at x=1 & 2.3. Discuss the continuity & diff. of f(x)=4. Show that f(x)= is not diff. at x=2.5.For which value of a &b the function f(x) = is diff. at x=0Solutions:1. R f’(1) = = =2L f’(1)= = =2 ⇨ L.H.D = R.H.D=22. R f’(1) = = =1-1/h=0{∵ [1+h]=1}L f’(1)= = = 0-1/-h=∞ ⇨R.H.D≠L.H.D ⇨ f is not diff. at x=1. At x=2, similarly f is not diff. at x=2, ∵ R.H.D= 0≠ L.H.D=∞ {[2+h]=2 &[2-h] = 1}3. L.H.lt =0 = R.H.lt ⇨ f is cts. At x=0
- 32. L.H.D = =0= R.H.D ⇨ diff.4. R.H.D. = = -1L.H.D = = 1 ⇨ not diff.5. Given L.H.D= R.H.D ⇨ = = a=2 & b=0 [ ∵ f is cts. ⇨b=0] ASSIGNMENT(continuity & differentiability) (XII)**Question 1 Determine a and b so that the function f given by f(x) = , x<п/2 =a, x=п/2 = , x>п/2 Is continuous at x=п/2. Answer [a = 1/3 , b = 8/3] **Question 2 Find k such that following functions are continuous atindicated point (i) f(x) = at x=0 (ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2 = k, x = 0 at x=2.
- 33. Answer [ (i) k=1,(ii) k=1/2] **Question 3 The function f is defined as If f(x) is continuous on [0,8], find the values of a and b. Answer [a=3,b=-2]** Question 4 If f(x) = is continuous inthe [-1,1], find p. Answer [p=-1]**Question 5 Find the value of a and b such that the f(x) defined as f(x) = is continuous for allvalues of x in [0,п]. ANSWER [a=п/6 , b=-п/12]** Question 6 Prove that = -4 [ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-sinx)(tanx+1)]/cosx Cosx-sinx = cos )]**Question 7 Prove that (i) = [ Hint:put x= sinѲ]
- 34. (ii) ) = -3/2. [Hint: = & useformula of ]Question 8 f(x) = , =1 & =1, then p.t.f(-2)=f(2)=1. [ Hint: =0]Question 9 [Dr. = 2|sinx/2| & =1 |sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist]Question 10 Show that the function f(x) is continuous at x=0. [Hint: use =1 , =1]Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-1≠1(R.h.d)QUESTION 12 Discuss the continuity of the fn. f(x) = |x+1|+|x+2|, at x= -1 & -2 [Hint:f(x) = yes cts. At x=-1,-2Question 13 Find the values of p and q so that f(x)= is diff. at x = 1. [ answer is p=3 , q=5]Question 14 For what choice of a, b, c if any , does the functionF(x) = becomes diff at x=1,2 & show that a=b=c=0.
- 35. Question15For what values a,b f(x)= is diff.at x=0[Hint: L.H.d= 2 =1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R]ASSIGNMENT WITH HINTS (XI)Question.1 Evaluate ** [Hint ] Question.2 [Hint =7/4] Question.3 [use sin3x=3sinx-4sin³x, put ( ) =h ,answer is -4] Question.4 [Hint = 2]Question.5 [hint: put tanx=sinx/cosx, answer is 1/2]
- 36. Question.6[Hint: use sinx-siny=2cos(x+y)/2.sin(x-y)/2 , answer is 0]Question.7 [ answer is ½]Question.8 [ answer is 1/8]Question.9 Let f(x) =Find a,b so that and exist.[Hint: exists(ax+b)= = x2/2 2a+b=2……(1)Similarly -2a+b = -1……(2) a=3/4,b=1/2]Question.10 and where [x] denotes theintegral part of x. Are they equal?[HINT: = 3/3=1,x>3 = 3/2 ,x<3]Question.11 and , are they equal? -5 as -x -6
- 37. **Question.12 Test the continuity of the function at x=0, f(x) = ,x ≠0 0, x=0 [Hint e-∞ =0,e∞ =∞]Derivatives (XI)BY FIRST PRINCIPLE o Use the limit definition to compute the derivative, f(x), for .SOLUTION :(Get a common denominator for the expression in the numerator.Recall that division by is the same as multiplication by .)(Algebraically and arithmetically simplify the expression in thenumerator. It is important to note that the denominator of this
- 38. expression should be left in factored form so that the term can beeasily eliminated later.)(The term now divides out and the limit can be calculated.) o Use the limit definition to compute the derivative, f(x), forSOLUTION 6 :(Recall a well-known trigonometry identity :
- 39. .)(Recall the following two well-known trigonometry limits : and .)SOLUTION 3 : Differentiate . Apply the quotient rule.Then
- 40. (Recall the well-known trigonometry identity .) .SOLUTION 4 : Differentiate . Apply the product rule.Then
- 41. SOLUTION 5 : Differentiate . This is NOT a product offunctions. Its a composition of functions. Apply the chain rule. Then .SOLUTION 6 : Differentiate . Apply the product rule first,followed by the chain rule. Then(Ncert)limits & derivatives (xi)Question 4:Evaluate the Given limit:Question 10:
- 42. Evaluate the Given limit:Question 12:Evaluate the Given limit:Question 14:Evaluate the Given limit:Question 15:Evaluate the Given limit:Question 16:Evaluate the given limit:Question 17:Evaluate the Given limit:Question 18:Evaluate the Given limit:
- 43. Question 20:Evaluate the Given limit:Question 21:Evaluate the Given limit:Question 22:Question 23:Find f(x) and f(x), where f(x) =Question 24:Find f(x), where f(x) =Question 25:Evaluate f(x), where f(x) =Question 26:
- 44. Find f(x), where f(x) =Question 28:Suppose f(x) = and if f(x) = f(1) what arepossible values of a and b?Question 30:If f(x) = .For what value (s) of a does f(x) exists?Question 31:If the function f(x) satisfies , evaluate .Question 32:
- 45. If . For what integers m and n does and exist?ANSWERS: 4. 19/2, 10. 2, 12. -1/4, 14. A/B, 15. 1/п, 16.1/п, 17. 4, 18. (a+1)/b, 20. 1, 21. 0, 22. 2, 23. 3,6, 24.Limit does not exist at x=1, 25. Limit does not exist atx=0, 26. Limit does not exist at x=0, 28. A=0,b=4, 30. f(x) exists for all a≠0, 31. 2, 32. we needm=n; and exists for any integral value of m & n.DERIVATIVESQuestion 1:Find the derivative of the following functions from firstprinciple:(i) –x (ii) (–x)–1 (iii) sin (x + 1)(iv)Solution of (iv) : use formula of first principle(ab-initio)
- 46. We have f’(x) = = = =sin(x-п/8) [ , cosA-cosB= -2sin(A+B)/2.sin(A-B)/2]Question 11:Find the derivative of the following functions:(i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x(iv) cosec x (v) 3cot x + 5cosec x(vi) 5sin x – 6cos x + 7 (vii) 2tan x – 7sec xSolution of (i) (2 sin x cos x)/2= sin2x/2 , derivative is ½(2cos2x) = cos2x or can use Leibnitz product rule.Question 16:Find the derivative of the following functions :Solution: by quotient rule or take derivative after simplificationIt can be written as = =tan(Derivative is - sec2( .Question 17:Find the derivative of the following functions:
- 47. Solution: multiple & divide by cosx, we get = -tan(п/4+x)Derivative is - sec2(п/4+x) [simplest method]Question 18:Find the derivative of the following functionsSolution: can be written as = tan2(x/2) Derivative is 2 tan(x/2) . sec2(x/2) . ½ = tan(x/2) . sec2(x/2).Question 23:Find the derivative of the following functions (x2 + 1) cos xQuestion 24:Find the derivative of the following functions: (ax2 + sin x)(p + q cos x)Question 25:Find the derivative of the following functionQuestion 26:Find the derivative of the following function :Question 27:
- 48. Find the derivative of the following function:Question 28:Find the derivative of the following functions :Question 29:Find the derivative of the following functions : (x + sec x)(x – tan x)Question 30:Find the derivative of the following functions:

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