CONTINUITY AND DIFFERENTIABILITY

1 Of a function at a pointLet f be a real function on a subset of the real numbers and let c be a point in thedomain of f. Then f is continuous at c iflim()()xcfxfc =More elaborately, if the left hand limit, right hand limit and the value of the functionat x = c exist and are equal to each other, ,lim()()lim()xcxcfxfcfx + ==then f is said to be continuous at x = in an interval(i)f is said to be continuous in an open interval (a, b) if it is continuous at everypoint in this interval.(ii)f is said to be continuous in the closed interval [a, b] if f is continuous in (a, b) limxa+ f (x) = f (a) limxb f (x) = f (b)Chapter5 CONTINUITY ANDDIFFERENTIABILITY20/04/2018 CONTINUITY AND DIFFERENTIABILITY meaning of CONTINUITY (i)Function f will be continuous at x = c if there is no break in the graph of thefunction at the point (),()cfc.(ii)In an interval, function is said to be continuous if there is no break in thegraph of the function in the entire function f will be discontinuous at x = a in any of the following cases :(i)limxa f (x) and limxa+ f (x) exist but are not equal.

2 (ii)limxa f (x) and limxa+ f (x) exist and are equal but not equal to f (a).(iii) f (a) is not of some of the common functionsFunction f (x)Interval in whichf is continuous1. The constant function, f (x) = c2. The identity function, f (x) = xR3. The polynomial function, (x)= a0 xn + a1 x n 1 + .. + an 1 x + an4.| x a |( , ) n, n is a positive integer( , ) {0} (x) / q (x), where p (x) and q (x) areR { x : q (x) = 0}polynomials in x, cos x, sec xR { (2 n + 1) 2: n Z} x, cosec xR { (n : n Z}20/04/201888 x(0, ) inverse trigonometric functions,In their , sin 1 x, cos 1 x of composite functionsLet f and g be real valued functions such that (fog) is defined at a. If g is continuousat a and f is continuous at g (a), then (fog) is continuous at function defined by f (x) = 0()()limhfxhfxh + , wherever the limit exists, isdefined to be the derivative of f at x.)

3 In other words, we say that a function f isdifferentiable at a point c in its domain if both 0()()limhfchfch + , called left handderivative, denoted by Lf (c), and 0()()limhfchfch+ + , called right hand derivative,denoted by R f (c), are finite and equal.(i)The function y = f (x) is said to be differentiable in an open interval (a, b) ifit is differentiable at every point of (a, b)(ii)The function y = f (x) is said to be differentiable in the closed interval [a, b]if R f (a) and L f (b) exist and f (x) exists for every point of (a, b).(iii)Every differentiable function is continuous, but the converse is not of derivativesIf u, v are functions of x, then(i)()duvdx = dudvdxdx(ii)()=+ddvduuvuvdxdxdx(iii)2dud vvududxdxdxvv = 20/04/2018 CONTINUITY AND DIFFERENTIABILITY Chain rule is a rule to differentiate composition of functions. Let f = vou. Ift = u (x) and both dtdx and dvdt exist then.

4 = Following are some of the standard derivatives (in appropriate domains)1. 121(sin)1= dxdxx2. 121(cos)1dxdxx = 3. 121(tan)1=+dxdxx4. 121(cot)1dxdxx =+5. 121(sec),11dxxdxxx=> 6. 121(cosec),11dxxdxxx => Exponential and logarithmic functions(i)The exponential function with positive base b > 1 is the functiony = f (x) = bx. Its domain is R, the set of all real numbers and range is the setof all positive real numbers. Exponential function with base 10 is called thecommon exponential function and with base e is called the natural exponentialfunction.(ii)Let b > 1 be a real number. Then we say logarithm of a to base b is x if bx=a,Logarithm of a to the base b is denoted by logb a. If the base b = 10, we sayit is common logarithm and if b = e, then we say it is natural logarithms. logxdenotes the logarithm function to base e. The domain of logarithm functionis R+, the set of all positive real numbers and the range is the set of all realnumbers.

5 (iii)The properties of logarithmic function to any base b > 1 are listed below:1. logb (xy) = logb x + logb y2. logb xy = logb x logb y20/04/201890 MATHEMATICS3. logb xn = n logb x4. logloglogcbcxxb= , where c > 15. logb x 1log=xb6. logb b = 1 and logb 1 = 0(iv)The derivative of ex , x is ex , ()xxdeedx=. The derivative of , x is 1x; 1(log)dxdxx=. differentiation is a powerful technique to differentiate functionsof the form f (x) = (u (x))v(x), where both f and u need to be positive functionsfor this technique to make of a function with respect to another functionLet u = f (x) and v = g (x) be two functions of x, then to find derivative of f (x) g (x), , to find dudv, we use the formuladududxdvdvdx=. order derivative22ddydydxdxdx = is called the second order derivative of y x. It is denoted by y ory2 , if y = f (x). s TheoremLet 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.

6 Then there exists at least one point c in(a, b) such that f (c) = AND DIFFERENTIABILITY 91 Geometrically Rolle s theorem ensures that there is at least one point on the curvey = f (x) at which tangent is parallel to x-axis (abscissa of the point lying in (a, b)). Value Theorem (Lagrange)Let f : [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Thenthere exists at least one point c in (a, b) such that f (c) = ()()fbfaba .Geometrically, Mean Value Theorem states that there exists at least one point c in(a, b) such that the tangent at the point (c, f (c)) is parallel to the secant joining thepoints (a, f (a) and (b, f (b)). Solved ExamplesShort Answer ( )Example 1 Find the value of the constant k so that the function f defined below iscontinuous at x = 0, where 21 cos4(),08,0xfxxxkx = = .Solution It is given that the function f is continuous at x = 0.)

7 Therefore,0limx f (x) = f (0) 201 cos4lim8xxkx = 2202sin2lim8xxkx = 20sin2lim2xxkx = k = 1 Thus, f is continuous at x = 0 if k = 2 Discuss the CONTINUITY of the function f(x) = sin x . cos Since sin x and cos x are continuous functions and product of two continuousfunction is a continuous function, therefore f(x) = sin x . cos x is a continuous MATHEMATICSE xample 3 If 322 1620,2()( 2),2xxxxfxxkx ++ = = is continuous at x = 2, findthe value of Given f (2) = , 322222 1620lim()lim()lim( 2)xxxxxxfxfxx+ ++=== 2222(5)( 2)limlim(5)7( 2)xxxxxx +=+=As f is continuous at x = 2, we have 2lim()(2)xfxf = k = 4 Show that the function f defined by1sin,0()0,0xxfxxx = = is continuous at x = Left hand limit at x = 0 is given by 001lim()limsinxxfxxx = = 0[since, 1 < sin1x < 1]Similarly 001lim()limsin0xxfxxx++ ==. Moreover f (0) = 00lim()lim()(0)xxfxfxf+ ==. Hence f is continuous at x = 0 Example 5 Given f(x) = 1 1x.

8 Find the points of discontinuity of the compositefunction y = f [f(x)].Solution We know that f (x) = 1 1x is discontinuous at x = 1 Now, for 1x ,20/04/2018 CONTINUITY AND DIFFERENTIABILITY 93f (f (x))= 1 1fx = 1 112 1 1xxx=,which is discontinuous at x = , the points of discontinuity are x = 1 and x = 6 Let f(x) = xx, for all x R. Discuss the derivability of f(x) at x = 0 Solution We may rewrite f as 22,if0(),if0xxfxxx = < Now Lf (0) = 2000(0) (0) 0limlimlim0hhhfhfhhhh +== =Now Rf (0) = 2000(0) (0) 0limlimlim0hhhfhfhhhh++ +===Since the left hand derivative and right hand derivative both are equal, hence f isdifferentiable at x = 7 Differentiate tanx xSolution Let y = tanx. Using chain rule, we have1.(tan)2tandydxdxdxx== ()2tandxxdxx= 211(sec)22tanxxx = 2(sec) 8 If y = tan(x + y), find Given y = tan (x + y). differentiating both sides x, we have20/04/201894 MATHEMATICS2sec()()dydxyxydxdx=++= sec2 (x + y) 1dydx + or[1 sec2 (x + y] dydx= sec2 (x + y)Therefore,22sec()1sec()dyxydxxy+= + = cosec2 (x + y).)

9 Example 9 If ex + ey = ex+y, prove thatyxdyedx = .Solution Given that ex + ey = ex+y. Differentiating both sides x, we haveex + eydydx = ex+y 1dydx + or (ey ex+y)dydx = ex+y ex,which implies that xyxxyxyxyxyyxydyeeeeeedxeeeee+ ++ === .Example 10 Find dydx, if y = tan 1 32311,1333xxxx << .Solution Put x = tan , where 66 < <.Therefore,y = tan 1 323tantan13tan = tan 1 (tan3 )= 3 (because 322 < <)= 3tan 1x20/04/2018 CONTINUITY AND DIFFERENTIABILITY 95 Hence, dydx= 231x+.Example 11 If y = sin 1{}211xxxx and 0 < x < 1, then find We have y = sin 1{}211xxxx , where 0 < x < = sinA and x = sinBTherefore, y = sin 1{}22sinA1sinBsinB1sinA = sin 1{}sinAcosBsinBcosA = sin 1{}sin(AB) = A BThusy = sin 1 x sin 1 xDifferentiating x, we get()() = 211211xxx .Example 12 If x = a sec3 and y = a tan3 , find dydx at 3 =.Solution We have x = a sec3 and y = a tan3.

10 Differentiating , we get233sec(sec)3sectandxdaadd= = and2223tan(tan)3tansecdydaadd= = .Thus2233tansectansinsec3sectandydyaddxd xad ==== .20/04/201896 MATHEMATICSH ence,33sin32atdydx = == .Example 13 If xy = ex y, prove that dydx = 2log(1log)xx+.Solution We have xy = ex y . Taking logarithm on both sides, we gety log x = x y y (1 + log x) = = 1logxx+Differentiating both sides x, we get221(1log).1log(1log)(1log)xxdyxxdxxx + ==++.Example 14 If y = tanx + secx, prove that 22dydx = 2cos(1sin)xx .Solution We have y = tanx + secx. Differentiating x, we getdydx = sec2x + secx tanx= 221sincoscosxxx+ = 21sincosxx+= 1sin(1sin)(1sin)xxx++ .thus dydx = 11 , differentiating again x, we get22dydx =()22 coscos(1 sin)(1 sin)xxxx=Example 15 If f (x) = |cos x|, find f 34 .20/04/2018 CONTINUITY AND DIFFERENTIABILITY 97 Solution When 2 < x < , cosx < 0 so that |cos x| = cos x, , f (x) = cos x f (x) = sin , f 34 = sin 34 = 12 Example 16 If f (x) = |cos x sinx|, find f 6.