3y ago

153 Views

8 Downloads

1.86 MB

124 Pages

Transcription

Engineering Mathematics – IDr. V. Lokesha2011Engineering Mathematics – I(10 MAT11)LECTURE NOTES(FOR I SEMESTER B E OF VTU)VTU-EDUSAT Programme-15Dr. V. LokeshaProfessor and HeadDEPARTMENT OF MATHEMATICSACHARYA INSTITUTE OF TECNOLOGYSoldevanahalli, Bangalore – 9010 MAT111

Engineering Mathematics – IDr. V. Lokesha2011ENGNEERING MATHEMATICS – IContentCHAPTERUNIT IDIFFERENTIAL CALCULUS – IUNIT IIDIFFERENTIAL CALCULUS – IIUNIT IIIDIFFERENTIAL CALCULUS – III10 MAT112

Engineering Mathematics – IDr. V. Lokesha2011UNIT - IDIFFERENTIAL CALCULUS – IIntroduction:The mathematical study of change like motion, growth or decay is calculus. The Rate of change ofgiven function is derivative or differential.The concept of derivative is essential in day to day life. Also applicable in Engineering, Science,Economics, Medicine etc.Successive Differentiation:Let y f (x) --(1) be a real valued function.dyThe first order derivative of y denoted byor y’ or y1 or 1dx 2The Second order derivative of y denoted by d yor y’’ or y2 or 2dx 2Similarly differentiating the function (1) n-times, successively,dnyththe n order derivative of y exists denoted by n or yn or yn or ndxThe process of finding 2nd and higher order derivatives is known as Successive Differentiation.nth derivative of some standard functions:1.Sol :y eaxy1 a eaxy2 a2 eaxDifferentiating Successivelyyn an eaxie.Dn[eax] an eaxFor, a 1Dn[ex] ex10 MAT113

Engineering Mathematics – IDr. V. Lokesha201110 MAT114

Engineering Mathematics – IDr. V. Lokesha201110 MAT115

Engineering Mathematics – IDr. V. Lokesha201110 MAT116

Engineering Mathematics – IDr. V. Lokesha201110 MAT117

Engineering Mathematics – IDr. V. Lokesha2011Leibnitz’s Theorem :It provides a useful formula for computing the nth derivative of a product of two functions.Statement : If u and v are any two functions of x with un and vn as their nth derivative. Then the nthderivative of uv is(uv)n u0vn nC1 u1vn-1 nC2u2vn-2 nCn-1un-1v1 unv0Note : We can interchange u & v (uv)n (vu)n,nC 11.n , nC2 n(n-1) /2! , nC3 n(n-1)(n-2) /3! Find the nth derivations of eax cos(bx c)Solution: y1 eax – b sin (bx c) a eax cos (b x c), by product rule.i.e, y1 eax [a cos (bx c ) b sin (bx c )]Let us put a r cos θ , and b r sin θ . a 2 b 2 r 2 and tan θ b / a.ie., r a 2 b 2 and θ tan-1 (b/a)Now, y1 e ax [r cos θ cos(bx c) r sin θ sin( bx c)]Ie., y1 r eax cos (θ bx c )where we have used the formula cos A cos B – sin A sin B cos (A B)Differentiating again and simplifying as before,y2 r2 eax cos (2θ bx c ) .Similarly y3 r3 e ax cos (3θ bx c ) . Thus y n r n e ax cos(nθ bx c )Where r a 2 b 2 and θ tan-1 (b/a).Thus Dn [eax cos (b x c)][[ ( a 2 b 2 ) n e ax cos n tan 1 (b / a ) bx c]]10 MAT118

Engineering Mathematics – IDr. V. Lokesha2. Find the nth derivative of logy 4x2 8x 34 x 2 8x 3 log (4x2 8x 3) ½Solution : Let y logie., y 20111log (4x2 8x 3) log xn n log x21log { (2x 3) (2x 1)}, by factorization.2 y 1{log (2x 3) log (2x 1)}21 ( 1) (n 1)!2 n ( 1) (n 1)!2 n Now y n 2 (2x 3)n(2x 1)n n 1n 1 11 Ie., yn 2n-1 (-1) n-1 (n-1) ! nn (2 x 3) (2 x 1) 3. Find the nth derivative of log 10 {(1-2x)3 (8x 1)5}Solution : Let y log 10 {1-2x)3 (8x 1)5}It is important to note that we have to convert the logarithm to the base e by the property:log10 x Thus y Ie., y yn Ie.,log e xlog e 10{135log e (1 2 x ) (8x 1)log e 10}1{3 log(1 2x ) 5 log(8x 1)}log e 10n 1n( 1)n 1 (n 1)!8n 1 ( 1) (n 1)!( 2)3.5 loge 10 (1 2x )n(8x 1)n n 1nn(5(4) 1) (n 1)!2 n 3( 1)yn nn log e 10 (1 2x ) (8x 1) 10 MAT119

Engineering Mathematics – IDr. V. Lokesha20114. Find the nth derivative of e2x cos2 x sin x 1 cos 2 x Solution : let y e2x cos2 x sin x e2x sin x2 ie., y e2x(sin x sin x cos 2x)2e2x21 sin x [sin 3x sin ( x )] 2 e2x(2 sin x sin 3x sin x ) sin (-x) -sin x 4e2x y (sin x sin 3x)4{ ()(Now y n 1 n 2xD e sin x D n e 2 x sin 3x4Thus y n 14 yn e2x4{( 5 ) en2x)}] ( 13 ) e[sin n tan 1 (1 2 ) x {( 5 ) sin[n tann 1(1 2) x ] (n2x[sin n tan 1 (3 2 ) 3x) [13 sin n tan 1 (3 2) 3xn]}]}5. Find the nth derivative of e2x cos 3xSolution : Let y e2x cos3 x e 2x.1(3 cos x cos 3x)41(3 e2x cos x e2x cos 3x)4Ie.,y yn 1{3Dn (e2x cos x) Dn (e2x cos 3x)}4yn {( )] ( 13 ) e[n13 5 e 2 x cos n tan 1 (1 2 ) x 4Thus y n {( )[n2x] ( )[]}[]}cos n tan 1 (3 2 ) 3 xnne2x3 5 cos n tan 1 (1 2) x 13 cos n tan 1 (3 2) 3x410 MAT1110

Engineering Mathematics – I6. Find the nth derivative ofy Solution :Dr. V. Lokesha2011x2(2x 1)(2x 3)x2is an improper fraction because; the degree of the(2x 1)(2x 3)numerator being 2 is equal to the degree of the denominator. Hence we must divide andrewrite the fraction.4x 2x21 for convenience.4 x 2 8x 3 4 4 x 2 8x 3y 14x4x2 8x 324 x 2 8x 3 8x 31 8x 3 1 2 4 4 x 8x 3 y Ie., y 1 1 8x 3 4 4 4 x 2 8x 3 The algebraic fraction involved is a proper fraction.NowLet1 8x 3 yn 0 Dn 2 4 4 x 8x 3 8x 3AB (2x 1)(2x 3) 2x 1 2x 3Multiplying by (2x 1) (2x 3) we have, 8x 3 A (2x 3) B (2x 1).(1)By setting 2x 1 0, 2x 3 0 we get x -1/2, x -3/2.Put x -1/2 in (1): -1 -1 A (2) A -1/2Put x -3/2 in (1): -9 B (-2) B 9/2 1 1 1 9 n 1 y n D n D 4 2 2x 1 2 2x 3 10 MAT1111

Engineering Mathematics – IDr. V. Lokesha2011( 1)n n!2n 9 ( 1)n n!2n 1 ( 1) 8 (2x 3)n 1 (2x 1)n 1ie., y nn 1( 1) n!2 n 87. Find the nth derivative ofSolution : y 19 n 1 n 1(2x 3) (2x 1)x4( x 1) ( x 2)x4is an improper fraction.( x 1) ( x 2)(deg of nr. 4 deg. of dr. 2)On dividing x4 by x2 3 x 2, We get 15 x 14 y ( x2 – 3x 7 ) 2 x 3x 2 15 x 14 yn Dn (x2-3x 7)-Dn 2 x 3x 2 But D ( x2 – 3x 7 ) 2x – 3, D2 ( x2 – 3x 7 ) 2D3( x2 – 3x 7 ) 0. Dn ( x2 – 3x 7 ) 0 if n 2 15 x 14 Hence yn -Dn ( x 1) ( x 2) Now, let DnAB15 x 14 2x 3x 2 ( x 1) ( x 2) 15x 14 A(x 2) B(x 1 )Put x - 1 ; - 1 A ( 1 ) or A - 1Put x - 2 ; - 16 B ( - 1 ) or B 16 1 1 16 D n Yn D n x 2 x 1 10 MAT1112

Engineering Mathematics – I Dr. V. Lokesha2011( 1) n n ! 1n( 1) n n ! 1n 16( x 1) n 1( x 2) n 1 116 yn ( 1) n n ! n 2n 1( x 2) n 1 ( x 1)8. Show thatd n log x ( 1) n n! 1 1 1 log x 1 n n 12 3 n dx x x Solution : Let y log x11 log x. and let u log x, v xxxWe have Leibnitz theorem,(uv)n uvn nC1 u1v n 1 nC2 u 2 v n 2 . u n vNow, u log x (1)( 1) n 1 (n 1)! un xn( 1) n n!1v vn x n 1xUsing these in (1) by taking appropriate values for n we get,( 1) n n!1 ( 1) n 1 (n 1)! log x Dn log x. n 1 n .xxxn x n(n 1) 1 ( 1) n 2 (n 2)! 1. 2 x 2 x n 1( 1) n 1 (n 1)! 1.xxn . Ie. log x ( 1) n n! ( 1) n 1 n! x n 1x n 1( 1) n 2 n!( 1) n 1 (n 1)! . x n 12 x n 1( 1) n 2 n! ( 1) 2( 1) 1 (n 1)! 1 . log x( 1) x n 1 2n1 Note : (-1)-1 11 1; ( 1) 2 1 1( 1) 210 MAT1113

Engineering Mathematics – IAlso Dr. V. Lokesha2011(n 1)! (n 1)! 1 n!n (n 1)! n1 11 d n log x ( 1) n n! log x 1 . n n 1 2 3n dx x x 9. If yn Dn (xn logx)Prove that yn n yn-1 (n-1)! and hence deduce that1 11 yn n log x 1 . 2 3n Solution : yn Dn(xn log x) Dn-1 {D (xn log x} n 1 n 1 Dn-1 x . nx log x x Dn-1(xn-1) nDn-1 (xn-1 log x} yn (n-1)! nyn-1. This proves the first part.Now Putting the values for n 1, 2, 3.we gety1 0! 1 y0 1 log x 1! (log x 1 )y2 l! 2y1 l 2 (l log x)1 ie., y2 21og x 3 2(log x 3/2) 2! log x 1 2 y3 2! 3y2 2 3(2 log x 3)1 1 ie., y3 61og x ll 6 (log x ll/6) 3! log x 1 2 3 .1 11 y n n! log x 1 . n 2 3 10. If y a cos (log x) b sin ( log x), show thatx2y2 xy1 y 0. Then apply Leibnitz theorem to differentiate this result n times.orIf y a cos (log x) b sin (log x ), show thatx2yn 2 (2n l)xyn l (n2 1)yn 0.[July-03]10 MAT1114

Engineering Mathematics – IDr. V. Lokesha2011Solution : y a cos (log x) b sin (log x)Differentiate w.r.t x y1 -a sin (log x)11 b cos (log x).xx(we avoid quotient rule to find y2) . xy1 - a sin (log x) b cos (log x)Differentiating again w.r.t x we have,xy2 1 y1 - a cos (log x) b sin ( log x)or1xx2y2 xy1 - [ a cos (log x) b sin (log x) ] -y x2y2 xy1 y 0Now we have to differentiate this result n times.ie., Dn (x2y2) Dn (xy1) Dn (y) 0We have to employ Leibnitz theoreom for the first two terms.Hence we have, 2 n n(n 1)n 1. 2 . D n 2 ( y 2 ) ) x . D ( y 2 ) n. 2 x. D ( y 2 ) 1. 2 {x. Dn}( y1 ) n. 1 . D n 1 ( y1 ) y n 0ie., {x2yn 2 2n x yn 1 n (n – 1)yn} {xyn 1 nyn} yn 0ie., x2yn 2 2n x yn 1 n2yn - nyn xyn 1 nyn yn 0ie., x2yn 2 (2n l)xy n l (n2 l)yn 011. If cos-1 (y/b ) log (x/n)n, then show thatx2yn 2 (2n l) xy n l 2n2yn 0Solution :By data, cos-1 (y/b) n log (x/n) log(am) m log a y cos [n log (x/n )]bor y b . cos [ n log (x/n)]Differentiating w.r.t x we get,10 MAT1115

Engineering Mathematics – Iy1 -b sin [n log (x/n)] n Dr. V. Lokesha201111 (x / n ) nor xy1 - n b sin [n log (x/n )]Differentiating w.r.t x again we get,xy2 1. y1 - n . b cos [ n log (x/n )] n11.( x / n) nor x (xy2 y1) n2b cos [n log (x/n) ] -n2y, by using (1).or x2y2 xyl n2y 0Differentiating each term n times we have,D(x2y2) Dn(xy1) n2Dn (y) 0Applying Leibnitz theorem to the product terms we have, 2 n(n 1). 2 . yn x y n 2 n. 2 x. y n 1 1. 2 2 {xy n 1 n. 1 . y n } n y n 0ie x2yn 2 2 x yn 1 n2yn xy n 1 nyn n2yn 0or x2 yn 2 (2n l) xyn 1 2n2yn 012. If y sin( log (x2 2 x 1)),or[Feb-03]If sin-1 y 2 log (x 1), show that(x l)2yn 2 (2n 1)(x 1)yn l (n2 4)yn 0Solution : By data y sin log (x2 2 x 1 ) y1 cos log (x2 2 x 1)ie., y1 cos log (x2 2 x 1)12x 2( x 1) 212 (x 1)x 2x 122 cos log( x 2 2 x 1 )ie., y1 ( x 1)or (x 1) y1 2 cos log (x2 2 x 1 )Differentiating w.r.t x again we get10 MAT1116

Engineering Mathematics – IDr. V. Lokesha(x 1)y2 1 y1 -2 sin log (x2 2x 1)20111. 2( x 1)( x 1) 2or (x 1)2y2 (x 1) y1 -4yor (x l )2y2 (x l) y1 4y 0 ,Differentiating each term n times we have,Dn [(x 1)2y2] Dn [(x 1)y1] Dn [y] 0Applying Leibnitz theorem to the product terms we have, n(n 1)2.2 . y n ( x 1) y n 2 n. 2( x 1). y n 1 1. 2 {(x l) yn 1 n. 1 .yn} 4yn 0ie.,(x l)2yn 2 2n (x 1)yn 1 n2yn-nyn (x l)yn l nyn 4yn 0ie.,(x l)2yn 2 (2n l) (x l) y n 1 (n2 4)yn 0)(13. If log x 1 x 2 prove that(1 x2) yn 2 (2n 1) xyn 1 n2yn 0( By data, y log x 1 x 2 y1 Ie., y1or) 11 .2x ( x 1 x 2 ) 2 1 x 2 11 x2 x1( x 1 x )21 x2 11 x21 x 2 y1 1Differentiating w.r.t.x again we get1 x 2 y2 12 1 x 2 ).2 x. y1 0or (1 x2)y2 xy1 0Now Dn [(l x2)y2] Dn[xy1] 0Applying Leibnitz theorem to each term we get,10 MAT1117

Engineering Mathematics – IDr. V. Lokesha2011 n(n 1)2.2 . y n (1 x ) y n 2 n. 2 x . y n 1 1 .2 [x . yn 1 n .1 yn] 0Ie., (1 x2) yn 2 2 n x yn 1 n2yn – nyn xyn l nyn 0or (l x2)yn 2 (2n l)xyn 1 n2yn 014. If x sin t and y cos mt, prove that(l-x2)yn 2-(2n 1)xyn l (m2-n2)yn 0.[Feb-04]Solution : By data x sin t and y cos mtx sin t t sin-1 x and y cos mt becomesy cos [ m sin-1x)Differentiating w.r.t.x we getmy1 - sin (m sin-1x)1 x2or 1 x 2 y1 - m sin (m sin-1x)Differentiating again w.r.f .x we get,1 x 2 y2 12 1 x2( 2 x ) y1 m cos (m sin 1 x ).m1 x2or (1 -x2)y2-xyl -m2yor (1 -x2)y2 –xy1 m2y 0Thus (1-x2)yn 2-(2n 1)xyn 1 (m2-n2)yn 015. If x tan ( log y), find the value of(l x2)yn 1 (2nx-l) yn n(n-1)yn-1[July-04]Solution : By data x tan(log y) tan-1 x log y or y etan-1 x Since the desired relation involvesyn 1, yn and yn-1 we can find y1 and differentiate n times the result associated with y1 and y.Consider y e tan 1x y. e tan 1x 11 x2or (1 x2)y1 yDifferentiating n times we have10 MAT1118

Engineering Mathematics – IDr. V. Lokesha2011Dn[(l x2)y1] Dn[y]Anplying Leibnitz theorem onto L.H.S, we have,{(l x2)Dn(y1) n .2x .Dn-1 (y1) n( n 1).2 .D n 2 ( y1 )} y n1 .2Ie., (1 x2)yn 1 2n x yn n (n-1) yn-1-yn 0Or (l x2)yn 1 (2nx-l)yn n(n-l)yn-1 010 MAT1119

Engineering Mathematics – IDr. V. Lokesha2011Continuity & DifferentiabilitySome Fundamental DefinitionsA function f (x) is defined in the interval I, then it is said to be continuous at a point x af ( x) f (a)if limx af ( x h) f (a) f '(a) existsa IA function f (x) is said to be differentiable at x a if limh 0hEx : Consider a function f (x) is defined in the interval [-1,1] by f (x) xx x 1 x 00 x 1It is continuous at x 0But not differentiable at x 0Note : If a function f (x) is differentiable then it is continuous, but converse need not be true.Geometrically :(1) If f (x) is Continuous at x a means, f (x) has no breaks or jumps at the point x aEx : 1f (x) x 1 x 00 x 1Is discontinuous at x 0(2) If f (x) is differentiable at x a means, the graph of f (x) has a unique tangent at the point or graphis smooth at x a1. Give the definitions of Continuity & Differentiability:Solution: A function f (x) is said to be continuous at x a, if corresponding to an arbitrary positivenumber ε, however small, their exists another positive number δ such that. f (x) – f (a) ε, where x - a δIt is clear from the above definition that a function f (x) is continuous at a point ‘a’.If (i) it exists at x a(ii) Lt f (x) f (a)x ai.e, limiting value of the function at x a is to the value of the function at x a10 MAT1120

Engineering Mathematics – IDr. V. Lokesha2011Differentiability:A function f (x) is said to be differentiable in the interval (a, b), if it is differentiable atevery point in the interval.In Case [a,b] the function should posses derivatives at every point and at the end points a & b i.e., Rf1(a) and Lf1 (a) exists.2. State Rolle’s Theorem with Geometric Interpretation.Statement: Let f (x) be a function is defined on [a,b] & it satisfies the following Conditions.(i)f (x) is continuous in [a,b](ii)f (x) is differentiable in (a,b)(iii)f (a) f (b)Then there exists at least a point C (a,b), Here a b such that f1 ( c ) 0Proof:Geometrical Interpretation of Rolle’s Theorem:Y y f (x)PRAf(a)BQABSOx a c1 c2 c3 c4 x b f(a)x a x cf(b)x bLet us consider the graph of the function y f (x) in xy – plane. A (a,.f(a)) andB (b, f( b ) ) be the two points in the curve f (x) and a, b are the corresponding end points of A & Brespectively. Now, explained the conditions of Rolle’s theorem as follows.(i)f (x) is continuous function in [a,b], Because from figure without breaks or jumps inbetween A & B on y f (x).(ii)f (x) is a differentiable in (a,b), that means let us joining the points A & B, weget a line AB. Slope of the line AB 0 then a point C at P and also the tangent at P (or Q or R or S) isParallel to x –axis. Slope of the tangent at P (or Q or R or S) to be Zero even the curve y f (x) decreases orincreases, i.e., f (x) is Constant.10 MAT1121

Engineering Mathematics – IDr. V. Lokesha2011f1 (x) 0 f1 (c) 0(iii) The Slope of the line AB is equal to Zero, i.e., the line AB is parallel to x – axis. f (a) f (b)3. Verify Rolle’s Theorem for the function f (x) x2 – 4x 8 in the internal [1,3]Solution: We know that every Poly nominal is continuous and differentiable for all points and hence f(x) is continuous and differentiable in the internal [1,3].Also f (1) 1 – 4 8 5, f (3) 32 – 43 8 5Hence f (1) f (3)Thus f (x) satisfies all the conditions of the Rolle’s Theorem. Now f1 (x) 2x – 4 and f1 (x) 0 2x – 4 0 or x 2. Clearly 1 2 3. Hence there exists 2t (1,3) such that f1 (2) 0. This showsthat Rolle’s Theorem holds good for the given function f (x) in the given interval.4. Verify Rolle’s Theorem for the function f (x) x (x 3) e x2in the interval [-3, 0]Solution: Differentiating the given function W.r.t ‘x’ we get x 1 xf 1 ( x) ( x 2 3 x) e 2 (2 x 3)e 2 2 1 x ( x 2 x 6) e 221 f (x) exists (i.e finite) for all x and hence continuous for all x.Also f (-3) 0, f (0) 0 so that f (-3) f (0) so that f (-3) f (0). Thus f (x) satisfies all the conditions ofthe Rolle’s Theorem.Now, f1 (x) 01 x ( x 2 x 6) e 2 02Solving this equation we get x 3 or x -2Clearly –3 -2 0. Hence there exists –2 (-3,0) such that f1 (-2) 0This proves that Rolle’s Theorem is true for the given function.10 MAT1122

Engineering Mathematics – I5.Dr. V. Lokesha2011Verify the Rolle’s Theorem for the function Sin x in [-π, π]Solution: Let f (x) Sin xClearly Sinx is continuous for all x.Also f1 (x) Cos x exists for all x in (-π, π) and f (-π) Sin (-π) 0; f (π) Sin (π) 0 so that f (-π) f(π)Thus f (x) satisfies all the conditions of the Rolle’s Theorem .Now f1 (x) 0 Cos x 0 so thatX π2Both these values lie in (-π,π). These exists C f1Such that ( c ) 0π2Hence Rolle’s theorem is vertified.6. Discuss the applicability of Rolle’s Theorem for the function f (x) ІxІ in [-1,1].Solution: Now f (x) x x for 0 x 1-x for –1 x 0f (x) being a linear function is continuous for all x in [-1, 1]. f(x) is differentiable for all x in (1,1) except at x 0. Therefore Rolle’s Theorem does not hold good for the function f (x) in [-1,1].Graph of this function is shown in figure. From which we observe that we cannot draw a tangent to thecurve at any point in (-1,1) parallel to the x – axis.Y-1y x 01x10 MAT1123

Engineering Mathematics – IDr. V. Lokesha2011Exercise:7. Verify Rolle’s Theorem for the following functions in the given intervals.a) x2 – 6x 8 in [2,4]b) (x – a)3 (x – b)3 in [a,b] x 2 ab c) log in [a,b] ( a b) x 8. Find whether Rolle’s Theorem is applicable to the following functions. Justify youranswer.a) f (x) x – 1 in [0,2]b) f (x) tan x in [0, π] .9. State & prove Lagrange’s (1st) Mean Value Theorem with Geometric meaning.Statement: Let f (x) be a function of x such that(i) If is continuous in [a,b](ii) If is differentiable in (a,b)Then there exists atleast a point (or value) C (a,b) such that.f (b) f (a )f 1 (c ) b ai.e., f (b) f (a) (b – a) f1 (c)Proof:y[b,f(b)][a,f(a)]xacb10 MAT1124

Engineering Mathematics – IDr. V. Lokesha2011Define a function g (x) so that g (x) f (x) – Ax ---------- (1)Where A is a Constant to be determined.So that g (a) g (b)Now, g (a) f (a) – AaG (b) f (b) – Ab g (a) g (b) f (a) – Aa f (b) – Ab.f (b) f ( a )---------------- (2)b aNow, g (x) is continuous in [a,b] as rhs of (1) is continuous in [a,b]G(x) is differentiable in (a,b) as r.h.s of (1) is differentiable in (a,b).i.e., A Further g (a) g (b), because of the choice oif A.Thus g (x) satisfies the conditions of the Rolle’s Theorem. These exists a value x c sothat a c b at which g1 ( c ) 0 Differentiate (1) W.r.t ‘x’ we getg1 (x) f 1 (x) – A g1 ( c ) f1 ( c )- A ( x c) f 1 ( c ) - A 0 ( g1 ( c ) 0) f1 ( c ) A-------------- (3)From (2) and (3) we getf 1 (c ) f (b) f (a )(or) f (b) f (a) (b – a) f1 (c) For a c bb aCorollary: Put b – a hi.e., b a h and c a θ hWhere 0 θ 1Substituting in f (b) f (a) (b – a) f1 ( c ) f (a h) f (a) h f (

Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 8 2011 Leibnitz’s Theorem : It provides a useful formula for computing the nth derivative of a product of two functions. Statement : If u and v are any two functions of x with u n and v n as their nth derivative. Then the nth derivative of uv is (uv)n u0vn nC

Related Documents: