2.Problems on Leibnitz Theorem

April 2, 2018 | Author: Shubham | Category: Trigonometric Functions, Quantity, Mathematical Concepts, Algebra, Numbers


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Successive Differentiation Leibnitz Rule Solved Problems Leibnitz’s Rule :- Q} If y = uv y = x2excosx Find yn = ? Soln:Let u = excosx , v = x2  yn = exrn cos(x+n∈) x2 + nexrn-1 cos(x+(n-1)∈) (2x) + ?(?−?) ? ?−? ? ? cos(x+(n-2)∈) ×2 ?! where r = √? ∈ = ???−? (?) = ??⁄? Q} y = ???? ? P.t. y5 = ?! ? ? ? ? ? ? ? + + ? { 1+ ? + − logx} ? Soln:(−?)?−? (?−?)! Let u = logx un = ?? v = ?⁄? y5 = u5v + 5u4v1 + 10u3v2 + 10u2v3 +5u1v4 + uv5 ?! = ?? × ? ? ?? ? − ? ?×?! −? ?? ? = ?? { ? + ? ? ?? ×?! ?? ?? −??? ( ?? ) + ???? { ? ?! + ? ?? ? } + ? + ? + ? − ????} ? 1 −? −? (?? ) + ?? ( ?? ) ( ?? ) + Successive Differentiation Leibnitz Rule Solved Problems ?? Q} If In = ??? {?? ????} P.t In=nIn-1 + (n-1)! Also P.t In=n! {logx + 1 + ?⁄? +. . . .+ ?⁄?} ?? Soln :- In = ??? {?? ????} ??−? ? = ???−? {?? (?? . ????)} ??−? ?? = ???−? { ? + (????)? ??−? } ??−? ??−? = ???−? {? + ? ???−? (??−? . ????) In= (? − ?)? + ???−? ??  ?! = ?−? } ? + ? ? (?−?)! ??−? Put n= 1,2,3,…. ?? ? = ? + ?? ?! (?) ?! ??−? (?−?) ? = (?−?) + ? (?−?)! ?? ?! ? ? ? ?! = + (??) ?? ??−? (? − ?) Adding ?? + ?! ?? + ?! ??−? (?−?)! ?? ? + ?? + ……+ ?! ??−? (?−?)! ??−? (?−?)! + ?? ?! = ?? + ?! ?? ?! + + ? ? ? ? ? + + +⋯+ + ? ? ?−? ? ? ? ? ? = ?!? + ? + ? + ? + ⋯ + ? ? ? ? = logx +? + ? + ? + ⋯ + ? 2 ?? ?! +⋯+ Successive Differentiation Leibnitz Rule Solved Problems Q) If y = ?? ???? ?! ??+? = ? Soln :y = ?? ???? ?? ?? = ? + ???−? ???? ??? = ?? + ??? ???? i.e. ??? = ?? + ?? Using leibintz theorem {??+? ? + ??? } = n! + ???  ??+? ? = ?! ?! ??+? = ? Q) If x+y = 1 ?? P.t. ??? (?? ?? ) = ?! {?? − (nC1)2??−? ? + (nC2)2??−? + … … + (nCn)2?? (−?)? } Let u = ?? v = ?? = (? − ?)? ?? = ?(? − ?)?−? (−?) = ???−? (−?) ?? = ?(? − ?)(? − ?)?−? (−?)? = ?(? − ?)??−? (−?)? ?? = ?! (−?)? ?? = ???−? ?? = ?(? − ?) ??−? 3 Successive Differentiation Leibnitz Rule Solved Problems  ?? = ?! ?? (?? . ?? ) = (nC0) n! ?? + (nC1)n! ????−? (−?) + (nC2) ??? ?! ? ? ? (−?)? ?(? − ?)??−? + … . . + (nCn)?? n!(−?)? = n!{?? −(nC1)2??−? ? +( nC2)2?? ??−? + … … + ?? (−?)? (nCn)2} Q) If ? = ???? ?. ?. ?? (?) − nC2??−? (?)+ nC4??−? (?) + … … = y= ????? ? ???? ????  ycosx = sinx By leibintz Rule ?? (?)???? + nC1 ??+? (?)(−????) + nC2??−? (?)(−????) + n C3??−? (?)(????) + n C4??−? (?)(????) + ⋯ = ???(? + ?? ? ) Put x = 0  (?) − nC2??−? (?)+ nC4??−? (?) + … … = 4 ????? ? Successive Differentiation Leibnitz Rule Solved Problems ?+? Q) If y = √?−? ?? = (? − ?? )?? Hence P.T (? − ?? )?? − (?(? − ?)? + ?)??−? − (? − ?)(? − ?)??−? = ? Soln:- ?? = (?−?)? ? 2y?? = = ?−? (?−?)(?)−(?+?)(−?) 2y?? = ?? = ?+? (?−?)? ? (?−?)? ? = ? (?−?)? ?? ? (?−?)? ( ?? = ?+? ) ?−? ? (?−?)? y = (? − ?? ) ?? i.e (? − ?? )?? − ? = ? By leibnitz rule taking (n-1)th derivative = { ?? (? − ?? ) + (? − ?)??−? (−??) + (?−?)(?−?) ?! ??−? (−?) − ??−? } = ? (? − ?? )?? − ?(? + ? + ?)??−? − (? − ?)(? − ?)??−? = ? 5 Successive Differentiation Leibnitz Rule Solved Problems Q) If y = cos(m???−? ?) P.T. (? − ?? )??−? − ?(?? + ?)??−? + (?? − ?? )?? = ? Soln:?? = sin(m???−? ?) ? √?−?? ?? (√? − ?? ) = −????(m???−? ?) ?? (√? − ?? ) + ?? (−??) √?−?? = −m cos(m???−? ?) ? √?−??  ?? (? − ?? ) − ??? = −?? ? ?? (? − ?? ) − ? ?? + ?? ? = ? By leibint’s theorem {??+? (? − ?? ) + ???+? (−??) + { ??+? ? + ??? } + ?? ?? = ? ?(?−?) ? ?? (−?)} −  ??+? (? − ?? ) + ??+? (−??? − ?) +(?? − ?? ) ?? = ?  ??+? (? − ?? ) − (?? + ?)???+? + (?? − ?? ) ?? = ? Q) If ? ?⁄ ? + ?− ?⁄ ? = ?? P.T (?? − ?)??+? + (?? + ?)???+? +(?? − ?? ) ?? = ? Soln:? ?⁄ ? + ?− ?⁄ ? = ?? 6 Successive Differentiation Leibnitz Rule (? ?⁄ ? ?) + ? = ??? (? ?⁄ ? ?) − ??. ? ?  ? ⁄? = ?⁄ ? Solved Problems ?⁄ ? + ? =0 ?? ± √??? −? ? = ? ± √?? − ? y = (? ± √?? − ?)? ?? = ?(? ± √?? − ?)?−? {? + ?−?  ?? = ?(? ± √?? − ?) { ? √?? −? } ?+√?? −? } √?? −? ? √??  ?? = √?? − ? = ?(? ± − ?)  ?? (?? − ?) + ??? − ?? ? = ? Using Leibnitz theorem {??+? (?? − ?) + ???+? (??) + ??? } − ?? ? = ? ?(?−?) ? ?? (?)} + {??+? ? +  (?? − ?)??+? + (2n+1)x??+? + (?? − ?? )?? = ? Q) If x = cos∈ ∈ = ?⁄? ???? (1-?? )??+? − (?? + ?)???+? − (?? + ?? )?? = ? logy = m∈ y = ??∈ 7 Successive Differentiation  y = ?? ??? ?? = ?? ??? Leibnitz Rule Solved Problems −? ? −? ? { −? √?−?? }  ?? √? − ?? = − ?? ? ?? (−??) √?−?? + ?? √? − ?? = − ???  ?? (? − ?? ) − ??? − ?? ? = ? Q) If y = ???−? ? ? P.T ? (? + ? )+(2n+3)x??+? + (? + ?+? ? √?+? ?)? ?? = ? y(√? + ?? ) = ???(? + √?? + ?) Take 2 derivatives and apply Leibnitz rule ?? (√? + ?? ) + ? ?? 8 Successive Differentiation Leibnitz Rule Solved Problems Q) If u = f(x) y = ??? ? P.T ?? ? = ??? (? + ?)? ? ?? ?−? ? ?? n ?? ? = nC0??? ??? + nC1??−? ? (?. ? ) + C2?? (? . ? ) + … + nCn ? (?? . ??? ) = ??? { nC0 ?? + nC1 ??−? ? + nC2 ??−? ?+ … + nCn ??−? ?? } ? = ??? (? + ?)? ? (1) If y = x2 – e2x P.T. yn = 2m–2 m(n – 1) Soln.: = x2 e2x y By Leibnitz's theorem, Note : Take ‘v’as that function whose derivative vanish after some derivative. = (x2e2x)m ym = (e2x)m x2 + (e2x)m–1 (x2)1 + … = 2me2x x2 + m2 m–1 e2x x+ m(m  1) 2 2n–2 ex 2+0  ym (0) = 0 + 0 + m(m  1) 2 9 2m–2, e0 2 Successive Differentiation  Leibnitz Rule Solved Problems = 2m–2 m (n – 1) ym (0) (2) If y = x log (x + 1) prove that : ym = (1) Soln.: m 2 y (m  2 )! ( x  m ) ( x  1 )m = x log(x  1) By Leibnitz's theorem, = x log(x  1) ym m ym = log(x  1) ym = m ym (1)m 1 (m  1)! m (1)m 2 (m  2)! x  1  0 (x  1)n (x  1)n1 = ym  x  m  log(x  1)m 1 (x )1  .... (1)m  2 (m  2)! (1)(m  1)x  m (x  1) (x  1)m = 10 (1)m  2 (m  2)!(x  m ) (x  1)m Successive Differentiation Leibnitz Rule (3) If y = xm log x P.T. ym+1 = Soln.:  Solved Problems m! x y = xm log x 1 x y1 = mxm–1 logx + xm Note : Since we want ym+1 so first find y1 and then take the mth derivative.  xy1 = m xm log x + xm  xy1 = my + xm Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have,  (x, y1)m = (ny)m + (xm)m  ym+1 x + m ym 1 = mym + m !  ym+1 = m! x (4) If f () = cot  show that n Soln.: f ()  cot   c1  f n 1 ()  nc 3 f n  3 ()  nc 5 f n  5 ()....  cos n 2 cos  sin  f () sin  = cos  Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have, f () sin   cos  n n f n () sin   nc f n 1 () cos   nc f n  2 ()( sin ) 1 2 11 Successive Differentiation Leibnitz Rule Solved Problems n    nc f n  3 ()( cos )  ..  cos    2   3 put  = 0, 0  nc1 f n 1 (0)  0  nc 3 f n 3 (0)...  cos  n 2 n 2 nc1 f n1 ()  n(3 f n3 ()....  cos (5) If w = tan (logy)or y = etan–1(w) prove that : (1 + w2)yn+1 + (2nw – 1)yn n (n – 1) yn–1 = 0 y = etan–1 (w) Soln.:  y1 = etan–1(w)  (1 + w2)y1 = y 1 1  w2 = y 1  w2 Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have,  (1  w)y1 n  y n  y n 1 (1  w 2 )  n  y n (2w)   1  w 2 )y n1  (2nw  1)y n  n(n  1)yn  1  (6) If x  sin , y  sin 2  prove Soln.: y1 =  that (1  x y  sin 2  2 xin  cos  y= Note : n(n  1) y n 1 (2)  0  y n 2 2x 1  x 2 2 1  x 2  2x cos  = (2 x ) 2 1 x2 1  sin 2  1  x 2 y1  2(1  x 2 )  2 x 2 12 2 ) y n  2  (2 n  1 ) xy n  1  (n 2  4 ) y n  0 Successive Differentiation  Leibnitz Rule Solved Problems 1  x 2 y1  2  4 x 2 on differentiating w.r.t. x, y 2 1  x 2  y1 2  8 x 2 1  x2  (1  x 2 )y 2  xy 1  8 x 1  x 2  (1  x 2 )y 2  xy 1  4 y  (1  x 2 )y 2  xy 1  4 y  0 Now on differentiating ‘n’ times by using Leibnitz’s theorem we have,  (1  x  1  x y n 2  n(2 x )y n1   1  x y n 2  2nxy n1  n(n  1)yn  xy n1  nyn  4 yn  0  1  x y n2  (2n  1)xy n1  [n 2  n  n  4 ]yn  0  1  x y n2  (2n  1)xy n1  (n 2  4 )yn  0 2  )y 2 n  xy 1 n  4 y n  0 2 2 2 2 n(n  1) (2)  yn  xy n1  2ny n  4 y n  0 2 13 Successive Differentiation Leibnitz Rule Solved Problems (7) If y = A cos (log x) + B sin (log x)then show that x2yn+2 + (2n +1)xy constant. Soln.: y= n+1+ (n2 +1)n = 0 where A and B are A cos(log x )  B sin(wgx ) y1 = A sin(log x )  B cos(log x ) x xy1 = x A sin(log x )  B cos(log x ) on differentiating w.r.t. x,  Aws (log x ) B sin(log x )  x x  xy2 + y1 =  x2y2 + xy1 = Aws (log x )  BSin (log x )  x2y2 + xy1 = –y … By (1) Now on differentiating ‘n’ times by using Leibnitz’s theorem we have, x y   x, y    y  2 2 n n x 2 y n 2  n(2 x )y n1  n n(n  1)  2 y n1  ny n  y n 2  x 2 y n2  (2n  1)xy n1  n(n  1)  n  1y n  0  xy n 2  (2n  1)xy n1  (n 2  1)y n  0 (8) If y = log( x  x 2  1 ) prove that y2m (0) = 0 and y2m+1 (0) = (–1)m, 12, 32, 52 … (2m – 1)2 Soln.: y  log(x  x 2  1 ) y1  y1    2x 1   2 x  x 1  2 x 1 1 2 1 x2 1 14 Successive Differentiation Leibnitz Rule Solved Problems x 2  1 y1  1 on differentiating w.r.t. ‘x’ we have, x 2  1 y 2  y1 2x 2 x2 1 =0 (x 2  1)y 2  xy 1  0 Now on differentiating ‘m’ times by using Leibnitz’s theorem we have, (x x at  1)y m  2  m (2 x )y m 1  2 2 m (m  1)  2 y m  xy m 1  my m  0 2   1 ym  2  (2m  1)xy n1  m 2 y m  0 x = 0, y (0) = 0, y1(0)= 1, y2(0)= 0 ym+2(0) = –m2ym(0) To find y3 (0)Put m = 1 in y12 (0)  y 3 (0)  1 2  y1 (10 )  (1)1  1 2 m  2, y 2 2 (0)  y 4 (0)  2 2 , y 2 (0)  0 m  3  y 3  2 (0)  y 5 (0)  3  y 3 (0)  3 2  1 2  (1)2  1 2  3 2 = (1) 2  1 2 (2  2  1)2 ) y m(0)  0 2 and (9) If y 2 m  1  (1)m  1 2 ,3 2....( 2m  1)2 or if p =sin–1x = sin–1y prove that  (2 n  1 ) xy  ( p  n ) y  0 and hence de duce yn(0) = y  sin( p sin 1 x ) (1  x 2 ) y n  2 2 n 1 2 n 0 if ‘n’ is even and yn(0) = ….(32 – p2)p (n–2)2 – p2 if ‘n’ is add where ‘p’ is any constant. Soln.: y  sin( p sin 1 x ) 15 Successive Differentiation Leibnitz Rule y1  cos( p sin 1 x )  Solved Problems p 1  x2 1  x 2 y 1  p(p sin 1 x ) [ cos2 + sin2 = (1  x 2 )y 2  p 2 (1  y 2 ) 1] Now on differentiating ‘n’ times by using Leibnitz’s theorem we have, (1  x 2 )2 y1 y 2  y1 (2 x )  p 2 (2 yy 1 )(1  x 2 )y 2  xy 1   p 2 y 2 (1  x 2 )y n2  ny n1 (2 x )  n(n  1)  y n (2)  y n1 , x  ny n   p 2 y n 2 (1  x 2 )yn  2  (2 x  1)xy n1  (p 2  n 2 )yn  0 at x = 0, y (o) = sin (p sin–1(o) = sin (p y1 (o) = cos (psin–1x p 1  02 y 2 (o)   p 2 y(0)  0 y n  2 (o )  (n 2  p 2 )yn(o ) If n=1, y 3 (o)  (12  p 2 )y, (o)  (12  p 2 )p n=2, y 4 (o)  (2 2  p 2 )y 2 (o)  (2 2  p 2 )o  0 n=3, y 5 (o)  (3 2  p 2 )y 3 (o)  (3 2  p 2 )(12  p 2 )p 16 o) = o =p Successive Differentiation Leibnitz Rule Solved Problems yn (o)  o   yn (o)  (n  2)2  p 2 ....( 3 2  p 2 )(12  p 2 )p (10) If y p  x p  x   tan–1 = ( p 2  x 2 ) yn  2  2 (n  1 ) xy n 1  n(n  1 ) y n  prove that where ‘p’ is any constant. p  x y  tan 1   p  x Soln.: put x = p tan  y=  p  p tan   1  tan   tan 1   tan 1   1  tan    p  p tan    y=    tan 1  tan       4 y   x     tan 1 ( ) 4 4 p 1 1 p   2 2 x p p  x2 1 2 p on differentiating w.r.t. ‘x’ we have,  (p2 +x2)y2 + y1 (2x)= 0 Now on differentiating ‘n’ times by wing Leibintz’s theorem we have, (p 2  x 2 )y 2   2 xy  n 1 n 0 n(n  1)  2  yn  2  x  y n1  n  2 y n  0 2  ( p 2  x 2 )y n 2 (2 x )  y n 2   ( p 2  x 2 )y n  2  2(n  1)x  y n 1  [n(n  1)  2n]y n  0  ( p 2  x 2 )y n2  2(n  1)xy n1  n(n  1)y n  0 17
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