2.Problems on Leibnitz Theorem
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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)n1 = 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 n1 () n(3 f n3 ().... 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 n1 (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 n1 1 x y n 2 2nxy n1 n(n 1)yn xy n1 nyn 4 yn 0 1 x y n2 (2n 1)xy n1 [n 2 n n 4 ]yn 0 1 x y n2 (2n 1)xy n1 (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 n1 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 n1 n n(n 1) 2 y n1 ny n y n 2 x 2 y n2 (2n 1)xy n1 n(n 1) n 1y n 0 xy n 2 (2n 1)xy n1 (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 n1 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 y12 (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 n2 ny n1 (2 x ) n(n 1) y n (2) y n1 , x ny n p 2 y n 2 (1 x 2 )yn 2 (2 x 1)xy n1 (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 n1 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 n2 2(n 1)xy n1 n(n 1)y n 0 17
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