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Published in: Mathematics
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DIFFERENTIATION, SUCCESSIVE DIFFERENTIATION

Hari B / Hyderabad

9 years of teaching experience

Qualification: M.Sc (jntuh campus - 2012)

Teaches: Algebra, Business Mathematics, Mathematics, Statistics, B.Tech Tuition, Polytechnic

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  1. 1. sol: 2. Sol: 11. 1. Sol: SUCCESSIVE DIFFERENTIATION -2 EXERCISE -- 7 (B) Find D (sin x. log x) when x = E Y = sin x. log x D sin x. log x = 3Co sin x. log x + D 2 sin x.D log x + 3c2 D sin x.D2 log x +3 c sin x.D3 log x. =l.sin x +3— . log x— 3 sin x.—+ 3 cos x. —#1. sin x. — .12 — 3 cos x..+ sin x.— = —cos x. log x— 3 sin x 0 3.1.2_0+1. 16 16 6 (D 3 sin x. log x) atx= Find the sum of the middle terms is Leibnitz expansion of D (ex cos x D (ex COSX) = 3Co (e .cosx+3q (ex) .Dcosx+3c2 (ex) D cosx+3qeXD3cosx=0 Middle terms are 2nd and 3rd terms. Sum of the middle terms =3q (ex) .Dcosx+3c2 (ex) D2cosx = 3 ex (—sinx)+3e ( x —cosx = —3ex (sin X + COS X) Find the nth derivatives of the following using Leibnitz theorem i) e2X.cos x ii) e2X.sin x.cos5x iii) x3 .eax iv) x2 cos 2x (i) let y = e2x cos2 x = —(1+ cos 2x) —(e2x + eh. cos 2X) Differentiate n times w.r.t. x, —(e2x + eh. cos 2X) dxn 2 - (Y.e2x .e2x tan-I l)) 2 n 1 +2 . COS 2X-k— 3n/2
  2. y = e2X.cos 5x.sin x = -L (2 cos 5x. sin x) = e— (sin 6x — sin 4x) = — ( 1 e x sin 6x—e 2 x. sin 4x) 2 2 Differentiate n times w.r.t. x, — e2x sin 6x—e2X .sin4x dxn 2 4+16) e2x sin (6x+n tan —(4(Y 2. sin (6x+ n tan 13) —2(Y 2 3 ax = ncoDneaxx3 +nqDl leax.Dx +nc2D 6n(n —1) = aneaxx3 +3nx .a .e + 2 ax 2 n X.an + 2 4+16 e sin 4x+ntan 2) iii) iv) v) sol: sin (4x+ntan 12) n—3 ax 3 3 e .Dx +nc3D e D x +0 n—2 ax 3 2 ax . x.a .e x +3nx .a .e + 6 an 3 eax 6 an F +3nFd-l Y = x cos 2x yn = IYx2.cos 2X = ncol)n cos2x.x + nqIY I cos 2x.Dx + nc21Y 2 cos 2x.D2x2 +0 = x2.2n COS 2x+n— +2nxcos 2x+n—1— 2n I + 2. cos 2 nTt = Y 1 2x cos 2x+— 2 ex. log 2X Let y = ex. log 2x ) . log 2x +nq.(ex ) . log 2 x +nq. e 1 n(n—l) = ex log2x+neX.—+ 2 x -1 —ex log 2X+—+ 2 2 x x 2 2 cos 2x+(n—2)— 2 .D log 2x + ne. 1 .—.2 +11c2. e n—1 2 X -1 2 D2 log 2x + O + -1 2 x x 2 .... +nc .eXI)n log 2x (-11 1 n—l)! x 2 x n-l (n—l)! x
  3. 2. Find the nth derivative of x2. cos2 (2x) for n > 2 . 2 1+ cos4x 1 -( x 2 + x 2 cos 4x) Sol: Let y=x2.cos2 (2x) = x 2 2 Differentiate n times using leitbnitz's theorem, 1 —(Dnx2 + Dnx2. cos 4x) 2 1 .x2 +n Cl (cos 4x)n D (x2)+n C2 (cos 4x)n D 2 1 [0+ (COS 4x)n .X2 +n Cl (COS4X) (2x) 1 nit — x2.4n COS 4X+— + 2nx.4n I COS 4X+ 2 2 2 no-I) C2 (cos 4x) 2n(n—1) cos 4n 2 COS 2 2 2 = 411 2 8x2 cos 4x+— -k 4nx cos 4x 2 2 3. Find the nth derivative of e 2x+3 . sin (2x) . 2x+3 Sol: let u = sin 2x,v=e n—r Ans: Encr.Y.sin 2x+— 2x+3 2 r=() 4. If f(x) = tan I (x) then show that (1+x2)fn f n Sol: Let y = tan x 1 (1+X2 Then = 2 Diff. w.r.t. x, (1+x2)+ 2XY1 = 0 ) +2XY1 = 0 Now diff. n times using Leibnitz's theorem, D [(1+x2)Y21+2.D Dn CPD n 1 (Y2) 2x+n q.1Y 2 (Yl )x+'l q.D'l I (Yl +0] 2
  4. 2n(n —1) (l -k X y n +2 -k 2xn.yn+l + yn + 2xyn+1 + 2nyn 2 (1 + X 2) y n +2 4-2x(n+1) y n +1 + (n + n) yn = O (I -k X 2 ) y n +2 + 2X(n +1) y n +1 + (n y n = O 2 1 (x) 4-11 (n +1) f n (x) If y = xn log x then show that xyn+l = [U • 5. sol: 111. 1. Sol: Y = xn log x Diff. w.r.t. x 1 (1) + n.xn I log x x Multiplying with x, we get xyl = xn + nxn log x XY'I = xn + usisng Leibnitz theorem Diff. n times w.r.t. x D (xyl) = Dnxn + nDy ncpn yrx + ncpn lyrDx+0 = n!+ ny xyn+l + nyn = n !+ ny XYn+l = n ! log x x > 0) then show that yn x log x _ 1 log x x x Diff. n times, 1 Y'n = Dn —.logx, let u=— and v= logx log x n 4-1 x 1 1 2 x x 1 1 log x + nq x x n—l 01-1) 1 1 x x n—2 2 x 2 + — (log x) n x (n-2)! . log x + x 1 1 1 2 x x n—l)! x x 1 x 1 n x log x —1 —— 2 x
  5. If y=cos 1m log xl , x > 0 , then show that x2Y2 + xyl +m y = 0 and hence deduce that 2. X yn+2 +112) y n = 0 Sol: y = cos (m log x ) 3. m = —sin (m log x). x xyl = —m. sin (m log x) Differentiating w.r.to x x. Y2 + h. 1 = —m cos (m log x). x x + xyl =—m cos(mlogx) = —m y x + xyl +m y = 0 Differentiating n times w.r.to x, using Leibnitz's theorem (x2Y2)n + (xyl)n +m2yn = 0 2 2 +1 (Y'l)n x+nCl y 2.yn +(n2 —n+n+m2 y n = 0 X .yn+2 + 2n.xyn+1 + 2 X 2 Jin +2 +1) -k 11'12 -1-112) y n = O Sinh x then show that (1+x2 +3XYl + y = 0 and hence deduce the If Y - (1+ x2) +(2n +3) xyn+l + (n +1) 2 yn = 0 Sinh x sol: Y - I—x — y I—x2 = sinh x Diff. w.r.t x, 1 .2x = 2 2 (1+ x2) + xy = 1------(2) Diff. w.r.t.x, +2XY'1 + xyl+y=0 (1+x2 + 3xY1+y 0 Differentiating n times w.r.to x [(1+x2)Y2 yn = 0 1 2
  6. +nC1 ) n 2x+nC1 (Y2)n ) n .x+nC1 n .11 -k y n = O (1 y n +2 -k 2nxyn+1 -k yn 2 +3xyn+1 +3.n.yn + yn = 0 2 x , x > 0 , then show that x2Y2 + xyl +n y = 0, and hence deduce 4. If cos — = n. log b n that X 2 y n +2 + 1) X)' n I + 2n2yn = O • Sol. Given cos-l = n log— cos = n (log x — log n) b n Differentiating w.r to x , 1 1 n x 1- 192 —Y2 n x 5. x 2.Y12 = n 2 (b2 — Differentiating w.r.to x, x2 (2YIY2) + Y12.2x = n2 (0 — 2 yyl ) h -k x Y24-XYl+n y = O Differentiating n times w.r to x using leibnitz's theorem (x2Y'2)n +n2yn = 0 X2)1 +nC2 (h) n 2 .x+nq (h) n .Dx]+n2 (y)n =0 y 2 ) n .x2 -k nC1 ( Y2 (2X) +nC2 (Y2)n 2.2+1 (Yl ) n .x+ÆC1 (h) n 1 .ll+n y n = 0 ) n . X 2 -k nC1 (Y2)n 1 n(n—l) X .yn+2 -k 2nxyn+1 -k yn.2+ xyn+l +nyn + n .yn = 0 2 X y n +2 -k X)' n +1 -k 2n2yn = O , then show that (I—x2 —xyl —y = 0 and hence deduce that If x 2) —(2n+1)xy 1 Sol: y =esm x Diff w.r.t. x, sin x 2 I—x 2 I—x
  7. Differentiating w.r to x 1 2 1-x2 (1 —x 2 —xyl = Yl 1 —x 2 = y (l — x2) Y 2 — XYI — Y O Differentiating n times w.r.to x using Leibnitz's theorem — (xyl — yn = 0 1—12) +nÇ.( (1—12) 1 +nÇ (1—12) 2 x+nq .18—yn = 0 (—2) — ) n . X + nq .l]-yn (—2x) + nC2 (Y2)n 2 (1—X2 yn+2 —2nxyn+l — .2yn — [xyn+l + = 0 2 (l — x2 Y 11+2 —n (n —1) y n — xyn+l —nyn — y n = 0 — 2nx. (n 2 —n+n+l) yn —(2n+1)xy 1 (1— X2 ) Yll+2 —(2n+1)xy 1 If y = ex log x, x > 0, then show that xy2 —(2x —1) + (x—l) y = 0 and hence deduce 6. that xyn+2 — (2x—1 —n) yn+l + (x—l— 2n) y n + nyn_l Sol: y = ex log x -----(1) Differentiating w.r.t. x, ex = ex — + log X.e x x XYI — XY e Differentiating w.r t. x, + — (xyl + y) = ex + (l —x) — y = xyl —xy —(2x—1) + = 0 using Leibnitz's Theorem Differentiate n times w.r t. x, (xy2)n —c (2x—1) + [ x— I +nÇ (Y'l)n ) n ) n 21 + ) n . X + IICI ( y 2 ) n I .1-
  8. X. y n +211 —2n.yn + (X —1) y n + n.yn_l + (nC2)2 yn 2.x2 7. Sol: — (2X— 1— n) yn+l (x—l — 2n) y n + n.yn_l xnyn) = n '(Y . n —(nq) xyn If x + y = 1, then prove that clxn Given x + y = 1 dx n—l Dyn = nyn— Yl = D2yn = Dnyn •YI — - Dnyn-l (-1) Now dn x n ) n y n +nq.(xn) Dyn +nÇ.(xn) D2yn + ....+ xn .D'I yn] clxn yn.(xn) +nq.(x n.yn n (n —l)n.yn dx = yn (n + nC1.n !.x.n.yn +nC2 —1) y n = yn (n !) + nq.n !.x.nq.yn-l (—1) + nÇ.n !.x2nqyn-2 + ...+ xn .n . n ) 2 .xyn + (nC2 ) 2 x n , then prove that yn+2 + 2xyn+1 +2(x+1) y n 8. If y=e- Sol: y = e (—2x) = —2xy + 2xy = O y 2 + 2 (XYI + y) using Leibnitz's Theorem Differentiate n times w.r t. x, .11 +2 y n = O yn+2 ) n .X+ nq.( + xyn+l + 2n.yn +2yn = 0 .... (-11]
  9. . yn+2 + 2xyn+1 + y n = O If y —1) n , then prove that (x2 9. Sol: y = (x 2 —1 = n(X2 —l) n 1 2X (X2 —1) Yl = 2nx(x2 —1) n = 2nxy (x 2 — = 2nxy Differentiating w.r.to x (x 2 —1) Y'2 +2XY1 = 2n (xyl + y) = 2nXY'1 + 2ny —n(n+l) y n —1 y n +2 + 2xy I (x 2 —1) +2(1 —n) xyl —2ny = 0----(3) using Leibnitz's Theorem ((x2 —1) Y) +20 —n) (xyl)n —2nyn = 0 Differentiate n times w.r t. x, (x2 —l) +nqyn+l —1) +nqyn.D2 (x2 ) n = O (x 2 —1) yn+2 + nqyn+12x + y n. 2 + 2 (1 — n) ( yn+1X + nq.yn ) — 2nyn 2 (x2 —1) + 2nxyn+1 + n (n —1) yn + —n) (xyn+l + xyn ) — 2nyn = 0 . —n(n+l) y n = 0 Y11+2 +2xy 1 m sin 10. If msln x Sol: y = e m sin x, then prove that (I—x Y 2 2) —(2n+1)xyn+1 x = my -----(2) Diff wrt x, 1 (—2x) = myl 1-x .Y2+Yl• 2 1-x2 (1—x2 —xyl = m. Yl 1 —x 2 •.•y'1 1—x2 = my (1 —x —xyl —m2y = 0 using Leibnitz's Theorem
  10. — x2) ) n —m2yn = 0 Differentiate n times w.r t. x, )n2.x+n (1 — X 2) +nq.(yn ) n 1 (—2x)+nC2 (Y2 2 —my = 0 n (I—x2 yn+2 —2nxyn+1 — (l — X2 Y Rl+2 Find the 4th derivative of 1. 2. 3. 4. 5. 6. sol: (-1) 7. yn.2—xyn+1 —nyn —m y n = 0 2 —(n2 —n+n+m2)yn = 0 n2+m2 yn = 0 . PROBLEMS FOR PRACTICE 1 2x+3 Find the 5th derivative of sin6x Find the 3rd derivative of ex cos x. Find nth derivative of coe x Find the nth derivative of e4x sin x Find the nth derivative of log (4-x2), Ixl
  11. 8. sol: 9. sol: 10. 11. Sol: Find the nth derivative of f(x) = 1 6x2-7x+2 12 VxeRR 1 1 6x2 -7x+2 - 3 2 3x-2 2x-1 Diff . ntimes w.r.t. x, 3 2 3x-2 2x-1 3(-1Y 2(-1Y (2x-1Y+l (3X - 2)n+l (2x-1Y+l Suppose f(x) = ax +— V x * 0, then show that x2 f" (x) + xf'(x) —f (x) x b x b 2 x 2b 3 x L.H.S (x) 2b b 3 x 2b x 2 x b ax — x b x x Show that f(x) = x+ cot x satisfies (sin 2 x) f" (x) + 2x = 2 f (x) 3 Find the nth derivative of f(x) = log(8x3+36x2+54x+27)for all x 2 3 8x3 +36x2 +54x+27 = (2x+3) f (x) = log (2x + 3) Diff n times, (x) = 3D (log(2x+3)) (2x+3Y 3 for x > —— and ne N. 2
  12. 12. Find the nth derivative of f(x) = sin 7x cos x Y xe 1 nit nit Ans: Yn — 8n sin 8X+ + ff. sin 6x+— 13. Sol: 14. Ans : 15. sol: 3b Let f(x) = ax2 +—Vx 0 , then show that — 2 f (x) 3b f (x) = ax2 + 3b = 2ax— 6b 2 3b f = 261-1- 3b ax2+— = 2f (x) =RHS Find the nth derivative of f(x) = e3x sin 2 4xVxe 0 sin (n +1) 0. n 4-1 Find the nth derivative of f (0) = coss 0. sin 0(0e R) Let x= cos 0+ i sin 0 (cos 0 + i sin 0) = cos 0— i sine x+— = 2cos 0, By De Moivres theorem = 2 cos no, xn —— = 2i sin no 25 cos50= x+_ , 2717 sin x— 21217 cos5 0. sin 0 = AND x10 —5X6 +10X 10 5 10 10 — 2i sin 0 -20 x2 12 :. —21 —2 x10 12 icos5 Osin 2i(sin120— -1-10 X6 2 sin 100)
  13. —211 coss Osin7 0 = sin 120 — 2sin 100—4. sin 80 + 10 sin 60 + 5 sin 40 — 20. sin 20) 1 nit Dn (COS5 Osin 0) sin 120+— 211 2 nit nit -I(Y2sin 10 0+— 2 • 20 +4115 sin —2.20 Sin -V4sin 80+— + 611.10 sin 2 2 2 2 Prove that the nth derivative of f(x) = tan I x(xe R) is sine where 16. 17. 18. 19. 20. 21. 22. 23. 24. sol: x Oe (—7t, It) is the angle satisfying cose = and e = 1 If u and v are functions of x having 4th order derivatives. Write the formula for Find D (sin xlog x), x > 0 . Find D (ex.cosx), at x = 0 Find D4 (x.logx) at x = e (x > 0) Find the nth derivative of x.ex. Find DI (x2.sin 3x) If y = sin-I xVxe (—1, 1) then show that (I—x2 —xyl =0 and hence deduce that (1 —x 2 Yn+2 —n2yn = 0 — (2n + 1) xyn+l If y = a cos (logx) + b sin (logx) (x >0), then prove that x . yn+2 + (2n +1) xyn+l + (n 2 +1) yn = 0 y = a cos (log x) + b sin (log x) 1 = —a sin (log x).—+bcos (log x). x x xyl = —a sin (log x) + b cos (log x) Differentiating w.r.to x .1 —b sin (log x). 1 x. Y2 + yr 1 = —a cos (log x) x x [a cos (log x) + b sin log x] x Y2 + xyl = — x + xyl + y = 0 Differentiating n times w.r.to x (x2Y2)n + Yn Differentiating n times w.r.to x (x2Y2)n + Yn (Y'2)n .x2 I .Dx2 =O
  14. 25. i) ii) 26. sol: 27. (y) n .X2 +nC1 (Y'2)n .2x+nC2 (Y2)n 2.2 .11 + y n X .yn+2 + 2nx.yn+l + .yn.2 +xyn+l + n. y n + yn = 0 2 x . % +2 + (2n +1) xyn+l + (n 2 +1) y n = 0. If f (x) = xn-l.logx and g(x) =xn.logx for x > O, then prove that n—l)! x x If log y = tan I x(y > 0) , then show that (I + x2) yn+2 + (2nx+ 2x—1) yn+l +n(n+l) yn log y = tan x Diff. wrt. X, 1 1 (1+X2)Y1 = Y 2 Y' Differentiating w.r.to x (l + x2) + 2XYl = Yl Differentiating n times w.r.to x ((1+x2)Y2) = 0 (Y2)n (Y2)n .2X +nc2 (Y2)n . (2X— (1 + x yn+2 + 2nxyn+1 + n .2.yn + n (n —1+2)yn = O 2 (1+ x2) + (2nx + 2x —1) +n(n +1) y n = 0 1)) ( Yl)n | 2) = O If y = sin msin x dn+2y clxn+2 at x=O 2 n Sol: y = sin m sIn x Diff wrt x, Yl = cos m sin x ; ne 1), then show that (yn+2 dny 2 clxn m 1 —x =mcos msin x Squaring on both sides,
  15. 28. 29. sol: =m2.cos2 (msin I x)] Differentiating w.r.to x (1 — x 2 ).2Y'1Y2 + "12 (—2x) = —2m YYI Dividing with 2 Yl , we have (1 —x 2 —xyl = —m y (I—x —xyl + m2y = 0 Differentiating n times w.r to x — (xyl)n + m2yn = 0 (Y2)n =() (—2) —C ( ) n . X + nq.(Y1 11+m y (Y2)n I (—2X) + ncr( 2 (l —X 2 yn+2 yn.2—xyn —xyn + m2yn = 0 — 2nx.yn+l — 2 (l — x2 ) Yn+2 —(2n+1)xyn+1 + m2 —n2)yn = 0 At x = 0, we get (y 1 3 Find the nth derivative of f(x) = xlog(4x+6) x > — 2x+5 2 If x = cost and y = sin pt, where p is a constant and t is not a multiple of TC, then prove that (I—x2) Yn+2 —(2n+1) xyn+l— x —p2) y n = eYne (—1, 1) Given x = cos t, y = sin pt dy = —sin t and — p cos pt dt dt dy — dt —p cos pt dx dt sin t dx 2 _ p2.cos2 pt _ P 2 (I—Y 2) sin t Differentiating w.r.to x (1— x2 ) .2YIY2 + (—2x) = p2 (—2 yyl ) —xyl + py 0 Using Leibnitz' s theorem, Differentiate n times w.r.to x
  16. -(XYl)n + p2yn = O D q +p2yn (Y2)n (1-x2)+n (—2x) + ( Y2 )n-2 x—l 2 2) —(2n+1)xyn +1. n —n+n—p y n = 0 2) —(2n+1)xy 112 -p -O x2 .2x = n2 ( > x y + xyl+n y = 0

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