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  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Ex 13.2, 9 Find the derivative of (i) 2๐‘ฅ โ€“ 3/4 Let f(x) = 2๐‘ฅ โ€“ 3/4 fโ€™(x) = ( ๐‘‘(2๐‘ฅ โˆ’ 3/4 ))/๐‘‘๐‘ฅ = 2 โ€“ 0 = 2 โˆด fโ€™(x) = 2 Ex 13.2, 9 Find the derivative of (ii) (5x3 + 3x โ€“ 1) (x โ€“ 1) Let f(x) = (5x3 + 3x โ€“ 1) (x- 1) Let u = 5x3 + 3x โ€“ 1 & v = x โ€“ 1 โˆด f(x) = uv So, fโ€™(x) = (uv)โ€™ fโ€™(x) = uโ€™v + vโ€™u (xn)โ€™ = nxn โ€“ 1 & (a)โ€™ = 0 where a is constant Finding uโ€™ & vโ€™ u = 5x3 + 3x โ€“ 1 uโ€™ = 5(3x2) + 3(1) โ€“ 0 = 15x2 + 3 v = x โ€“ 1 vโ€™ = 1 โ€“ 0 = 1 fโ€™(x) = uโ€™v + vโ€™u = (15x2 + 3) (x โ€“ 1) + (1) (5x3 + 3x โ€“ 1) = 15x2 (x โ€“ 1) + 3 (x โ€“ 1) + 5x3 + 3x โ€“ 1 = 15x3 โ€“ 15x2 + 3x โ€“ 3 + 5x3 + 3x โ€“ 1 (xn)โ€™ = nxn โ€“ 1 & (a)โ€™ = 0 where a is constant = 15x3 + 5x3 โ€“ 15x2 + 3x + 3x โ€“ 3 โ€“ 1 = 20x3 โ€“ 15x2 + 6x โ€“ 4 Hence fโ€™(x) = 20x3 โ€“ 15x2 + 6x โ€“ 4 Ex 13.2, 9 (Method 1) Find the derivative of (iii) xโ€“3 (5 + 3x) Let f(x) = x โ€“3 (5 + 3x) Let u = xโ€“3 & v = 5 + 3x โˆด f(x) = uv So, fโ€™(x) = (uv)โ€™ fโ€™(x) = uโ€™v + vโ€™u Finding uโ€™ & vโ€™ u = x โ€“ 3 uโ€™ = โ€“3xโ€“3 โ€“ 1 = โ€“3xโ€“4 v = 5 + 3x vโ€™ = 0 + 3 = 3 Now, fโ€™(x) = (uv)โ€™ = uโ€™v + vโ€™u = โ€“3xโ€“ 4 (5 + 3x) + 3 (xโ€“3) = โ€“15xโ€“4 โ€“ 9xโ€“4 + 1 + 3x โ€“3 (xn)โ€™ = nxn โ€“ 1 & (a)โ€™ = 0 where a is constant = โ€“15xโ€“4 โ€“ 9xโ€“3 + 3x โ€“3 = โ€“15xโ€“4 โ€“ 6xโ€“3 = (โˆ’15)/๐‘ฅ^4 โ€“ 6/๐‘ฅ^3 = โˆ’3 [5/๐‘ฅ^4 +2/๐‘ฅ^3 ] = โˆ’3 [(5 + 2๐‘ฅ)/๐‘ฅ^4 ] = (โˆ’๐Ÿ‘)/๐’™^๐Ÿ’ (5 + 2x) Hence, fโ€™(x) = (โˆ’3)/๐‘ฅ^4 (5 + 2x) Ex 13.2, 9 (Method 2) Find the derivative of (iii) xโ€“3 (5 + 3x) ๐‘ฅ^(โˆ’3) (5+3๐‘ฅ) = (5 + 3๐‘ฅ)/๐‘ฅ^3 = 5/๐‘ฅ^3 + 3๐‘ฅ/๐‘ฅ^3 = 5/๐‘ฅ^3 + 3/๐‘ฅ^2 = ใ€–5๐‘ฅใ€—^(โˆ’3) + ใ€–3๐‘ฅใ€—^(โˆ’2) Differentiating w.r.t.x (ใ€–5๐‘ฅใ€—^(โˆ’3) " + " ใ€–3๐‘ฅใ€—^(โˆ’2) )^โ€ฒ = 5 [โˆ’3๐‘ฅ^(โˆ’4) ] + 3 [โˆ’2๐‘ฅ^(โˆ’3) ] = โˆ’3 [5/๐‘ฅ^4 +2/๐‘ฅ^3 ] = โˆ’3 [(5 + 2๐‘ฅ)/๐‘ฅ^4 ] = (โˆ’๐Ÿ‘)/๐’™^๐Ÿ’ (5 + 2x) Ex 13.2,9 (Method 1) Find the derivative of (iv) x5 (3 โˆ’ 6xโˆ’9 ). Let f (x) = x5 (3 โˆ’ 6xโˆ’9 ) Let u = x5 & v = 3 โ€“ 6xโ€“9 So, f(x) = uv โˆด fโ€™(x) = (uv)โ€™ fโ€™(x) = uโ€™v + vโ€™ u Finding uโ€™ & vโ€™ u = x5 uโ€™ = 5x5 โ€“ 1 uโ€™ = 5x4 v = 3 โˆ’ 6xโˆ’9 vโ€™ = 0 โ€“ 6( โ€“9)xโ€“10 vโ€™ = 54xโ€“10 Now, fโ€™(x) = (uv)โ€™ = uโ€™v + vโ€™ u = 5x4 (3 โ€“ 6xโ€“9) + 54xโ€“10 (x5) = 15x4 โ€“ 30xโ€“9 + 4 + 54xโ€“10 + 5 = 15x4 โ€“ 30x โ€“5 + 54x โ€“5 = 15x4 + 24x โ€“5 = 15x4 + 24x โ€“5 = 15x4 + 24/๐‘ฅ^5 Hence fโ€™(x) = 15x4 + ๐Ÿ๐Ÿ’/๐’™^๐Ÿ“ Ex 13.2, 9 (Method 2) Find the derivative of (iv) x5 (3 โˆ’ 6xโˆ’9 ). x5 (3 โˆ’ 6xโˆ’9 ). = ใ€–3๐‘ฅใ€—^5 โˆ’ ใ€–6๐‘ฅใ€—^(โˆ’4) Differentiating w.r.t.x ใ€–15๐‘ฅใ€—^4 โˆ’6 [โˆ’4๐‘ฅ^(โˆ’5)] =ใ€–15๐‘ฅใ€—^4+24๐‘ฅ^(โˆ’5) = 15x4 + ๐Ÿ๐Ÿ’/๐’™^๐Ÿ“ Ex 13.2, 9 Find the derivative of (v) f (x) = xโ€“4 (3 โ€“ 4xโ€“5) Let f(x) = xโ€“4 (3 โ€“ 4x โ€“5) = 3xโ€“4 โ€“ 4xโ€“5 ร— xโ€“4 = 3xโ€“4 โ€“ 4xโ€“5 + (โ€“4) = 3xโ€“4 โ€“ 4xโ€“9 Differentiating w.r.t. x fโ€™(x) = (3xโ€“4 โ€“ 4xโ€“9)โ€™ = (3(โˆ’4๐‘ฅ^(โˆ’4 โˆ’ 1) )โˆ’4(โˆ’9๐‘ฅ^(โˆ’9โˆ’1) ) = (3(โˆ’4๐‘ฅ^(โˆ’5) )โˆ’4(โˆ’9๐‘ฅ^(โˆ’10) ) = โˆ’12๐‘ฅ^(โˆ’5)+36๐‘ฅ^(โˆ’10) = (โˆ’12)/๐‘ฅ^5 + 36/๐‘ฅ^10 Hence fโ€™(x) = (โˆ’๐Ÿ๐Ÿ)/๐’™^๐Ÿ“ + ๐Ÿ‘๐Ÿ”/๐’™^๐Ÿ๐ŸŽ Ex 13.2, 9 Find the derivative of (vi) f(x) = 2/(x + 1) โ€“ x2/(3x โˆ’ 1) Let f (x) = 2/(x + 1) โ€“ x2/(3x โˆ’ 1) Let f1 (x) = 2/(x + 1) & f2 (x) = x2/(3x โˆ’ 1) โˆด f(x) = f1(x) โ€“ f2 (x) So, fโ€™(x) = (f1(x) โ€“ f2(x))โ€™ fโ€™(x) = fโ€™1(x) โ€“ fโ€™2(x) Finding f1โ€˜(x) f1 (x) = 2/(๐‘ฅ + 1) Let u = 2 & v = x + 1 โˆด f1(x) = ๐‘ข/๐‘ฃ Now, f1โ€™(x) = (๐‘ข/๐‘ฃ)^โ€ฒ f1โ€™(x) = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 u = 2 uโ€™ = 0 v = x + 1 vโ€™ = 1 + 0 = 1 fโ€™1(x) = (๐‘ข/๐‘ฃ)^โ€ฒ = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 = (0(๐‘ฅ + 1) โˆ’1 (2))/(๐‘ฅ + 1)2 = (โˆ’2)/ใ€–(๐‘ฅ + 1)ใ€—^2 Hence, f1โ€™ (x) = (โˆ’2)/(๐‘ฅ + 1)2 Finding f2โ€˜(x) f2 (x) = ๐‘ฅ2/(3๐‘ฅ โˆ’ 1) Let u = x2 & v = 3x โ€“ 1 Now, f2โ€™(x) = (๐‘ข/๐‘ฃ)^โ€ฒ f2โ€™(x) = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 Finding uโ€™ & vโ€™ u = x2 uโ€™ = 2x2 โ€“ 1 = 2x & v = 3x โ€“ 1 vโ€™ = 3(1) โ€“ 0 = 3 fโ€™2(x) = (๐‘ข/๐‘ฃ)^โ€ฒ (xn)โ€™ = nxn โ€“ 1 & (a)โ€™ = 0 where a is constant = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 = (2๐‘ฅ(3๐‘ฅ โˆ’ 1) โˆ’ 3 (๐‘ฅ2))/(3๐‘ฅ โˆ’ 1)2 = (6๐‘ฅ2 โˆ’ 2๐‘ฅ โˆ’ 3๐‘ฅ2)/ใ€–(3๐‘ฅ โˆ’ 1)ใ€—^2 = (3๐‘ฅ2 โˆ’ 2๐‘ฅ )/ใ€–(3๐‘ฅ โˆ’ 1)ใ€—^2 = (๐‘ฅ(3๐‘ฅ โˆ’ 2))/ใ€–(3๐‘ฅ โˆ’ 1)ใ€—^2 Hence fโ€™2(x) = (๐‘ฅ (3๐‘ฅ โˆ’ 2))/(3๐‘ฅ โˆ’ 1)2 Now fโ€™ (x) = f1โ€™(x) โ€“ f2โ€™ (x) = (โˆ’๐Ÿ)/(๐’™ + ๐Ÿ)๐Ÿ โ€“ (๐’™(๐Ÿ‘๐’™ โˆ’ ๐Ÿ))/(๐Ÿ‘๐’™ โˆ’ ๐Ÿ)๐Ÿ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.