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Ex 13.2, 9 - Find derivative of (i) 2x - 3/4 - Chapter 13 - Derivatives by formula - x^n formula

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  1. Class 11
  2. Important Question for exams Class 11
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Ex 13.2, 9 Find the derivative of (i) 2š‘„ ā€“ 3ļ·®4ļ·Æ Let f(x) = 2š‘„ ā€“ 3ļ·®4ļ·Æ fā€™(x) = š‘‘(2š‘„ āˆ’ 3ļ·®4ļ·Æ )ļ·®š‘‘š‘„ļ·Æ = 2 ā€“ 0 = 2 āˆ“ fā€™(x) = 2 Ex 13.2, 9 Find the derivative of (ii) (5x3 + 3x ā€“ 1) (x ā€“ 1) Let f(x) = (5x3 + 3x ā€“ 1) (x- 1) Let u = 5x3 + 3x ā€“ 1 & v = x ā€“ 1 āˆ“ f(x) = uv So, fā€™(x) = (uv)ā€™ fā€™(x) = uā€™v + vā€™u Finding uā€™ & vā€™ u = 5x3 + 3x ā€“ 1 uā€™ = 5.3x3ā€“1 + 3.1.x1ā€“1 ā€“ 0 = 15x2 + 3x0 = 15x2 + 3 v = x ā€“ 1 vā€™ = 1 .x1ā€“1 ā€“ 0 = 1. x0 = 1 fā€™(x) = uā€™v + vā€™u = (15x2 + 3) (x ā€“ 1) + (1) (5x3 + 3x ā€“ 1) = 15x2 (x ā€“ 1) + 3 (x ā€“ 1) + 5x3 + 3x ā€“ 1 = 15x3 ā€“ 15x2 + 3x ā€“ 3 + 5x3 + 3x ā€“ 1 = 15x3 + 5x3 ā€“ 15x2 + 3x + 3x ā€“ 3 ā€“ 1 = 20x3 ā€“ 15x2 + 6x ā€“ 4 Hence fā€™(x) = 20x3 ā€“ 15x2 + 6x ā€“ 4 Ex 13.2, 9 Find the derivative of (iii) xā€“3 (5 + 3x) Let f(x) = x ā€“ 3 (5 + 3x) Let u = x ā€“ 3 & v = 5 + 3x āˆ“ f(x) = uv So, fā€™(x) = (uv)ā€™ fā€™(x) = uā€™v + vā€™u Finding uā€™ & vā€™ u = x ā€“ 3 uā€™ = ā€“ 3x ā€“ 3 ā€“ 1 = ā€“ 3x ā€“ 4 v = 5 + 3x vā€™ = 0 + 3 = 3 Now, fā€™(x) = (uv)ā€™ = uā€™v + vā€™u = ā€“ 3x ā€“ 4 (5 + 3x) + 3 (x ā€“ 3) = ā€“ 15x ā€“ 4 ā€“ 9x ā€“ 4+ 1 + 3x ā€“ 3 = ā€“ 15x ā€“ 4 ā€“ 6x ā€“ 3 = āˆ’15ļ·® š‘„ļ·®4ļ·Æļ·Æ ā€“ 6ļ·® š‘„ļ·®3ļ·Æļ·Æ = āˆ’3ļ·® š‘„ļ·®4ļ·Æļ·Æ (5 + 2x) Hence fā€™(x) = āˆ’šŸ‘ļ·® š’™ļ·®šŸ’ļ·Æļ·Æ (5 + 2x) Ex 13.2,9 Find the derivative of (iv) x5 (3 āˆ’ 6xāˆ’9 ). Let f (x) = x5 (3 āˆ’ 6xāˆ’9 ) Let u = x5 & v = 3 ā€“ 6x ā€“ 9 So, f(x) = uv āˆ“ fā€™ (x) = (uv)ā€™ fā€™ (x) = uā€™v + vā€™ u Finding uā€™ & vā€™ u = x5 uā€™ = 5x5 ā€“ 1 = 5x4 v = 3 āˆ’ 6xāˆ’9 vā€™ = 0 ā€“ 6( ā€“ 9)x ā€“ 9 ā€“ 1 vā€™ = 54x ā€“10 Now, fā€™(x) = (uv)ā€™ = uā€™v + vā€™ u = 5x4 (3 ā€“ 6x ā€“ 9) + 54x ā€“ 10) (x5) = 15x4 ā€“ 30x ā€“ 9 + 4 + 54 ā€“ 10 + 5 = 15x4 ā€“ 30x ā€“5 + 54x ā€“ 5 = 15x4 + 24x ā€“ 5 = 15x4 + 24ļ·® š‘„ļ·®5ļ·Æļ·Æ Hence fā€™(x) = 15x4 + šŸšŸ’ļ·® š’™ļ·®šŸ“ļ·Æļ·Æ Ex 13.2, 9 Find the derivative of (v) f (x) = xā€“4 (3 ā€“ 4xā€“5) Let f(x) = x ā€“ 4 (3 ā€“ 4x ā€“ 5) Let u = x ā€“ 4 & v = 3 ā€“ 4x ā€“ 5 āˆ“ f(x) = uv So, fā€™(x) = (uv)ā€™ fā€™(x) = uā€™v + vā€™u Finding uā€™ & vā€™ u = x ā€“ 4 uā€™ = ā€“ 4x ā€“ 4 ā€“ 1 = ā€“ 4x ā€“ 5 v = 3 ā€“ 4x ā€“ 5 vā€™ = 0 ā€“ 4 ( ā€“ 5) x ā€“ 5 ā€“ 1 = 20x ā€“ 6 Now, fā€™(x) = (uv)ā€™ = uā€™v + vā€™u = (ā€“ 4xā€“5 ) (3 ā€“ 4xā€“5) + (20xā€“6) (xā€“4) = ā€“ 12xā€“5 + 16x ā€“5 ā€“ 5 + 20 ā€“6 ā€“ 4 = ā€“ 12x ā€“5 + 16x ā€“10 + 20x ā€“ 10 = ā€“ 12x ā€“5 + 36x ā€“ 10 = āˆ’12ļ·® š‘„ļ·®5ļ·Æļ·Æ + 36ļ·® š‘„ļ·®10ļ·Æļ·Æ Hence fā€™(x) = āˆ’šŸšŸļ·® š’™ļ·®šŸ“ļ·Æļ·Æ + šŸ‘šŸ”ļ·® š’™ļ·®šŸšŸŽļ·Æļ·Æ Ex 13.2, 9 Find the derivative of (vi) f(x) = 2ļ·®x + 1ļ·Æ ā€“ x2ļ·®3x āˆ’ 1ļ·Æ Let f (x) = 2ļ·®x + 1ļ·Æ ā€“ x2ļ·®3x āˆ’ 1ļ·Æ Let f1 (x) = 2ļ·®x + 1ļ·Æ & f2 (x) = x2ļ·®3x āˆ’ 1ļ·Æ āˆ“ f(x) = f1(x) ā€“ f2 (x) So, fā€™(x) = (f1(x) ā€“ f2(x))ā€™ fā€™(x) = fā€™1(x) ā€“ fā€™2(x) Finding f1ā€˜(x) f1 (x) = 2ļ·®š‘„ + 1ļ·Æ Let u = 2 & v = x + 1 āˆ“ f1(x) = š‘¢ļ·®š‘£ļ·Æ Now, f1ā€™(x) = š‘¢ļ·®š‘£ļ·Æļ·Æļ·®ā€²ļ·Æ f1ā€™(x) = š‘¢ļ·®ā€²ļ·Æ š‘£ āˆ’ š‘£ļ·®ā€²ļ·Æ š‘¢ļ·® š‘£ļ·®2ļ·Æļ·Æ u = 2 uā€™ = 0 v = x + 1 vā€™ = 1 + 0 = 1 fā€™1(x) = š‘¢ļ·®š‘£ļ·Æļ·Æļ·®ā€²ļ·Æ = š‘¢ļ·®ā€²ļ·Æ š‘£ āˆ’ š‘£ļ·®ā€²ļ·Æ š‘¢ļ·® š‘£ļ·®2ļ·Æļ·Æ = 0 š‘„ + 1ļ·Æ āˆ’1 (2)ļ·® š‘„ + 1ļ·Æ2ļ·Æ = āˆ’2ļ·® (š‘„ + 1)ļ·®2ļ·Æļ·Æ Hence, f1ā€™ (x) = āˆ’2ļ·® š‘„ + 1ļ·Æ2ļ·Æ Finding f2ā€˜(x) f2 (x) = š‘„2ļ·®3š‘„ āˆ’ 1ļ·Æ Let u = x2 & v = 3x ā€“ 1 āˆ“ f2 (x) = š‘¢ļ·®š‘£ļ·Æ Now, f2ā€™(x) = š‘¢ļ·®š‘£ļ·Æļ·Æļ·®ā€²ļ·Æ f2ā€™(x) = š‘¢ļ·®ā€²ļ·Æ š‘£ āˆ’ š‘£ļ·®ā€²ļ·Æ š‘¢ļ·® š‘£ļ·®2ļ·Æļ·Æ Finding uā€™ & vā€™ u = x2 uā€™ = 2x2 ā€“ 1 = 2x & v = 3x ā€“ 1 vā€™ = 3(1) ā€“ 0 = 3 fā€™2(x) = š‘¢ļ·®š‘£ļ·Æļ·Æļ·®ā€²ļ·Æ = š‘¢ļ·®ā€²ļ·Æ š‘£ āˆ’ š‘£ļ·®ā€²ļ·Æ š‘¢ļ·® š‘£ļ·®2ļ·Æļ·Æ = 2š‘„ 3š‘„ āˆ’ 1ļ·Æ āˆ’ 3 (š‘„2)ļ·® 3š‘„ āˆ’ 1ļ·Æ2ļ·Æ = 6š‘„2 āˆ’ 2š‘„ āˆ’ 3š‘„2ļ·® (3š‘„ āˆ’ 1)ļ·®2ļ·Æļ·Æ = 3š‘„2 āˆ’ 2š‘„ ļ·® (3š‘„ āˆ’ 1)ļ·®2ļ·Æļ·Æ = š‘„(3š‘„ āˆ’ 2)ļ·® (3š‘„ āˆ’ 1)ļ·®2ļ·Æļ·Æ Hence fā€™2(x) = š‘„ (3š‘„ āˆ’ 2)ļ·® 3š‘„ + 1ļ·Æ2ļ·Æ Now fā€™ (x) = f1ā€™(x) ā€“ f2ā€™ (x) = āˆ’šŸļ·® š’™ + šŸļ·ÆšŸļ·Æ ā€“ š’™(šŸ‘š’™ āˆ’ šŸ)ļ·®(šŸ‘š’™ āˆ’ šŸ)ļ·Æ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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