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Last updated at Nov. 30, 2019 by Teachoo
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Ex 13.2, 9 Find the derivative of (i) 2๐ฅ โ 3/4 Let f(x) = 2๐ฅ โ 3/4 fโ(x) = ( ๐(2๐ฅ โ 3/4 ))/๐๐ฅ = 2 โ 0 = 2 โด 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 โด f(x) = uv So, fโ(x) = (uv)โ fโ(x) = uโv + vโu (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) = uโv + vโu = (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) xโ3 (5 + 3x) Let f(x) = x โ3 (5 + 3x) Let u = xโ3 & v = 5 + 3x โด f(x) = uv So, fโ(x) = (uv)โ fโ(x) = uโv + vโu Finding uโ & vโ u = x โ 3 uโ = โ3xโ3 โ 1 = โ3xโ4 v = 5 + 3x vโ = 0 + 3 = 3 Now, fโ(x) = (uv)โ = uโv + vโu = โ3xโ 4 (5 + 3x) + 3 (xโ3) = โ15xโ4 โ 9xโ4 + 1 + 3x โ3 (xn)โ = nxn โ 1 & (a)โ = 0 where a is constant = โ15xโ4 โ 9xโ3 + 3x โ3 = โ15xโ4 โ 6xโ3 = (โ15)/๐ฅ^4 โ 6/๐ฅ^3 = โ3 [5/๐ฅ^4 +2/๐ฅ^3 ] = โ3 [(5 + 2๐ฅ)/๐ฅ^4 ] = (โ๐)/๐^๐ (5 + 2x) Hence, fโ(x) = (โ3)/๐ฅ^4 (5 + 2x) Ex 13.2, 9 (Method 2) Find the derivative of (iii) xโ3 (5 + 3x) ๐ฅ^(โ3) (5+3๐ฅ) = (5 + 3๐ฅ)/๐ฅ^3 = 5/๐ฅ^3 + 3๐ฅ/๐ฅ^3 = 5/๐ฅ^3 + 3/๐ฅ^2 = ใ5๐ฅใ^(โ3) + ใ3๐ฅใ^(โ2) Differentiating w.r.t.x (ใ5๐ฅใ^(โ3) " + " ใ3๐ฅใ^(โ2) )^โฒ = 5 [โ3๐ฅ^(โ4) ] + 3 [โ2๐ฅ^(โ3) ] = โ3 [5/๐ฅ^4 +2/๐ฅ^3 ] = โ3 [(5 + 2๐ฅ)/๐ฅ^4 ] = (โ๐)/๐^๐ (5 + 2x) Ex 13.2,9 (Method 1) Find the derivative of (iv) x5 (3 โ 6xโ9 ). Let f (x) = x5 (3 โ 6xโ9 ) Let u = x5 & v = 3 โ 6xโ9 So, f(x) = uv โด fโ(x) = (uv)โ fโ(x) = uโv + vโ u Finding uโ & vโ u = x5 uโ = 5x5 โ 1 uโ = 5x4 v = 3 โ 6xโ9 vโ = 0 โ 6( โ9)xโ10 vโ = 54xโ10 Now, fโ(x) = (uv)โ = uโv + vโ u = 5x4 (3 โ 6xโ9) + 54xโ10 (x5) = 15x4 โ 30xโ9 + 4 + 54xโ10 + 5 = 15x4 โ 30x โ5 + 54x โ5 = 15x4 + 24x โ5 = 15x4 + 24x โ5 = 15x4 + 24/๐ฅ^5 Hence fโ(x) = 15x4 + ๐๐/๐^๐ Ex 13.2, 9 (Method 2) Find the derivative of (iv) x5 (3 โ 6xโ9 ). x5 (3 โ 6xโ9 ). = ใ3๐ฅใ^5 โ ใ6๐ฅใ^(โ4) Differentiating w.r.t.x ใ15๐ฅใ^4 โ6 [โ4๐ฅ^(โ5)] =ใ15๐ฅใ^4+24๐ฅ^(โ5) = 15x4 + ๐๐/๐^๐ Ex 13.2, 9 Find the derivative of (v) f (x) = xโ4 (3 โ 4xโ5) Let f(x) = xโ4 (3 โ 4x โ5) = 3xโ4 โ 4xโ5 ร xโ4 = 3xโ4 โ 4xโ5 + (โ4) = 3xโ4 โ 4xโ9 Differentiating w.r.t. x fโ(x) = (3xโ4 โ 4xโ9)โ = (3(โ4๐ฅ^(โ4 โ 1) )โ4(โ9๐ฅ^(โ9โ1) ) = (3(โ4๐ฅ^(โ5) )โ4(โ9๐ฅ^(โ10) ) = โ12๐ฅ^(โ5)+36๐ฅ^(โ10) = (โ12)/๐ฅ^5 + 36/๐ฅ^10 Hence fโ(x) = (โ๐๐)/๐^๐ + ๐๐/๐^๐๐ Ex 13.2, 9 Find the derivative of (vi) f(x) = 2/(x + 1) โ x2/(3x โ 1) Let f (x) = 2/(x + 1) โ x2/(3x โ 1) Let f1 (x) = 2/(x + 1) & f2 (x) = x2/(3x โ 1) โด f(x) = f1(x) โ f2 (x) So, fโ(x) = (f1(x) โ f2(x))โ fโ(x) = fโ1(x) โ fโ2(x) Finding f1โ(x) f1 (x) = 2/(๐ฅ + 1) Let u = 2 & v = x + 1 โด f1(x) = ๐ข/๐ฃ Now, f1โ(x) = (๐ข/๐ฃ)^โฒ f1โ(x) = (๐ข^โฒ ๐ฃ โใ ๐ฃใ^โฒ ๐ข)/๐ฃ^2 u = 2 uโ = 0 v = x + 1 vโ = 1 + 0 = 1 fโ1(x) = (๐ข/๐ฃ)^โฒ = (๐ข^โฒ ๐ฃ โใ ๐ฃใ^โฒ ๐ข)/๐ฃ^2 = (0(๐ฅ + 1) โ1 (2))/(๐ฅ + 1)2 = (โ2)/ใ(๐ฅ + 1)ใ^2 Hence, f1โ (x) = (โ2)/(๐ฅ + 1)2 Finding f2โ(x) f2 (x) = ๐ฅ2/(3๐ฅ โ 1) Let u = x2 & v = 3x โ 1 Now, f2โ(x) = (๐ข/๐ฃ)^โฒ f2โ(x) = (๐ข^โฒ ๐ฃ โใ ๐ฃใ^โฒ ๐ข)/๐ฃ^2 Finding uโ & vโ u = x2 uโ = 2x2 โ 1 = 2x & v = 3x โ 1 vโ = 3(1) โ 0 = 3 fโ2(x) = (๐ข/๐ฃ)^โฒ (xn)โ = nxn โ 1 & (a)โ = 0 where a is constant = (๐ข^โฒ ๐ฃ โใ ๐ฃใ^โฒ ๐ข)/๐ฃ^2 = (2๐ฅ(3๐ฅ โ 1) โ 3 (๐ฅ2))/(3๐ฅ โ 1)2 = (6๐ฅ2 โ 2๐ฅ โ 3๐ฅ2)/ใ(3๐ฅ โ 1)ใ^2 = (3๐ฅ2 โ 2๐ฅ )/ใ(3๐ฅ โ 1)ใ^2 = (๐ฅ(3๐ฅ โ 2))/ใ(3๐ฅ โ 1)ใ^2 Hence fโ2(x) = (๐ฅ (3๐ฅ โ 2))/(3๐ฅ โ 1)2 Now fโ (x) = f1โ(x) โ f2โ (x) = (โ๐)/(๐ + ๐)๐ โ (๐(๐๐ โ ๐))/(๐๐ โ ๐)๐
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