Ex 13.2, 9 - Chapter 13 Class 11 Limits and Derivatives

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Ex 13.2, 9 Find the derivative of (i) 2 3 4 Let f(x) = 2 3 4 f (x) = (2 3 4 ) = 2 0 = 2 f (x) = 2 Ex 13.2, 9 Find the derivative of (ii) (5x3 + 3x 1) (x 1) Let f(x) = (5x3 + 3x 1) (x- 1) Let u = 5x3 + 3x 1 & v = x 1 f(x) = uv So, f (x) = (uv) f (x) = u v + v u Finding u & v u = 5x3 + 3x 1 u = 5.3x3 1 + 3.1.x1 1 0 = 15x2 + 3x0 = 15x2 + 3 v = x 1 v = 1 .x1 1 0 = 1. x0 = 1 f (x) = u v + v u = (15x2 + 3) (x 1) + (1) (5x3 + 3x 1) = 15x2 (x 1) + 3 (x 1) + 5x3 + 3x 1 = 15x3 15x2 + 3x 3 + 5x3 + 3x 1 = 15x3 + 5x3 15x2 + 3x + 3x 3 1 = 20x3 15x2 + 6x 4 Hence f (x) = 20x3 15x2 + 6x 4 Ex 13.2, 9 Find the derivative of (iii) x 3 (5 + 3x) Let f(x) = x 3 (5 + 3x) Let u = x 3 & v = 5 + 3x f(x) = uv So, f (x) = (uv) f (x) = u v + v u Finding u & v u = x 3 u = 3x 3 1 = 3x 4 v = 5 + 3x v = 0 + 3 = 3 Now, f (x) = (uv) = u v + v u = 3x 4 (5 + 3x) + 3 (x 3) = 15x 4 9x 4+ 1 + 3x 3 = 15x 4 6x 3 = 15 4 6 3 = 3 4 (5 + 2x) Hence f (x) = (5 + 2x) Ex 13.2,9 Find the derivative of (iv) x5 (3 6x 9 ). Let f (x) = x5 (3 6x 9 ) Let u = x5 & v = 3 6x 9 So, f(x) = uv f (x) = (uv) f (x) = u v + v u Finding u & v u = x5 u = 5x5 1 = 5x4 v = 3 6x 9 v = 0 6( 9)x 9 1 v = 54x 10 Now, f (x) = (uv) = u v + v u = 5x4 (3 6x 9) + 54x 10) (x5) = 15x4 30x 9 + 4 + 54 10 + 5 = 15x4 30x 5 + 54x 5 = 15x4 + 24x 5 = 15x4 + 24 5 Hence f (x) = 15x4 + Ex 13.2, 9 Find the derivative of (v) f (x) = x 4 (3 4x 5) Let f(x) = x 4 (3 4x 5) Let u = x 4 & v = 3 4x 5 f(x) = uv So, f (x) = (uv) f (x) = u v + v u Finding u & v u = x 4 u = 4x 4 1 = 4x 5 v = 3 4x 5 v = 0 4 ( 5) x 5 1 = 20x 6 Now, f (x) = (uv) = u v + v u = ( 4x 5 ) (3 4x 5) + (20x 6) (x 4) = 12x 5 + 16x 5 5 + 20 6 4 = 12x 5 + 16x 10 + 20x 10 = 12x 5 + 36x 10 = 12 5 + 36 10 Hence f (x) = + Ex 13.2, 9 Find the derivative of (vi) f(x) = 2 x + 1 x2 3x 1 Let f (x) = 2 x + 1 x2 3x 1 Let f1 (x) = 2 x + 1 & f2 (x) = x2 3x 1 f(x) = f1(x) f2 (x) So, f (x) = (f1(x) f2(x)) f (x) = f 1(x) f 2(x) Finding f1 (x) f1 (x) = 2 + 1 Let u = 2 & v = x + 1 f1(x) = Now, f1 (x) = f1 (x) = 2 u = 2 u = 0 v = x + 1 v = 1 + 0 = 1 f 1(x) = = 2 = 0 + 1 1 (2) + 1 2 = 2 ( + 1) 2 Hence, f1 (x) = 2 + 1 2 Finding f2 (x) f2 (x) = 2 3 1 Let u = x2 & v = 3x 1 f2 (x) = Now, f2 (x) = f2 (x) = 2 Finding u & v u = x2 u = 2x2 1 = 2x & v = 3x 1 v = 3(1) 0 = 3 f 2(x) = = 2 = 2 3 1 3 ( 2) 3 1 2 = 6 2 2 3 2 (3 1) 2 = 3 2 2 (3 1) 2 = (3 2) (3 1) 2 Hence f 2(x) = (3 2) 3 + 1 2 Now f (x) = f1 (x) f2 (x) = + ( ) ( )