

Ex 12.2
Last updated at Dec. 16, 2024 by Teachoo
Ex 12.2, 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 (xn)’ = nxn – 1 & (a)’ = 0 where a is constant Finding u’ & v’ u = 5x3 + 3x – 1 u’ = 5(3x2) + 3(1) – 0 = 15x2 + 3 v = x – 1 v’ = 1 – 0 = 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 (xn)’ = nxn – 1 & (a)’ = 0 where a is constant = 15x3 + 5x3 – 15x2 + 3x + 3x – 3 – 1 = 20x3 – 15x2 + 6x – 4 Hence f’(x) = 20x3 – 15x2 + 6x – 4