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Ex 13.2, 9 (ii) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at March 16, 2023 by Teachoo
Ex 13.2
Ex 13.2, 1
Ex 13.2, 2
Ex 13.2, 3
Ex 13.2, 4 (i) Important
Ex 13.2, 4 (ii)
Ex 13.2, 4 (iii) Important
Ex 13.2, 4 (iv)
Ex 13.2, 5
Ex 13.2, 6
Ex 13.2, 7 (i) Important
Ex 13.2, 7 (ii)
Ex 13.2, 7 (iii) Important
Ex 13.2, 8
Ex 13.2, 9 (i)
Ex 13.2, 9 (ii) Important You are here
Ex 13.2, 9 (iii)
Ex 13.2, 9 (iv) Important
Ex 13.2, 9 (v)
Ex 13.2, 9 (vi)
Ex 13.2, 10 Important
Ex 13.2, 11 (i)
Ex 13.2, 11 (ii) Important
Ex 13.2, 11 (iii) Important
Ex 13.2, 11 (iv)
Ex 13.2, 11 (v) Important
Ex 13.2, 11 (vi)
Ex 13.2, 11 (vii) Important
Transcript
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 (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
