Ex 12.2
Ex 12.2, 2
Ex 12.2, 3
Ex 12.2, 4 (i) Important
Ex 12.2, 4 (ii)
Ex 12.2, 4 (iii) Important
Ex 12.2, 4 (iv)
Ex 12.2, 5
Ex 12.2, 6
Ex 12.2, 7 (i) Important You are here
Ex 12.2, 7 (ii)
Ex 12.2, 7 (iii) Important
Ex 12.2, 8
Ex 12.2, 9 (i)
Ex 12.2, 9 (ii) Important
Ex 12.2, 9 (iii)
Ex 12.2, 9 (iv) Important
Ex 12.2, 9 (v)
Ex 12.2, 9 (vi)
Ex 12.2, 10 Important
Ex 12.2, 11 (i)
Ex 12.2, 11 (ii) Important
Ex 12.2, 11 (iii) Important
Ex 12.2, 11 (iv)
Ex 12.2, 11 (v) Important
Ex 12.2, 11 (vi)
Ex 12.2, 11 (vii) Important
Last updated at May 7, 2024 by Teachoo
Ex 12.2, 7 (Method 1) For some constants a and b, find the derivative of (i) (x – a) (x – b) Let f (x) = (x – a) (x – b) = x (x – b) – a (x – b) = x2 – xb – ax + ab = x2 – (b + a)x + ab f’(x) = (x2 – (b + a)x + ab)’ = 2.x2–1 – (b + a) .1x1–1 + 0 = 2x1 – (b + a) . x0 = 2x – (b + a) × 1 = 2x – (b + a) Ex 12.2, 7 (Method 2) For some constants a and b, find the derivative of (i) (x – a) (x – b) Let f (x) = (x – a) (x – b) Let u = (x – a) & v = (x – b) ∴ f(x) = uv So, f’(x) = (uv)’ f’(x) = u’v + v’u Finding u’ & v’ u = x – a u’ = 1. x1–1 – 0 = 1. x0 = 1 v = (x – b) v’ = 1 .x1–1 – 0 = 1 . x0 = 1 Now, f’ (x) = (uv)’ = u’ v + v’u = (1) (x – b) + 1(x – a) = x – b + (x – a) = 2x – b – a = 2x – (b + a)