Ex 12.2, 7 (i) - Chapter 12 Class 11 Limits and Derivatives
Last updated at May 7, 2024 by Teachoo
Ex 12.2
Ex 12.2, 1
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
Transcript
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)
