Ex 12.2
Ex 12.2, 2
Ex 12.2, 3
Ex 12.2, 4 (i) Important
Ex 12.2, 4 (ii) You are here
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
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, 4 Find the derivative of the following functions from first principle. (ii) (x – 1) (x – 2) Let f (x) = (x – 1) (x – 2) = x(x – 2) – 1 (x – 2) = x2 – 2x – x + 2 = x2 – 3x + 2 We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = (𝑙𝑖𝑚)┬(ℎ→0) 𝑓〖(𝑥 + ℎ) − 𝑓(𝑥)〗/ℎ Here, f(x) = x2 – 3x + 2 So, f(x + h) = (x + h)2 – 3 (x + h) + 2 Putting values f’(x) = (𝑙𝑖𝑚)┬(ℎ→0)〖([(𝑥 + ℎ)2 − 3(𝑥 + ℎ) + 2] − (𝑥2 − 3 + 2))/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖((𝑥 + ℎ)2 − 3𝑥 − 3ℎ + 2 − 𝑥2 + 3𝑥 − 2)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖((𝑥 + ℎ)2 −3ℎ − 𝑥2 + 3𝑥 − 3𝑥 + 2 − 2)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖((𝑥 + ℎ)2 − 𝑥2 −3ℎ )/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(𝑥2 + ℎ2+ 2𝑥ℎ − 𝑥2 −3ℎ )/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(ℎ2 + 2𝑥ℎ − 3ℎ − 𝑥2 + 𝑥2)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(ℎ (ℎ + 2𝑥 − 3) )/ℎ〗 = lim┬(h→0)〖ℎ+2𝑥 −3〗 Putting h = 0 = 0 + 2x – 3 = 2x – 3 Hence f’(x) = 2x – 3