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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) 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 29, 2023 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