---2nd-part.jpg)

Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
Ex 13.2
Ex 13.2, 2
Ex 13.2, 3
Ex 13.2, 4 (i) Important
Ex 13.2, 4 (ii) You are here
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
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
Last updated at Aug. 28, 2021 by Teachoo
Ex 13.2, 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