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Ex 13.2

Ex 13.2, 1

Ex 13.2, 2

Ex 13.2, 3

Ex 13.2, 4 (i) Important

Ex 13.2, 4 (ii)

Ex 13.2, 4 (iii) Important

Ex 13.2, 4 (iv) You are here

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

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 13.2, 4 Find the derivative of the following functions from first principle. (iv) (𝑥 + 1)/(𝑥 − 1) Let f (x) = (𝑥 + 1)/(𝑥 − 1) 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) = (𝑥 + 1)/(𝑥 − 1) So, f (x + h) = ((𝑥 + ℎ) + 1)/((𝑥 + ℎ) − 1) Ex 13.2, 4 Find the derivative of the following functions from first principle. (iv) (𝑥 + 1)/(𝑥 − 1) Let f (x) = (𝑥 + 1)/(𝑥 − 1) 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) = (𝑥 + 1)/(𝑥 − 1) So, f (x + h) = ((𝑥 + ℎ) + 1)/((𝑥 + ℎ) − 1) Putting values f’(x) = lim┬(h→0)〖([((𝑥 + ℎ) + 1)/(𝑥 + ℎ − 1)] −[ (𝑥 + 1)/(𝑥 − 1)])/h〗 = lim┬(h→0)〖((𝑥 + ℎ + 1)/(𝑥 + ℎ − 1) − (𝑥 + 1)/(𝑥 − 1))/h〗 = lim┬(h→0)〖(((𝑥 − 1)(𝑥 + ℎ + 1) − (𝑥 + 1)( 𝑥 + ℎ − 1))/(( 𝑥 + ℎ − 1 ) (𝑥 − 1)))/ℎ〗 = lim┬(h→0)〖((𝑥 − 1) ((𝑥 + 1) + ℎ) − (𝑥 + 1)( (𝑥 − 1) + ℎ))/(ℎ( 𝑥 + ℎ − 1 ) (𝑥 − 1))〗 = lim┬(h→0)〖((𝑥 − 1)(𝑥 + 1) + (𝑥 −1)ℎ − (𝑥 + 1)(𝑥 − 1) − (𝑥 + 1) ℎ)/(ℎ( 𝑥 + ℎ − 1 ) (𝑥 − 1))〗 = lim┬(h→0)〖((𝑥2 − 1) + 𝑥ℎ − (𝑥2 − 1) − 𝑥ℎ − ℎ)/(ℎ( 𝑥 + ℎ − 1 ) (𝑥 − 1))〗 = (𝑙𝑖𝑚)┬(ℎ→0)〖(− 2ℎ )/(ℎ (𝑥 + ℎ − 1) (𝑥 − 1))〗 = lim┬(h→0)〖(−2)/((𝑥 + ℎ − 1) (𝑥 − 1))〗 Putting h = 0 = (−2)/((𝑥 + 0 − 1)(𝑥 − 1)) = (−2)/((𝑥 − 1) (𝑥 − 1)) = (−2)/(𝑥 − 1)^2 Hence, f’(x) = (−𝟐)/(𝒙 − 𝟏)^𝟐