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  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Ex 13.2, 4 Find the derivative of the following functions from first principle. (i) x3 – 27 Let f(x) = x3 – 27 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 f (x) = x3 – 27 f (x + h) = (x + h)3 – 27 Putting values f’(x) = lim┬(h→0)⁡〖(((x + h)3 − 27) − (x3 − 27))/h〗 = lim┬(h→0)⁡〖((x + h)3 − 27− x3 + 27)/h〗 = lim┬(h→0)⁡〖((x + h)3 − x3 − 27 + 27)/h〗 = lim┬(h→0)⁡〖((x + h)3 − x3 )/h〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖(𝑥3 + ℎ3 + 3𝑥2 ℎ + 3𝑥ℎ2 − 𝑥3)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖(ℎ3 + 3𝑥2 ℎ + 3𝑥ℎ2 − 𝑥3 + 𝑥3)/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖(ℎ ( ℎ2 +3𝑥2 + 3𝑥ℎ) )/ℎ〗 = lim┬(h→0)⁡〖ℎ2+3𝑥2+3𝑥ℎ〗 Putting h = 0 = (0)2 + 3x2 + 3x(0) = 0 + 3x2 + 0 = 3x2 Hence, f’(x) = 3x2 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 Ex 13.2, 4 Find the derivative of the following functions from first principle. (iii) 1/𝑥^2 Let f (x) = 1/𝑥^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) = 1/𝑥^2 So, f (x + h) = 1/〖(𝑥 + ℎ)〗^2 Putting values f’(x) = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖(1/〖(𝑥 + ℎ)〗^2 − 1/𝑥^2 )/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖(〖〖𝑥 〗^2 − (𝑥 + ℎ)〗^2/(〖(𝑥 + ℎ)〗^2 𝑥^2 ))/ℎ〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖( 𝑥2 − (𝑥 + ℎ)2)/(ℎ𝑥2 (𝑥 + ℎ)2)〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖((𝑥 − ( 𝑥 + ℎ )) (𝑥 + (𝑥 + ℎ)))/(ℎ𝑥2 (𝑥 + ℎ)2)〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖( (𝑥 −𝑥 − ℎ) (𝑥 + 𝑥 + ℎ))/(ℎ.𝑥2 (𝑥 + ℎ)2)〗 (As a2 – b2 = (a – b) (a + b)) = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖((−ℎ)(2𝑥 + ℎ))/(ℎ𝑥2 (𝑥 + ℎ)2)〗 = (𝑙𝑖𝑚)┬(ℎ→0)⁡〖((−1) 2𝑥 + ℎ)/(𝑥2 (𝑥 + ℎ)2)〗 Putting h = 0 = ((−1) 2𝑥 + 0)/(𝑥2 (𝑥 + 0)2) = ((−1) 2𝑥)/(𝑥2 (𝑥)2) = (−2𝑥)/𝑥4 = (−2)/𝑥^3 Thus, f’(x) = (−𝟐)/𝒙^𝟑 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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.