Ex 6.4, 1 (x) - Find approximate value of (401)^1/2 (Differentials)

Ex 6.4, 1 (x) - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.4, 1 (x) - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.4, 1 (x) - Chapter 6 Class 12 Application of Derivatives - Part 4

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (x) 〖(401)〗^(1/2)Let 𝑦=𝑥^( 1/2) where 𝑥=400 & ∆𝑥=1 Now, 𝑑𝑦/𝑑𝑥=1/(2√𝑥) Using ∆𝑦=𝑑𝑦/𝑑𝑥 ∆𝑥 ∆𝑦=𝑑𝑦/(2√𝑥) ∆𝑥 Putting Values ∆𝑦=1/(2√400) ×(1) ∆𝑦=1/(2√400) ×(1) ∆𝑦=1/(2√(20^2 )) ∆𝑦=1/(2 × 20) ∆𝑦=1/40 ∆𝑦=0. 025 We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦= (𝑥+∆𝑥)^( 1/2)−(𝑥)^( 1/2) Putting Values 0. 025=(400+1)^( 1/(2 ))−〖(400) 〗^(1/2) 0. 025=(401)^( 1/2)−(20)^( 2 × 1/2) 0. 025=(401)^( 1/2)−20 0. 025+20=(401)^( 1/2) 20. 025=(401)^( 1/2) We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦= (𝑥+∆𝑥)^( 1/2)−(𝑥)^( 1/2) Putting Values 0. 025=(400+1)^( 1/(2 ))−〖(400) 〗^(1/2) 0. 025=(401)^( 1/2)−(20)^( 2 × 1/2) 0. 025=(401)^( 1/2)−20 0. 025+20=(401)^( 1/2) 20. 025=(401)^( 1/2)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo