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Ex 6.4, 1 (xi) - Find approximate value of (0.0037)^1/2 - Teachoo

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

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Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xi) 〖(0.0037)〗^(1/2)Let y = √𝑥 Let x = 0.0036 & △ x = 0.0001 Since y = √𝑥 𝑑𝑦/𝑑𝑥 = (𝑑(√𝑥))/𝑑𝑥 = 1/(2√𝑥) Now, ∆𝑦 = 𝑑𝑦/𝑑𝑥 △x = 1/(2√𝑥) (0.0001) = 1/(2√0.0036) (0.0001) = 1/(2 × 0.06) × (0.0001) = 0.0001/0.12 = 1/1200 = 0.000833 Also, ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦=(𝑥+∆𝑥)^( 1/2)−(𝑥)^( 1/2) Putting Values 0. 000833=(0. 0036+0. 0001)^( 1/2)−(0. 0036)^( 1/2) 0. 000833=(0. 0037)^( 1/2)−(36/10000)^(1/2) 0. 000833=(0. 0037)^( 1/2)−(6/100)^( 2 × 1/2) 0. 000833=(0. 0037)^( 1/2)−(0. 06)^(2 × 1/2 ) 0. 000833=(0. 0037)^( 1/2)−(0. 06) 0. 000833+0. 06=(0. 0037)^( 1/2) 0. 060833=(0. 0037)^( 1/2) (0. 0037)^( 1/2)=0. 060833 Thus, the Approximate Value of (0. 0037)^( 1/2)=𝟎. 𝟎𝟔𝟎𝟖𝟑𝟑

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.