Ex 6.4, 1 (xv) - Find approximate value of (32.15)^1/5 (upto 3 decimal

Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.4, 1 (xv) - 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. (xv) 〖(32.15)〗^(1/5)Let 𝑦=(𝑥)^( 1/5) where 𝑥=32 & ∆𝑥=0. 15 Now, 𝑦=(𝑥)^( 1/5) Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^( 1/5) )/𝑑𝑥=1/5 𝑥^( 1/5 − 1) =1/5 𝑥^( (− 4)/( 5) )=1/(5〖 𝑥〗^( 4/5 ) ) Using ∆𝑦=𝑑𝑦/𝑑𝑥 ∆𝑥 ∆𝑦=1/(5 〖𝑥 〗^(4/5) ) ∆𝑥 Putting Values ∆𝑦=1/(5(32)^( 4/5) ) × (0. 15) ∆𝑦=1/(5(2)^( 5 × 4/5) ) × (0. 15) ∆𝑦=1/(5(2)^( 4) ) × (0. 15) ∆𝑦=1/(5 × 16) × 0. 15 ∆𝑦=(0. 15" " )/80 ∆𝑦=0. 00187 We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦=(𝑥+∆𝑥)^( 1/5)−〖𝑥 〗^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)−(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)−(2)^(5 × 1/5 ) 0. 00187=(32. 15)^( 1/5)−2 0. 00187+2=(32. 15)^( 1/5) 2. 00187=(32. 15)^( 1/5) (32. 15)^( 1/5)=2. 00187 Hence, Approximate Values of (32. 15)^( 1/5) is 𝟐. 𝟎𝟎𝟏𝟖𝟕 Hence, Approximate Values of (32. 15)^( 1/5) is 𝟐. 𝟎𝟎𝟏𝟖𝟕

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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