# Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) ใ(32.15)ใ^(1/5)Let ๐ฆ=(๐ฅ)^( 1/5) where ๐ฅ=32 & โ๐ฅ=0. 15 Now, ๐ฆ=(๐ฅ)^( 1/5) Differentiating w.r.t.๐ฅ ๐๐ฆ/๐๐ฅ=๐(๐ฅ^( 1/5) )/๐๐ฅ=1/5 ๐ฅ^( 1/5 โ 1) =1/5 ๐ฅ^( (โ 4)/( 5) )=1/(5ใ ๐ฅใ^( 4/5 ) ) Using โ๐ฆ=๐๐ฆ/๐๐ฅ โ๐ฅ โ๐ฆ=1/(5 ใ๐ฅ ใ^(4/5) ) โ๐ฅ Putting Values โ๐ฆ=1/(5(32)^( 4/5) ) ร (0. 15) โ๐ฆ=1/(5(2)^( 5 ร 4/5) ) ร (0. 15) โ๐ฆ=1/(5(2)^( 4) ) ร (0. 15) โ๐ฆ=1/(5 ร 16) ร 0. 15 โ๐ฆ=(0. 15" " )/80 โ๐ฆ=0. 00187 We know that โ๐ฆ=๐(๐ฅ+โ๐ฅ)โ๐(๐ฅ) So, โ๐ฆ=(๐ฅ+โ๐ฅ)^( 1/5)โใ๐ฅ ใ^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)โ(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)โ(2)^(5 ร 1/5 ) 0. 00187=(32. 15)^( 1/5)โ2 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 ๐. ๐๐๐๐๐ Hence, Approximate Values of (32. 15)^( 1/5) is ๐. ๐๐๐๐๐

Ex 6.4

Ex 6.4, 1 (i)
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.