Ex 6.4, 1 (xiii) - Find approximate value upto 3 decimals - (81.5)^1/4

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

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiii) ใ€–(81.5)ใ€—^(1/4)Let ๐‘ฆ=(๐‘ฅ)^( 1/4) where ๐‘ฅ=81 & โˆ†๐‘ฅ=0. 5 Now, ๐‘ฆ=๐‘ฅ^( 1/4) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/4) )/๐‘‘๐‘ฅ=1/4 ๐‘ฅ^( 1/4 โˆ’ 1)=1/4 ๐‘ฅ^( (โˆ’ 3)/( 4)) =1/(4๐‘ฅ^(3/4) ) โˆ† ๐‘ฅ Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=๐‘‘๐‘ฆ/(4๐‘ฅ^(3/4) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(4(81)^( 3/4) ) ร— (0. 5) โˆ†๐‘ฆ=(0. 5)/(4 ร— (3^( 4) )^( 3/4) ) โˆ†๐‘ฆ=(0. 5)/(4 ร— 3^( 3) ) โˆ†๐‘ฆ=(0. 5)/(4 ร— 27) โˆ†๐‘ฆ=(0. 5)/108 โˆ†๐‘ฆ=0. 0046 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/4)โˆ’๐‘ฅ^( 1/4) Putting Values 0. 0046=(81+0. 5)^( 1/4)โˆ’(81)^( 1/4) 0. 0046=(81. 5)^( 1/4)โˆ’(3^4 )^( 1/4) 0. 0046=(81. 5)^( 1/4)โˆ’3 0. 0046+3=(81. 5)^( 1/4) 3. 0046=(81. 5)^( 1/4) Thus, Approximate Value of (81. 5)^( 1/4) is ๐Ÿ‘. ๐ŸŽ๐ŸŽ๐Ÿ’๐Ÿ”

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