# Misc 1 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Misc 1 Using differentials, find the approximate value of each of the following: (a) (17/81)^(1/4) (17/81)^(1/4) = (17)^(1/4)/(81)^(1/4) = (17)^(1/4)/3 Let x = 16 and โ๐ฅ=1 Since y = ๐ฅ^(1/4) ๐๐ฆ/๐๐ฅ = ๐(๐ฅ^(1/4) )/๐๐ฅ = 1/4 ๐ฅ^((โ3)/4) = 1/(4๐ฅ^(3/4) ) Now, โ๐ฆ = ๐๐ฆ/๐๐ฅ โ๐ฅ โ๐ฆ = (1/(4๐ฅ^(3/4) ))โ๐ฅ Putting values โ๐ฆ = 1/(4(16)^(3/4) ) (1) = 1/(4((16)^(1/4) )^3 ) = 1/(4(2)^3 ) = 1/32 Also, โ๐ฆ = f(x + โ๐ฅ) โ f(x) โ๐ฆ = ("x + " โ๐ฅ)^(1/4) โ ๐ฅ^(1/4) โ๐ฆ = ("16 + " 1)^(1/4) โ ใ(16)ใ^(1/4) โ๐ฆ = (17)^(1/4) โ ใ(16)ใ^(1/4) (17)^(1/4) = โ๐ฆ+(16)^(1/4) (17)^(1/4) = โ๐ฆ+2 (17)^(1/4) = 1/32+2 (17)^(1/4) = 0.03125 + 2 (17)^(1/4) = 2.03125 Now, Approximate value of (17/81)^(1/4) (17/81)^(1/4) = ((17)1/4)/3 = 2.03125/3 = 0.677 Hence approximate value of (17/81)^(1/4) is 0.677 (As approximate value of (17)^(1/4) = 2.03125) Misc 1 Using differentials, find the approximate value of each of the following: (b) ใ(33)ใ^(โ 1/5) ใ(33)ใ^(โ 1/5) = 1/(33)^(1/5) Let y = ๐ฅ^(1/5) & also let ๐ฅ = 32 & โ ๐ฅ = 1 Now, ๐ฆ = ๐ฅ^(1/5) (As (32)^(1/5)=2) Diff w.r.t x ๐๐ฆ/๐๐ฅ= (๐ (๐ฅ^(1/5)))/๐๐ฅ ๐๐ฆ/๐๐ฅ= 1/5 ๐ฅ^((1 )/5 โ1) ๐๐ฆ/๐๐ฅ = 1/5 ๐ฅ^((โ4)/5) ๐๐ฆ/๐๐ฅ= 1/ใ5๐ฅใ^(4/5) Using โy = ๐๐ฆ/๐๐ฅ โ๐ฅ โy = 1/ใ5๐ฅใ^(4/5) โ๐ฅ Putting values โy = 1/ใ5(32)ใ^(4/5) ร (1) โy = 1/ใ5(32)ใ^(4/5) โy = 1/ใ5(2)ใ^(5 ร 4/5) โy = 1/5(2^4 ) โy = 1/(5 ร 16) โy = 1/80 We know that โy = f(x + โx) โ f(x) So, โy = ใ(x+ ฮx)ใ^(1/5) ใโ๐ฅใ^(1/5) Putting values 1/80= ใ(32+1)ใ^(1/5) โ (32)^(1/5) 1/80= ใ(33)ใ^(1/5) โ ใ(2) ใ^(5 ร 1/5) 1/80= ใ(33)ใ^(1/5) โ 2 1/80+2= ใ(33)ใ^(1/5) (1 + 160)/80=ใ(33)ใ^(1/5) 161/80= ใ(33)ใ^(1/5) ใ(33)ใ^(1/5) = 161/80 But we need 1/(33)^(1/5) So 1/(33)^(1/5) = 1/(161/80) 1/(33)^(1/5) = 80/161 1/(33)^(1/5) = 0.497 (33)^((โ1)/5)= 0.497 Thus, the approximate value of (33)^((โ1)/5) ๐๐ ๐.๐๐๐

Miscellaneous

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Misc 4

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Misc 6 Important

Misc 7

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Misc 12 Important

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.