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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Example 47 For a positive constant a find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ , where ๐‘ฆ = ๐‘Ž^(๐‘ก+1/๐‘ก) , and ๐‘ฅ =(๐‘ก+1/๐‘ก)^2 We need to find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ก/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ฆ/๐‘‘๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐’…๐’š/๐’…๐’™ = (๐’…๐’š/๐’…๐’•)/(๐’…๐’™/๐’…๐’•) Calculating ๐’…๐’š/๐’…๐’• ๐‘ฆ=๐‘Ž^(๐‘ก+1/๐‘ก) Let (๐‘ก+1/๐‘ก) = ๐‘ ๐‘ฆ=๐‘Ž^๐‘ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘‘(๐‘Ž^๐‘ )/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘‘(๐‘Ž^๐‘ )/๐‘‘๐‘ก . ๐‘‘๐‘/๐‘‘๐‘ ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘‘(๐‘Ž^๐‘ )/๐‘‘๐‘ . ๐‘‘๐‘/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^๐‘ .logโก๐‘Ž. ๐‘‘(๐‘)/๐‘‘๐‘ก "As " ๐‘‘(๐‘Ž^๐‘ฅ )/๐‘‘๐‘ฅ " = " ๐‘Ž^๐‘ฅ.๐‘™๐‘œ๐‘”โก๐‘Ž Calculated in eg31 Putting values of p = ๐‘ก+1/๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^((๐‘ก + 1/๐‘ก) ) .logโก๐‘Ž. ๐‘‘(๐‘ก + 1/๐‘ก)/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^((๐‘ก + 1/๐‘ก) ) .logโก๐‘Ž. (๐‘‘(๐‘ก)/๐‘‘๐‘ก + ๐‘‘(1/๐‘ก)/๐‘‘๐‘ก) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^((๐‘ก + 1/๐‘ก) ) .logโก๐‘Ž. (1+ ๐‘‘(๐‘ก^(โˆ’1) )/๐‘‘๐‘ก) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^((๐‘ก + 1/๐‘ก) ) .logโก๐‘Ž.(1+(โˆ’1) ๐‘ก^(โˆ’1โˆ’1) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = ๐‘Ž^((๐‘ก + 1/๐‘ก) ) .logโก๐‘Ž.(1+(โˆ’1) ๐‘ก^(โˆ’2) ) ๐’…๐’š/๐’…๐’• = ๐’‚^((๐’• + ๐Ÿ/๐’•) ) .๐’๐’๐’ˆโก๐’‚.(๐Ÿโˆ’๐Ÿ/๐’•^๐Ÿ ) Calculating ๐’…๐’™/๐’…๐’• ๐‘ฅ=(๐‘ก+1/๐‘ก)^๐‘Ž Let (๐‘ก+1/๐‘ก)=๐‘ ๐‘ฅ=๐‘^๐‘Ž Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฅ/๐‘‘๐‘ก = ๐‘‘(๐‘^๐‘Ž )/๐‘‘๐‘ก ๐‘‘๐‘ฅ/๐‘‘๐‘ก = ๐‘‘(๐‘^๐‘Ž )/๐‘‘๐‘ก . ๐‘‘๐‘/๐‘‘๐‘ ๐‘‘๐‘ฅ/๐‘‘๐‘ก = ๐‘‘(๐‘^๐‘Ž )/๐‘‘๐‘ . ๐‘‘๐‘/๐‘‘๐‘ก ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a ๐‘^(๐‘Ž โˆ’1 ) . ๐‘‘(๐‘)/๐‘‘๐‘ก "As " ๐‘‘(๐‘ฅ^๐‘› )/๐‘‘๐‘ฅ " = " ใ€–๐‘›๐‘ฅใ€—^(๐‘› โˆ’ 1) ๐‘‘(๐‘^๐‘Ž )/๐‘‘๐‘ = ใ€–๐‘Ž๐‘ใ€—^(๐‘Ž โˆ’ 1) Putting value of p = ๐‘ก+1/๐‘ก ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a (๐‘ก+1/๐‘ก)^(๐‘Ž โˆ’1 ) . ๐‘‘(๐‘ก + 1/๐‘ก)/๐‘‘๐‘ก ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a (๐‘ก+1/๐‘ก)^(๐‘Ž โˆ’1 ) . (๐‘‘(๐‘ก)/๐‘‘๐‘ก + ๐‘‘(1/๐‘ก)/๐‘‘๐‘ก) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a (๐‘ก+1/๐‘ก)^(๐‘Ž โˆ’1 ) . (1+ ๐‘‘(๐‘ก^(โˆ’1) )/๐‘‘๐‘ก) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a (๐‘ก+1/๐‘ก)^(๐‘Ž โˆ’1 ) . (1+(โˆ’1) ใ€– ๐‘กใ€—^(โˆ’2) ) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = a (๐‘ก+1/๐‘ก)^(๐‘Ž โˆ’1 ) . (1โˆ’ 1/๐‘ก^2 ) Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘๐‘ฆ/๐‘‘๐‘ก)/(๐‘‘๐‘ฅ/๐‘‘๐‘ก) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘Ž^(๐‘ก + 1/๐‘ก) . logโกใ€–๐‘Ž ใ€— (1 โˆ’ 1/๐‘ก^2 ))/(๐‘Ž(๐‘ก + 1/๐‘ก)^(๐‘Ž โˆ’ 1) (1 โˆ’ 1/๐‘ก^2 ).) ๐’…๐’š/๐’…๐’™ = (๐’‚^(๐’• + ๐Ÿ/๐’•) . ๐’๐’๐’ˆโกใ€–๐’‚ ใ€—)/(๐’‚(๐’• + ๐Ÿ/๐’•)^(๐’‚ โˆ’ ๐Ÿ) )

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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.