Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

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Example 42 For a positive constant a find ๐๐ฆ/๐๐ฅ , where ๐ฆ = ๐^(๐ก+1/๐ก) , and ๐ฅ =(๐ก+1/๐ก)^2 Here ๐๐/๐๐ = (๐๐/๐๐)/(๐๐/๐๐) Calculating ๐๐/๐๐ ๐ฆ=๐^(๐ก + 1/๐ก) Differentiating ๐ค.๐.๐ก. t ๐๐/๐๐ = ๐(๐^((๐ + ๐/๐) ) )/๐๐ ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐.๐(๐ก + 1/๐ก)/๐๐ก ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐.(1+(โ1) ๐ก^(โ2) ) ๐๐/๐๐ = ๐^((๐ + ๐/๐) ) .๐๐๐โก๐.(๐โ๐/๐^๐ ) "As " ๐(๐^๐ฅ )/๐๐ฅ " = " ๐^๐ฅ.๐๐๐โก๐ Calculating ๐๐/๐๐ ๐ฅ=(๐ก+1/๐ก)^๐ Differentiating ๐ค.๐.๐ก. t ๐๐ฅ/๐๐ก = ๐((๐ก + 1/๐ก)^(๐ ) )/๐๐ก ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . ๐(๐ก + 1/๐ก)/๐๐ก ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ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 โ1 ) . (1+ ๐(๐ก^(โ1) )/๐๐ก) ๐๐ฅ/๐๐ก = a ๐^(๐ โ1 ) . ๐(๐)/๐๐ก ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (1+(โ1) ใ ๐กใ^(โ2) ) ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (1โ 1/๐ก^2 ) Calculating ๐๐/๐๐ 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^(๐ก + 1/๐ก) . logโกใ๐ ใ ร (1 โ 1/๐ก^2 ))/(๐(๐ก + 1/๐ก)^(๐ โ 1) (1 โ 1/๐ก^2 ).) ๐๐/๐๐ = (๐^(๐ + ๐/๐) . ๐๐๐โกใ๐ ใ)/(๐(๐ + ๐/๐)^(๐ โ ๐) )