Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12





Last updated at Jan. 3, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
Transcript
Example 47 For a positive constant a find ๐๐ฆ/๐๐ฅ , where ๐ฆ = ๐^(๐ก+1/๐ก) , and ๐ฅ =(๐ก+1/๐ก)^2 We need to find ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐๐ฆ/๐๐ฅ ร ๐๐ก/๐๐ก ๐๐ฆ/๐๐ฅ = ๐๐ฆ/๐๐ก ร ๐๐ก/๐๐ฅ ๐ ๐/๐ ๐ = (๐ ๐/๐ ๐)/(๐ ๐/๐ ๐) Calculating ๐ ๐/๐ ๐ ๐ฆ=๐^(๐ก+1/๐ก) Let (๐ก+1/๐ก) = ๐ ๐ฆ=๐^๐ Differentiating ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ก = ๐(๐^๐ )/๐๐ก ๐๐ฆ/๐๐ก = ๐(๐^๐ )/๐๐ก . ๐๐/๐๐ ๐๐ฆ/๐๐ก = ๐(๐^๐ )/๐๐ . ๐๐/๐๐ก ๐๐ฆ/๐๐ก = ๐^๐ .logโก๐. ๐(๐)/๐๐ก "As " ๐(๐^๐ฅ )/๐๐ฅ " = " ๐^๐ฅ.๐๐๐โก๐ Calculated in eg31 Putting values of p = ๐ก+1/๐ก ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐. ๐(๐ก + 1/๐ก)/๐๐ก ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐. (๐(๐ก)/๐๐ก + ๐(1/๐ก)/๐๐ก) ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐. (1+ ๐(๐ก^(โ1) )/๐๐ก) ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐.(1+(โ1) ๐ก^(โ1โ1) ) ๐๐ฆ/๐๐ก = ๐^((๐ก + 1/๐ก) ) .logโก๐.(1+(โ1) ๐ก^(โ2) ) ๐ ๐/๐ ๐ = ๐^((๐ + ๐/๐) ) .๐๐๐โก๐.(๐โ๐/๐^๐ ) Calculating ๐ ๐/๐ ๐ ๐ฅ=(๐ก+1/๐ก)^๐ Let (๐ก+1/๐ก)=๐ ๐ฅ=๐^๐ Differentiating ๐ค.๐.๐ก.๐ฅ. ๐๐ฅ/๐๐ก = ๐(๐^๐ )/๐๐ก ๐๐ฅ/๐๐ก = ๐(๐^๐ )/๐๐ก . ๐๐/๐๐ ๐๐ฅ/๐๐ก = ๐(๐^๐ )/๐๐ . ๐๐/๐๐ก ๐๐ฅ/๐๐ก = a ๐^(๐ โ1 ) . ๐(๐)/๐๐ก "As " ๐(๐ฅ^๐ )/๐๐ฅ " = " ใ๐๐ฅใ^(๐ โ 1) ๐(๐^๐ )/๐๐ = ใ๐๐ใ^(๐ โ 1) Putting value of p = ๐ก+1/๐ก ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . ๐(๐ก + 1/๐ก)/๐๐ก ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (๐(๐ก)/๐๐ก + ๐(1/๐ก)/๐๐ก) ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (1+ ๐(๐ก^(โ1) )/๐๐ก) ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (1+(โ1) ใ ๐กใ^(โ2) ) ๐๐ฅ/๐๐ก = a (๐ก+1/๐ก)^(๐ โ1 ) . (1โ 1/๐ก^2 ) Now, ๐๐ฆ/๐๐ฅ = (๐๐ฆ/๐๐ก)/(๐๐ฅ/๐๐ก) ๐๐ฆ/๐๐ฅ = (๐^(๐ก + 1/๐ก) . logโกใ๐ ใ (1 โ 1/๐ก^2 ))/(๐(๐ก + 1/๐ก)^(๐ โ 1) (1 โ 1/๐ก^2 ).) ๐ ๐/๐ ๐ = (๐^(๐ + ๐/๐) . ๐๐๐โกใ๐ ใ)/(๐(๐ + ๐/๐)^(๐ โ ๐) )
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