Examples

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

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) )

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.