Ex 5.5

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.5, 6 Differentiate the functions in, (๐ฅ+1/๐ฅ)^๐ฅ+ ๐ฅ^((1 + 1/๐ฅ) ) Let ๐ฆ= (๐ฅ+1/๐ฅ)^๐ฅ+ ๐ฅ^((1 + 1/๐ฅ) ) Let ๐ข = (๐ฅ+1/๐ฅ)^๐ฅ , ๐ฃ = ๐ฅ^((1 + 1/๐ฅ) ) ๐ฆ = ๐ข+๐ฃ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ฅ = (๐ (๐ข + ๐ฃ))/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ Calculating ๐๐/๐๐ ๐ข = (๐ฅ+1/๐ฅ)^๐ฅ Taking log both sides logโก๐ข = log (๐ฅ+1/๐ฅ)^๐ฅ logโก๐ข = ๐ฅ log (๐ฅ+1/๐ฅ) Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ข )/๐๐ฅ = (๐ (๐ฅ log" " (๐ฅ + 1/๐ฅ)))/๐๐ฅ ๐(logโก๐ข )/๐๐ฅ (๐๐ข/๐๐ข) = (๐ (๐ฅ log" " (๐ฅ + 1/๐ฅ)))/๐๐ฅ (As ๐๐๐โก(๐^๐ )=๐ . ๐๐๐โก๐) ๐(logโก๐ข )/๐๐ข (๐๐ข/๐๐ฅ)" = " (๐ (๐ฅ log" " (๐ฅ + 1/๐ฅ)))/๐๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" = " (๐ (๐ฅ log" " (๐ฅ + 1/๐ฅ)))/๐๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" = " ๐(๐ฅ)/๐๐ฅ . log (๐ฅ + 1/๐ฅ) + ๐(log" " (๐ฅ + 1/๐ฅ))/๐๐ฅ . ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" 1. log (๐ฅ + 1/๐ฅ) + ((1/(๐ฅ + 1/๐ฅ)).๐/๐๐ฅ (๐ฅ + 1/๐ฅ)) . ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ + 1/๐ฅ) + (1/(๐ฅ + 1/๐ฅ) . (๐(๐ฅ)/๐๐ฅ+(๐ (1/๐ฅ))/๐๐ฅ)) . ๐ฅ Using product rule in ๐ฅ ๐๐๐" " (๐ฅ + 1/๐ฅ) As (uv)โ = uโ v + vโ u 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + (1/(๐ฅ + 1/๐ฅ) . (1+(โ1)/๐ฅ^2 " " )) . ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + (๐ฅ/(๐ฅ^2 + 1) . (1โ1/๐ฅ^2 " " )) . ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + (๐ฅ/(๐ฅ^2 + 1) ((๐ฅ^2 โ 1)/๐ฅ^2 ).๐ฅ) 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + (๐ฅ/(๐ฅ^2 + 1) ((๐ฅ^2 โ 1)/๐ฅ^2 ).๐ฅ) 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + (๐ฅ^2/๐ฅ^2 ((๐ฅ^2 โ 1)/(๐ฅ^2 + 1))) 1/๐ข (๐๐ข/๐๐ฅ)" =" log (๐ฅ+1/๐ฅ) + ((๐ฅ^2 โ 1)/(๐ฅ^2+ 1)) ๐๐ข/๐๐ฅ "= " ๐ข (ใlog ใโก(๐ฅ+1/๐ฅ)+((๐ฅ^2 โ 1)/(๐ฅ^2+ 1))) ๐๐ข/๐๐ฅ "=" (๐ฅ+1/๐ฅ)^๐ฅ (ใlog ใโก(๐ฅ+1/๐ฅ)+((๐ฅ^2 โ 1)/(๐ฅ^2+ 1))) ๐๐/๐๐ "=" (๐+๐/๐)^๐ ((๐^๐ โ ๐)/(๐^๐+ ๐)โกใ+ ใ๐๐๐ ใโก(๐+๐/๐) ใ ) Calculating ๐๐/๐๐ ๐ฃ = ๐ฅ^(1 + 1/๐ฅ)" " Taking log both sides log ๐ฃ = log ๐ฅ^(1 + 1/๐ฅ)" " log ๐ฃ = (1 + 1/๐ฅ)log ๐ฅ^" " Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ฃ )/๐๐ฅ = (๐ ((1 + 1/๐ฅ)" . " log ๐ฅ))/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฅ (๐๐ฃ/๐๐ฃ) = (๐ ((1 + 1/๐ฅ)" . " log ๐ฅ))/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฃ (๐๐ฃ/๐๐ฅ) = (๐ ((1 + 1/๐ฅ)" . " log ๐ฅ))/๐๐ฅ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (๐ ((1 + 1/๐ฅ)" . " log ๐ฅ))/๐๐ฅ Using product rule in (๐ฅ+ 1/๐ฅ)" . " ๐๐๐ ๐ฅ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ๐(1 + 1/๐ฅ)/๐๐ฅ . logโก๐ฅ + ๐(logโก๐ฅ )/๐๐ฅ . (1 + 1/๐ฅ) 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (๐(1)/๐๐ฅ+๐(1/๐ฅ)/๐๐ฅ) . logโก๐ฅ + 1/๐ฅ (1 + 1/๐ฅ) 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (0+((โ1)/๐ฅ^2 )) . logโก๐ฅ + 1/๐ฅ (1 + 1/๐ฅ) 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (โ1)/๐ฅ^2 . logโก๐ฅ + 1/๐ฅ (1 + 1/๐ฅ) 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (โlogโก๐ฅ)/๐ฅ^2 + 1/๐ฅ + 1/๐ฅ^2 1/๐ฃ (๐๐ฃ/๐๐ฅ) = (โlogโก๐ฅ)/๐ฅ^2 + 1/๐ฅ + 1/๐ฅ^2 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ((โlogโก๐ฅ + ๐ฅ + 1)/๐ฅ^2 ) ๐๐ฃ/๐๐ฅ = ๐ฃ ((โlogโก๐ฅ + ๐ฅ + 1)/๐ฅ^2 ) ๐๐ฃ/๐๐ฅ = ๐ฅ^((1 + 1/๐ฅ) ) ((๐ฅ + 1 โ logโก๐ฅ )/๐ฅ^2 ) Now ๐๐ฃ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ Putting values of ๐๐ข/๐๐ฅ & ๐๐ฃ/๐๐ฅ ๐๐/๐๐ = (๐+๐/๐)^๐ ((๐^๐ โ ๐)/(๐^๐+ ๐)+๐ฅ๐จ๐ โก(๐+ ๐/๐) ) + ๐^((๐ + ๐/๐) ) ((๐ + ๐ โ ๐๐๐โก๐ )/๐^๐ )