Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

### Transcript

Ex 5.5, 12 Find ๐๐ฆ/๐๐ฅ of the functions in, ๐ฅ^๐ฆ + ๐ฆ^๐ฅ = 1 ๐ฅ^๐ฆ + ๐ฆ^๐ฅ = 1 Let ๐ข = ๐ฅ^๐ฆ , ๐ฃ = ๐ฆ^๐ฅ Hence, ๐ข+๐ฃ=1 Differentiating both sides ๐ค.๐.๐ก.๐ฅ. (๐(๐ฃโกใ+ ๐ขใ))/๐๐ฅ = ๐(1)/๐๐ฅ ๐๐ฃ/๐๐ฅ + ๐๐ข/๐๐ฅ = 0 (Derivative of constant is 0) Calculating ๐๐/๐๐ ๐ฃ=๐ฅ^๐ฆ Taking log both sides logโก๐ฃ=logโกใ (๐ฅ^๐ฆ)" " ใ logโก๐ฃ=ใ๐ฆ. logใโก๐ฅ" " Differentiating both sides ๐ค.๐.๐ก.๐ฅ (๐(logโก๐ฃ))/๐๐ฅ = (๐(๐ฆ . logโก๐ฅ))/๐๐ฅ (๐(logโก๐ฃ))/๐๐ฅ (๐๐ฃ/๐๐ฃ) = ๐(๐ฆ logโก๐ฅ )/๐๐ฅ (๐(logโก๐ฃ))/๐๐ฃ (๐๐ฃ/๐๐ฅ) = ๐(ใ๐ฆ logใโก๐ฅ )/๐๐ฅ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ( ๐(ใ๐ฆ logใโก๐ฅ ))/๐๐ฅ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ( ๐(๐ฆ))/๐๐ฅ . logโก๐ฅ + (๐ (logโก๐ฅ))/๐๐ฅ . ๐ฆ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ( ๐๐ฆ)/๐๐ฅ . logโก๐ฅ + 1/๐ฅ . ๐ฆ 1/๐ฃ (๐๐ฃ/๐๐ฅ) = ( ๐๐ฆ)/๐๐ฅ logโก๐ฅ + ๐ฆ/๐ฅ ๐๐ฃ/๐๐ฅ = v (log ๐ฅ+๐ฆ/๐ฅ) Putting value of ๐ฃ = ๐ฅ^๐ฆ ๐๐ฃ/๐๐ฅ = ๐ฅ^๐ฆ (logโกใ๐ฅ+ ๐ฆ/๐ฅใ ) ๐๐ฃ/๐๐ฅ = ๐ฅ^๐ฆ.log ๐ฅ . ๐๐ฆ/๐๐ฅ + ๐ฅ^๐ฆ. ๐ฆ/๐ฅ Using product Rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข Calculating ๐๐/๐๐ ๐ข = ๐ฆ^๐ฅ Taking log both sides logโก๐ข=logโกใ (๐ฆ^๐ฅ)" " ใ logโก๐ข=ใ๐ฅ . logใโก๐ฆ" " Differentiating both sides ๐ค.๐.๐ก.๐ฅ (๐(logโก๐ข))/๐๐ฅ = (๐(๐ฅ . logโก๐ฆ))/๐๐ฅ (๐(logโก๐ข))/๐๐ฅ (๐๐ข/๐๐ข) = ๐(๐ฅ.logโก๐ฆ )/๐๐ฅ (๐ด๐  logโกใ(๐^๐)ใ=๐ logโก๐ ) (๐(logโก๐ข))/๐๐ข (๐๐ข/๐๐ฅ) = (๐ (๐ฅ . logโก๐ฆ))/๐๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = (๐ (๐ฅ . logโก๐ฆ ))/๐๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = (๐ (๐ฅ))/๐๐ฅ . logโก๐ฆ + (๐(logโก๐ฆ))/๐๐ฅ . ๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = 1 . logโก๐ฆ + ๐ฅ. ๐(logโก๐ฆ )/๐๐ฅ . ๐๐ฆ/๐๐ฆ 1/๐ข . ๐๐ข/๐๐ฅ = logโก๐ฆ + ๐ฅ. ๐(logโก๐ฆ )/๐๐ฆ . ๐๐ฆ/๐๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = logโก๐ฆ + ๐ฅ. 1/๐ฆ . ๐๐ฆ/๐๐ฅ Using product Rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข ๐๐ข/๐๐ฅ = ๐ข(logโก๐ฆ " + " ๐ฅ/๐ฆ " . " (๐๐ฆ )/๐๐ฅ) ๐๐ข/๐๐ฅ " = " ๐ฆ^๐ฅ (logโก๐ฆ " + " ๐ฅ/๐ฆ " . " (๐๐ฆ )/๐๐ฅ) ๐๐ข/๐๐ฅ = ๐ฆ^๐ฅ logโก๐ฆ+๐ฆ^(๐ฅ โ1). ๐ฅ ๐๐ฆ/๐๐ฅ Now, ๐๐ฃ/๐๐ฅ + ๐๐ข/๐๐ฅ = 0 Putting value of ๐๐ฃ/๐๐ฅ & ๐๐ข/๐๐ฅ (๐ฅ^๐ฆ logโกใ๐ฅ ใ ๐๐ฆ/๐๐ฅ +๐ฅ^๐ฆ ๐ฆ/๐ฅ) + (๐ฆ^๐ฅ logโกใ๐ฆ+๐ฆ^(๐ฅ โ1) ใ.๐ฅ ๐๐ฆ/๐๐ฅ) = 0 ๐ฅ^๐ฆ logโกใ๐ฅ ใ ๐๐ฆ/๐๐ฅ +๐ฅ^(๐ฆโ1) ๐ฆ + ๐ฆ^๐ฅ logโกใ๐ฆ+๐ฆ^(๐ฅ โ1) ใ.๐ฅ ๐๐ฆ/๐๐ฅ = 0 ๐ฅ^๐ฆ logโกใ๐ฅ ใ ๐๐ฆ/๐๐ฅ+๐ฆ^(๐ฅ โ1).๐ฅ ๐๐ฆ/๐๐ฅ=โ(๐ฅ^(๐ฆโ1) ๐ฆ+๐ฆ^๐ฅ logโก๐ฆ ) ๐๐ฆ/๐๐ฅ(๐ฅ^๐ฆ logโกใ๐ฅ ใ+๐ฆ^(๐ฅ โ1).๐ฅ)=โ(๐ฅ^(๐ฆโ1) ๐ฆ+๐ฆ^๐ฅ logโก๐ฆ ) ๐๐/๐๐ = (โ(๐^(๐โ๐) ๐ + ๐^๐ ๐๐๐โก๐ ))/((๐^๐ ๐๐๐โกใ๐ ใ+ ๐^(๐ โ๐).๐))