Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

### Transcript

Example 30 Find ๐๐ฆ/๐๐ฅ , if ๐ฆ^๐ฅ+๐ฅ^๐ฆ+๐ฅ^๐ฅ=๐^๐. Let u = ๐ฆ๐ฅ, v = ๐ฅ๐ฆ & w = ๐ฅ^๐ฅ Now, ๐ + ๐ + ๐ = ๐^๐ Differentiating ๐ค.๐.๐ก.๐ฅ (๐ (๐ข + ๐ฃ + ๐ค))/๐๐ฅ = (๐(๐^๐))/๐๐ฅ (๐(๐ข))/๐๐ฅ + (๐(๐ฃ))/๐๐ฅ + (๐(๐ค))/๐๐ฅ = 0 We will calculate derivative of u, v & w separately . Finding Derivative of 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 1/๐ข . ๐๐ข/๐๐ฅ = (๐ (๐ฅ . logโก๐ฆ ))/๐๐ฅ (๐ด๐  logโกใ(๐^๐)ใ=๐ logโก๐) By product Rule (uv)โ = uโv + vโu 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ฅ . 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 1/๐ข . ๐๐ข/๐๐ฅ = logโก๐ฆ + ๐ฅ/๐ฆ . ๐๐ฆ/๐๐ฅ ๐๐ข/๐๐ฅ = ๐ข (logโก๐ฆ "+ " ๐ฅ/๐ฆ " " ๐๐ฆ/๐๐ฅ) ๐๐/๐๐ = ๐^๐ (๐๐๐โก๐ "+ " ๐/๐ " " ๐๐/๐๐) Finding derivative of v v = xy 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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 By product Rule (uv)โ = uโv + vโu 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 + (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๐ฆ)/๐๐ฅ ๐ฅ+๐ฆ/๐ฅ) Putting values of ๐ฃ = ๐ฅ^๐ฆ ๐๐/๐๐ = ๐^๐ (๐๐/๐๐ ๐๐๐โกใ๐+ ๐/๐ใ ) Calculating derivative of ๐ ๐ค = ๐ฅ^๐ฅ Taking log both sides logโก๐ค=logโกใ (๐ฅ^๐ฅ)" " ใ logโก๐ค=ใ๐ฅ. logใโก๐ฅ" " Differentiating both sides ๐ค.๐.๐ก.๐ฅ (๐(logโก๐ค))/๐๐ฅ = (๐(๐ฅ . logโก๐ฅ))/๐๐ฅ (๐(logโก๐ค))/๐๐ฅ (๐๐ค/๐๐ค) = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ (๐(logโก๐ค))/๐๐ค . ๐๐ค/๐๐ฅ = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ (๐ด๐  logโกใ(๐^๐)ใ=๐ logโก๐) 1/๐ค . ๐๐ค/๐๐ฅ = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ logโก๐ค=ใ๐ฅ. logใโก๐ฅ" " Differentiating both sides ๐ค.๐.๐ก.๐ฅ (๐(logโก๐ค))/๐๐ฅ = (๐(๐ฅ . logโก๐ฅ))/๐๐ฅ (๐(logโก๐ค))/๐๐ฅ (๐๐ค/๐๐ค) = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ (๐(logโก๐ค))/๐๐ค . ๐๐ค/๐๐ฅ = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ 1/๐ค . ๐๐ค/๐๐ฅ = ๐(๐ฅ logโก๐ฅ )/๐๐ฅ By product Rule (uv)โ = uโv + vโu 1/๐ค (๐๐ค/๐๐ฅ) = ( ๐(๐ฅ))/๐๐ฅ . logโก๐ฅ + (๐ (logโก๐ฅ))/๐๐ฅ . ๐ฅ 1/๐ค (๐๐ค/๐๐ฅ) = 1 . logโก๐ฅ + 1/๐ฅ . ๐ฅ 1/๐ค (๐๐ค/๐๐ฅ) = (logโกใ๐ฅ+1ใ) ๐๐ค/๐๐ฅ = ๐ค(logโกใ๐ฅ+1ใ) ๐๐/๐๐ = ๐^๐ (๐๐๐โกใ๐+๐ใ ) From (1) ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ + ๐๐ค/๐๐ฅ = 0 Putting values from (2), (3) & (4) (๐ฆ^๐ฅ logโกใ๐ฆ+๐ฆ^(๐ฅโ1). ๐ฅ ๐๐ฆ/๐๐ฅ ใ ) + (๐ฅ^๐ฆ logโกใ๐ฅ.๐๐ฆ/๐๐ฅ+๐ฅ^๐ฆ.๐ฆ/๐ฅ ใ ) + (๐ฅ^๐ฅ (logโกใ๐ฅ+1ใ))=0(๐ฆ^๐ฅ logโกใ๐ฆ+๐ฅ^๐ฆ. ๐ฆ/๐ฅ+๐ฅ^๐ฅ (logโกใ๐ฅ+1ใ)ใ ) + (๐ฆ^(๐ฅโ1) .โกใ๐ฅ ๐๐ฆ/๐๐ฅ+๐ฅ^๐ฆ logโกใ๐ฅ ๐๐ฆ/๐๐ฅใ ใ ) = 0 (๐ฆ^(๐ฅโ1) .โกใ๐ฅ ๐๐ฆ/๐๐ฅ+๐ฅ^๐ฆ logโกใ๐ฆ ๐๐ฆ/๐๐ฅใ ใ ) = โ (๐ฆ^๐ฅ logโกใ๐ฆ+๐ฅ^๐ฆ. ๐ฆ/๐ฅ+๐ฅ^๐ฅ (logโกใ๐ฅ+1ใ)ใ ) (๐ฆ^(๐ฅโ1) .โกใ๐ฅ +๐ฅ^๐ฆ logโกใ๐ฅ ใ ใ ) ๐๐ฆ/๐๐ฅ = โ (๐ฆ^๐ฅ logโกใ๐ฆ+๐ฅ^๐ฆ. ๐ฆ/๐ฅ+๐ฅ^๐ฅ (logโกใ๐ฅ+1ใ)ใ ) ๐๐ฆ/๐๐ฅ = "โ" (๐ฆ^๐ฅ ๐๐๐โกใ๐ฆ + ๐ฅ^๐ฆ. ๐ฆ/๐ฅ + ๐ฅ^๐ฅ (1 + ๐๐๐โก๐ฅ)ใ )/((ใ๐ฅ๐ฆใ^(๐ฅโ1) +โกใ๐ฅ^๐ฆ ๐๐๐โกใ๐ฅ ใ ใ)) ๐๐/๐๐ = "โ" (๐^๐ ๐๐๐โกใ๐ + ๐^(๐ โ ๐) ๐ + ๐^๐ (๐ + ๐๐๐โก๐)ใ )/((ใ๐๐ใ^(๐โ๐) +โกใ๐^๐ ๐๐๐โกใ๐ ใ ใ))