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

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Example 30 Differentiate √(((π‘₯βˆ’3) (π‘₯2+4))/( 3π‘₯2+ 4π‘₯ + 5)) 𝑀.π‘Ÿ.𝑑. π‘₯. Let y =√(((π‘₯ βˆ’ 3) (π‘₯^2 + 4))/( 3π‘₯2+ 4π‘₯ + 5)) Taking log on both sides log⁑𝑦 = log √(((π‘₯ βˆ’ 3) (π‘₯2 + 4))/( 3π‘₯2+ 4π‘₯ + 5)) log⁑𝑦 = log⁑ (((π‘₯ βˆ’ 3) (π‘₯2 + 4))/( 3π‘₯2+ 4π‘₯ + 5))^(1/2) log⁑𝑦 = 1/2 log ((π‘₯ βˆ’ 3) (π‘₯2 + 4))/((3π‘₯2+ 4π‘₯ + 5) ) (π‘ˆπ‘ π‘–π‘›π‘” logβ‘γ€–π‘Ž^𝑏 γ€—=𝑏 logβ‘π‘Ž) π‘ˆπ‘ π‘–π‘›π‘” π‘™π‘œπ‘”β‘π‘Žπ‘=π‘™π‘œπ‘”β‘π‘Ž+π‘π‘œπ‘ β‘π‘ π‘Žπ‘›π‘‘ π‘™π‘œπ‘”β‘γ€–π‘Ž/𝑏〗=π‘™π‘œπ‘”β‘γ€–π‘Žβˆ’π‘™π‘œπ‘”β‘π‘ γ€— log⁑𝑦 = 1/2 (log⁑〖 (π‘₯βˆ’3)γ€—+γ€–log 〗⁑(π‘₯2 + 4)βˆ’log⁑〖 (3π‘₯2 + 4π‘₯ + 5)γ€— ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑 (log⁑𝑦))/𝑑π‘₯ = 1/2 ((𝑑(log⁑(π‘₯βˆ’3)+γ€–log 〗⁑〖(π‘₯^2+4)βˆ’log⁑〖 (3π‘₯^2+4π‘₯+5)γ€— γ€— ) )/𝑑π‘₯) (𝑑 (log⁑𝑦))/𝑑π‘₯ = 1/2 ((𝑑(log⁑(π‘₯βˆ’3)))/𝑑π‘₯ " + " 𝑑(log⁑(π‘₯^2+4) )/𝑑π‘₯ " βˆ’ " 𝑑(log⁑(3π‘₯^2+4π‘₯+5) )/𝑑π‘₯) 1/𝑦 (𝑑𝑦/𝑑π‘₯) = 1/2 [1/(π‘₯ βˆ’ 3) " . " 𝑑(π‘₯ βˆ’3)/𝑑π‘₯ " + " 1/(π‘₯^2 + 4) " . " (𝑑 (π‘₯^2 + 4))/𝑑π‘₯ " – " 1/((3π‘₯^2+ 4π‘₯+ 5)) " . " (𝑑 (3π‘₯^2+ 4π‘₯+ 5))/(𝑑π‘₯ )] 1/𝑦 (𝑑𝑦/𝑑π‘₯) = 1/2 [1/((π‘₯ βˆ’ 3) ) " " (1βˆ’0) "+ " 1/(π‘₯^2+ 4) " " (2π‘₯ + 0)" βˆ’ " 1/(3π‘₯^2+ 4π‘₯ + 5) " " (6π‘₯ +4+0)" " ] 1/𝑦 (𝑑𝑦/𝑑π‘₯) = 1/2 (1/((π‘₯ βˆ’ 3) )+2π‘₯/(π‘₯^2+ 4)βˆ’(6π‘₯ + 4)/(3π‘₯^2 + 4π‘₯ + 5)) 𝑑𝑦/𝑑π‘₯ = 𝑦/2 (1/((π‘₯ βˆ’ 3) )+2π‘₯/(π‘₯^2+ 4)βˆ’(6π‘₯ + 4)/(3π‘₯^2 + 4π‘₯ + 5)) π’…π’š/𝒅𝒙 = 𝟏/𝟐 √(((𝒙 βˆ’ πŸ‘)(𝒙^𝟐+ πŸ’))/(πŸ‘π’™^𝟐+ πŸ’π’™ + πŸ“)) (𝟏/((𝒙 βˆ’ πŸ‘) )+πŸπ’™/(𝒙^𝟐+ πŸ’)βˆ’(πŸ”π’™ + πŸ’)/(πŸ‘π’™^𝟐 + πŸ’π’™ + πŸ“))

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.