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Misc  11 - Differentiate the function - x^(x^2-3) + (x-3)^(x^2)

Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc  11 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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Misc 11 Differentiate w.r.t. x the function, π‘₯^(π‘₯^2βˆ’ 3)+(π‘₯βˆ’3)π‘₯^2, for π‘₯ > 3Let 𝑦=π‘₯^(π‘₯^2βˆ’ 3)+(π‘₯βˆ’3)^(π‘₯^2 ) And let 𝑒=π‘₯^(π‘₯^2βˆ’ 3) , 𝑣 =(π‘₯βˆ’3)^(π‘₯^2 ) Now, π’š = 𝒖+𝒗 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 = π‘₯^(π‘₯^2βˆ’ 3) Taking log on both sides log 𝑒=log⁑〖π‘₯^(π‘₯^2βˆ’ 3) γ€— log 𝑒=γ€–(π‘₯γ€—^2βˆ’ 3). log⁑π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑π‘₯ = 𝑑(γ€–(π‘₯γ€—^2βˆ’ 3) log⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(γ€–(π‘₯γ€—^2 βˆ’ 3) log⁑π‘₯ )/𝑑π‘₯ " " (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(γ€–(π‘₯γ€—^2βˆ’ 3) log⁑π‘₯ )/𝑑π‘₯ " " 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(γ€–(π‘₯γ€—^2βˆ’ 3) log⁑π‘₯ )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑〖(π‘₯γ€—^2βˆ’ 3) )/𝑑π‘₯ . γ€– log〗⁑π‘₯ + 𝑑(log⁑π‘₯ )/𝑑π‘₯ . γ€–(π‘₯γ€—^2βˆ’ 3) 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = (2π‘₯ βˆ’0) γ€– log〗⁑π‘₯ + 1/π‘₯ Γ— γ€–(π‘₯γ€—^2βˆ’ 3) 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 2π‘₯ . log⁑π‘₯ + (π‘₯^2βˆ’ 3)/π‘₯ 𝑑𝑒/𝑑π‘₯ = u (2π‘₯ "." log⁑π‘₯ "+ " (π‘₯^2βˆ’ 3)/π‘₯) Using Product rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝒅𝒖/𝒅𝒙 = 𝒙^(𝒙^πŸβˆ’ πŸ‘) (πŸπ’™ "." π’π’π’ˆβ‘π’™ "+ " (𝒙^πŸβˆ’ πŸ‘)/𝒙) Calculating 𝒅𝒗/𝒅𝒙 𝑣 = (π‘₯βˆ’3)π‘₯^2 Taking log on both sides log 𝑣=log⁑〖(π‘₯βˆ’3)^(π‘₯^2 ) γ€— log 𝑣=γ€–π‘₯^2 . log〗⁑〖 (π‘₯βˆ’3)γ€— Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑣 )/𝑑π‘₯ = (𝑑(γ€–π‘₯^2. log〗⁑〖 (π‘₯ βˆ’ 3)γ€— ) )/𝑑π‘₯ (As log⁑(π‘Ž^𝑏) = 𝑏 logβ‘π‘Ž) 𝑑(log⁑𝑣 )/𝑑π‘₯ . 𝑑𝑣/𝑑𝑣 = (𝑑(γ€–π‘₯^2. log〗⁑〖 (π‘₯βˆ’3)γ€— ) )/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑𝑣 . 𝑑𝑣/𝑑π‘₯ = (𝑑(γ€–π‘₯^2. log〗⁑〖 (π‘₯βˆ’3)γ€— ) )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = (𝑑(γ€–π‘₯^2. log〗⁑〖 (π‘₯βˆ’3)γ€— ) )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(π‘₯^2 )/𝑑π‘₯ . log (π‘₯βˆ’3) + 𝑑(log" " (π‘₯ βˆ’ 3))/𝑑π‘₯ . π‘₯^2 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 2π‘₯ . log (π‘₯βˆ’3) + 1/((π‘₯ βˆ’ 3) ). (𝑑(π‘₯ βˆ’ 3)" " )/𝑑π‘₯ . π‘₯^2 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 2π‘₯ . log (π‘₯βˆ’3) + 1/((π‘₯ βˆ’ 3) ) . π‘₯^2 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 2π‘₯. log (π‘₯βˆ’3) + π‘₯^2/(π‘₯ βˆ’3)Using product rule (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑𝑣/𝑑π‘₯ = 𝑣 (2π‘₯". " log" " (π‘₯βˆ’3)" + " π‘₯^2/(π‘₯ βˆ’3)) 𝒅𝒗/𝒅𝒙 = (π’™βˆ’πŸ‘)𝒙^𝟐 (πŸπ’™". " π₯𝐨𝐠" " (π’™βˆ’πŸ‘)" + " 𝒙^𝟐/(𝒙 βˆ’πŸ‘)) Now, 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ = 𝒙^(𝒙^πŸβˆ’ πŸ‘) ((𝒙^πŸβˆ’ πŸ‘)/𝒙+πŸπ’™ π₯𝐨𝐠⁑𝒙 ) + (π’™βˆ’πŸ‘)𝒙^𝟐 (𝒙^𝟐/(𝒙 βˆ’πŸ‘)+πŸπ’™ .π₯𝐨𝐠⁑(𝒙 βˆ’πŸ‘) )

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