Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity      1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Miscellaneous

Transcript

Misc 11 Differentiate w.r.t. x the function, 𝑥^(𝑥^2− 3)+(𝑥−3)𝑥^2, for 𝑥 > 3 Let 𝑦=𝑥^(𝑥^2− 3)+(𝑥−3)^(𝑥^2 ) Let 𝑢=𝑥^(𝑥^2− 3) , 𝑣 =(𝑥−3)^(𝑥^2 ) ∴ 𝑦 = 𝑢+𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = (𝑑 (𝑢 + 𝑣))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑𝑢/𝑑𝑥 + 𝑑𝑣/𝑑𝑥 Calculating 𝒅𝒖/𝒅𝒙 𝑢 = 𝑥^(𝑥^2− 3) Taking log on both sides log 𝑢=log⁡〖𝑥^(𝑥^2− 3) 〗 log 𝑢=〖(𝑥〗^2− 3). log⁡𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑(log⁡𝑢 )/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log⁡𝑥 )/𝑑𝑥 𝑑(log⁡𝑢 )/𝑑𝑥 . 𝑑𝑢/𝑑𝑢 = 𝑑(〖(𝑥〗^2 − 3) log⁡𝑥 )/𝑑𝑥 " " (As 𝑙𝑜𝑔⁡(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔⁡𝑎) 𝑑(log⁡𝑢 )/𝑑𝑢 . 𝑑𝑢/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log⁡𝑥 )/𝑑𝑥 " " 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log⁡𝑥 )/𝑑𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = (𝑑〖(𝑥〗^2− 3) )/𝑑𝑥 . 〖 log〗⁡𝑥 + 𝑑(log⁡𝑥 )/𝑑𝑥 . 〖(𝑥〗^2− 3) 1/𝑢 . 𝑑𝑢/𝑑𝑥 = (2𝑥 −0) 〖 log〗⁡𝑥 + (〖(𝑥〗^2− 3)" " )/𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 2𝑥 . log⁡𝑥 + (𝑥^2− 3)/𝑥 Using product rule in 〖(𝑥〗^2− 3). 𝑙𝑜𝑔⁡𝑥 As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 𝑑𝑢/𝑑𝑥 = u (2𝑥 "." log⁡𝑥 "+ " (𝑥^2− 3)/𝑥) 𝑑𝑢/𝑑𝑥 = 𝑥^(𝑥^2− 3) (2𝑥 "." log⁡𝑥 "+ " (𝑥^2− 3)/𝑥) Calculating 𝒅𝒗/𝒅𝒙 𝑣= (𝑥−3)𝑥^2 Taking log on both sides log 𝑣=log⁡〖(𝑥−3)𝑥^2 〗 log 𝑣=〖𝑥^2 . log〗⁡〖 (𝑥−3)〗 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑(log⁡𝑣 )/𝑑𝑥 = (𝑑(〖𝑥^2. log〗⁡〖 (𝑥−3)〗 ) )/𝑑𝑥 (As log⁡(𝑎^𝑏) = 𝑏 log⁡𝑎) 𝑑(log⁡𝑣 )/𝑑𝑥 . 𝑑𝑣/𝑑𝑣 = (𝑑(〖𝑥^2. log〗⁡〖 (𝑥−3)〗 ) )/𝑑𝑥 𝑑(log⁡𝑣 )/𝑑𝑣 . 𝑑𝑣/𝑑𝑥 = (𝑑(〖𝑥^2. log〗⁡〖 (𝑥−3)〗 ) )/𝑑𝑥 1/𝑣 . 𝑑𝑣/𝑑𝑥 = (𝑑(〖𝑥^2. log〗⁡〖 (𝑥−3)〗 ) )/𝑑𝑥 1/𝑣 . 𝑑𝑣/𝑑𝑥 = 𝑑(𝑥^2 )/𝑑𝑥 . log (𝑥−3) + 𝑑(log" " (𝑥 − 3))/𝑑𝑥 . 𝑥^2 1/𝑣 . 𝑑𝑣/𝑑𝑥 = 2𝑥 . log (𝑥−3) + 1/((𝑥 − 3) ). (𝑑(𝑥 − 3)" " )/𝑑𝑥 . 𝑥^2 1/𝑣 . 𝑑𝑣/𝑑𝑥 = 2𝑥 . log (𝑥−3) + 1/((𝑥 − 3) ) . 𝑥^2 1/𝑣 . 𝑑𝑣/𝑑𝑥 = 2𝑥. log (𝑥−3) + 𝑥^2/(𝑥 −3) Using product rule (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 𝑑𝑣/𝑑𝑥 = 𝑣 (2𝑥". " log" " (𝑥−3)" + " 𝑥^2/(𝑥 −3)) 𝑑𝑣/𝑑𝑥 = (𝑥−3)𝑥^2 (2𝑥". " log" " (𝑥−3)" + " 𝑥^2/(𝑥 −3)) Now, 𝑑𝑦/𝑑𝑥 = 𝑑𝑢/𝑑𝑥 + 𝑑𝑣/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑥^(𝑥^2− 3) (2𝑥 "." log⁡𝑥 "+ " (𝑥^2− 3)/𝑥) + (𝑥−3)𝑥^2 (2𝑥". " log" " (𝑥−3)" + " 𝑥^2/(𝑥 −3)) 𝒅𝒚/𝒅𝒙 = 𝒙^(𝒙^𝟐− 𝟑) ((𝒙^𝟐− 𝟑)/𝒙+𝟐𝒙 𝐥𝐨𝐠⁡𝒙 ) + (𝒙−𝟑)𝒙^𝟐 (𝒙^𝟐/(𝒙 −𝟑)+𝟐𝒙 .𝐥𝐨𝐠⁡(𝒙 −𝟑) )

Miscellaneous 