Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10

Misc 11 Important You are here

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18

Misc 19

Misc 20

Misc 21

Misc 22 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at June 5, 2023 by Teachoo

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 ) And let 𝑢=𝑥^(𝑥^2− 3) , 𝑣 =(𝑥−3)^(𝑥^2 ) Now, 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). log𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑢 )/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log𝑥 )/𝑑𝑥 𝑑(log𝑢 )/𝑑𝑥 . 𝑑𝑢/𝑑𝑢 = 𝑑(〖(𝑥〗^2 − 3) log𝑥 )/𝑑𝑥 " " 𝑑(log𝑢 )/𝑑𝑢 . 𝑑𝑢/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log𝑥 )/𝑑𝑥 " " 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 𝑑(〖(𝑥〗^2− 3) log𝑥 )/𝑑𝑥 1/𝑢 . 𝑑𝑢/𝑑𝑥 = (𝑑〖(𝑥〗^2− 3) )/𝑑𝑥 . 〖 log〗𝑥 + 𝑑(log𝑥 )/𝑑𝑥 . 〖(𝑥〗^2− 3) 1/𝑢 . 𝑑𝑢/𝑑𝑥 = (2𝑥 −0) 〖 log〗𝑥 + 1/𝑥 × 〖(𝑥〗^2− 3) 1/𝑢 . 𝑑𝑢/𝑑𝑥 = 2𝑥 . log𝑥 + (𝑥^2− 3)/𝑥 𝑑𝑢/𝑑𝑥 = u (2𝑥 "." log𝑥 "+ " (𝑥^2− 3)/𝑥) 𝒅𝒖/𝒅𝒙 = 𝒙^(𝒙^𝟐− 𝟑) (𝟐𝒙 "." 𝒍𝒐𝒈𝒙 "+ " (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(log𝑣 )/𝑑𝑥 = (𝑑(〖𝑥^2. log〗〖 (𝑥 − 3)〗 ) )/𝑑𝑥 𝑑(log𝑣 )/𝑑𝑥 . 𝑑𝑣/𝑑𝑣 = (𝑑(〖𝑥^2. log〗〖 (𝑥−3)〗 ) )/𝑑𝑥 𝑑(log𝑣 )/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) 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 = 𝒙^(𝒙^𝟐− 𝟑) ((𝒙^𝟐− 𝟑)/𝒙+𝟐𝒙 𝐥𝐨𝐠𝒙 ) + (𝒙−𝟑)𝒙^𝟐 (𝒙^𝟐/(𝒙 −𝟑)+𝟐𝒙 .𝐥𝐨𝐠(𝒙 −𝟑) )