# Misc 11 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 11 (Method 1) Differentiate w.r.t. x the function, 𝑥 𝑥2− 3+(𝑥−3) 𝑥2, for 𝑥 > 3 Calculating 𝒅𝒖𝒅𝒙 𝑢 = 𝑥 𝑥2− 3 Taking log on both sides log 𝑢= log 𝑥 𝑥2− 3 log 𝑢= (𝑥2− 3). log𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑 log𝑢𝑑𝑥 = 𝑑 (𝑥2− 3) log𝑥𝑑𝑥 𝑑 log𝑢𝑑𝑥 . 𝑑𝑢𝑑𝑢 = 𝑑 (𝑥2 − 3) log𝑥𝑑𝑥 𝑑𝑢𝑑𝑥 = 𝑥 𝑥2− 3 2𝑥 . log𝑥+ 𝑥2− 3𝑥 Calculating 𝒅𝒗𝒅𝒙 𝑣= (𝑥−3) 𝑥2 Taking log on both sides log 𝑣= log(𝑥−3) 𝑥2 log 𝑣= 𝑥2 . log (𝑥−3) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑 log𝑣𝑑𝑥 = 𝑑 𝑥2. log (𝑥−3) 𝑑𝑥 𝑑 log𝑣𝑑𝑥 . 𝑑𝑣𝑑𝑣 = 𝑑 𝑥2. log (𝑥−3) 𝑑𝑥 𝑑 log𝑣𝑑𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 𝑥2. log (𝑥−3) 𝑑𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 𝑥2. log (𝑥−3) 𝑑𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 𝑥2𝑑𝑥 . log (𝑥−3) + 𝑑 log (𝑥 − 3)𝑑𝑥 . 𝑥2 1𝑣 . 𝑑𝑣𝑑𝑥 = 2𝑥 . log (𝑥−3) + 1 𝑥 − 3. 𝑑(𝑥 − 3) 𝑑𝑥 . 𝑥2 1𝑣 . 𝑑𝑣𝑑𝑥 = 2𝑥 . log (𝑥−3) + 1 𝑥 − 3 . 𝑥2 1𝑣 . 𝑑𝑣𝑑𝑥 = 2𝑥. log (𝑥−3) + 𝑥2𝑥 −3 𝑑𝑣𝑑𝑥 = 𝑣 2𝑥. log (𝑥−3) + 𝑥2𝑥 −3 𝑑𝑣𝑑𝑥 = (𝑥−3) 𝑥2 2𝑥. log (𝑥−3) + 𝑥2𝑥 −3 Now, 𝑑𝑦𝑑𝑥 = 𝑑𝑢𝑑𝑥 + 𝑑𝑣𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑥 𝑥2− 3 2𝑥 . log𝑥+ 𝑥2− 3𝑥 + (𝑥−3) 𝑥2 2𝑥. log (𝑥−3) + 𝑥2𝑥 −3 𝒅𝒚𝒅𝒙 = 𝒙 𝒙𝟐− 𝟑 𝒙𝟐− 𝟑𝒙+𝟐𝒙 𝐥𝐨𝐠𝒙 + (𝒙−𝟑) 𝒙𝟐 𝒙𝟐𝒙 −𝟑+𝟐𝒙 . 𝐥𝐨𝐠 𝒙 −𝟑

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.