Ex 5.5, 2 - Differentiate root (x-1)(x-2)/(x-3)(x-4)(x-5)

Ex 5.5, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Transcript

Ex 5.5, 2 Differentiate the functions in, √(((𝑥 − 1)(𝑥 − 2))/((𝑥 − 3)(𝑥 − 4)(𝑥 − 5))) Let 𝑦=√(((𝑥 − 1)(𝑥 − 2))/((𝑥 − 3)(𝑥 − 4)(𝑥 − 5))) 𝑦= (((𝑥 − 1)(𝑥 − 2))/((𝑥 − 3)(𝑥 − 4)(𝑥 − 5)))^(1/2) Taking log both sides log⁡𝑦 = log (((𝑥 − 1)(𝑥 − 2))/((𝑥 − 3)(𝑥 − 4)(𝑥 − 5)))^(1/2) log⁡𝑦 = 1/2 log (((𝑥 − 1)(𝑥 − 2))/((𝑥 − 3)(𝑥 − 4)(𝑥 − 5))) (As 𝑙𝑜𝑔⁡(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔⁡𝑎) log⁡𝑦 = 1/2 [log⁡〖((𝑥−1)(𝑥−2))−log⁡((𝑥−3)(𝑥−4)(𝑥−5)) 〗 ] log⁡𝑦 = 1/2 . [("log " (𝑥+1)" + log " (𝑥−2))" − " (log⁡(𝑥−3)+log⁡〖(𝑥−4)+log⁡(𝑥−5) 〗 )] log⁡𝑦 = 1/2 . ["log " (𝑥+1)" + log " (𝑥+2)" − " log⁡(𝑥−3)−log⁡〖(𝑥−4)−log⁡(𝑥−5) 〗 ] Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log⁡𝑦)/𝑑𝑥 = (𝑑 (1/2 " ." log⁡〖(𝑥 + 1)" +" log⁡〖 (𝑥 + 2)" − " log⁡(𝑥 − 3)−log⁡〖(𝑥 − 4)−log⁡(𝑥 − 5) 〗 〗 〗 ))/𝑑𝑥 𝑑(log⁡𝑦)/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 1/2 (𝑑(〖log 〗⁡〖(𝑥 + 1)" +" log⁡〖(𝑥 + 2)" −" log⁡(𝑥 − 3) − log⁡〖(𝑥 − 4) − log⁡(𝑥 − 5) 〗 〗 〗 )/𝑑𝑥) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 1/2 (1/(𝑥 + 1)+1/(𝑥 + 2)−1/(𝑥 − 3)−1/(𝑥 − 4)−1/(𝑥 − 5)) 𝑑𝑦/𝑑𝑥 = 1/2 𝑦(1/(𝑥 − 1)+1/(𝑥 − 2)−1/(𝑥 − 3)−1/(𝑥 − 4)−1/(𝑥 − 5)) 𝒅𝒚/𝒅𝒙 = 𝟏/𝟐 √(((𝒙 − 𝟏)(𝒙 − 𝟐))/((𝒙 − 𝟑)(𝒙 − 𝟒)(𝒙 − 𝟓))) (𝟏/(𝒙 − 𝟏)+𝟏/(𝒙 − 𝟐)−𝟏/(𝒙 − 𝟑)−𝟏/(𝒙 − 𝟒)−𝟏/(𝒙 − 𝟓))

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.