Ex 5.5, 2 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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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))