Example 27 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 27 Differentiate √(((𝑥−3) (𝑥2+4))/( 3𝑥2+ 4𝑥 + 5)) 𝑤.𝑟.𝑡. 𝑥. Let y =√(((𝑥 − 3) (𝑥^2 + 4))/( 3𝑥2+ 4𝑥 + 5)) Taking log on both sides 𝒍𝒐𝒈𝒚 = 𝒍𝒐𝒈 √(((𝒙 − 𝟑) (𝒙𝟐 + 𝟒))/( 𝟑𝒙𝟐+ 𝟒𝒙 + 𝟓)) log𝑦 = log (((𝑥 − 3) (𝑥2 + 4))/( 3𝑥2+ 4𝑥 + 5))^(1/2) 𝒍𝒐𝒈𝒚 = 𝟏/𝟐 log ((𝑥 − 3) (𝑥2 + 4))/((3𝑥2+ 4𝑥 + 5) ) (𝑈𝑠𝑖𝑛𝑔 log〖𝑎^𝑏 〗=𝑏 log𝑎) 𝒍𝒐𝒈𝒚 = 𝟏/𝟐 log ((𝑥 − 3) (𝑥2 + 4))/((3𝑥2+ 4𝑥 + 5) ) 𝑙𝑜𝑔𝑦 = 1/2 (log〖 (𝑥−3)〗+〖log 〗(𝑥2 + 4)−log〖 (3𝑥2 + 4𝑥 + 5)〗 ) Differentiating 𝑤.𝑟.𝑡.𝑥 (𝑑 (log𝑦))/𝑑𝑥 = 1/2 ((𝑑(log(𝑥−3)+〖log 〗〖(𝑥^2+4)−log〖 (3𝑥^2+4𝑥+5)〗 〗 ) )/𝑑𝑥) (𝑑 (log𝑦))/𝑑𝑥 = 1/2 ((𝑑(log(𝑥−3)))/𝑑𝑥 " + " 𝑑(log(𝑥^2+4) )/𝑑𝑥 " − " 𝑑(log(3𝑥^2+4𝑥+5) )/𝑑𝑥) 1/𝑦 (𝑑𝑦/𝑑𝑥) = 1/2 [1/(𝑥 − 3) " . " 𝑑(𝑥 −3)/𝑑𝑥 " + " 1/(𝑥^2 + 4) " . " (𝑑 (𝑥^2 + 4))/𝑑𝑥 " – " 1/((3𝑥^2+ 4𝑥+ 5)) " . " (𝑑 (3𝑥^2+ 4𝑥+ 5))/(𝑑𝑥 )] 𝑈𝑠𝑖𝑛𝑔 𝑙𝑜𝑔𝑎𝑏=𝑙𝑜𝑔𝑎+𝑐𝑜𝑠𝑏 &𝑙𝑜𝑔〖𝑎/𝑏〗=𝑙𝑜𝑔〖𝑎−𝑙𝑜𝑔𝑏 〗 1/𝑦 (𝑑𝑦/𝑑𝑥) = 1/2 [1/((𝑥 − 3) ) " " (1−0) "+ " 1/(𝑥^2+ 4) " " (2𝑥 + 0)" − " 1/(3𝑥^2+ 4𝑥 + 5) " " (6𝑥 +4+0)" " ] 1/𝑦 (𝑑𝑦/𝑑𝑥) = 1/2 (1/((𝑥 − 3) )+2𝑥/(𝑥^2+ 4)−(6𝑥 + 4)/(3𝑥^2 + 4𝑥 + 5)) 𝑑𝑦/𝑑𝑥 = 𝑦/2 (1/((𝑥 − 3) )+2𝑥/(𝑥^2+ 4)−(6𝑥 + 4)/(3𝑥^2 + 4𝑥 + 5)) 𝒅𝒚/𝒅𝒙 = 𝟏/𝟐 √(((𝒙 − 𝟑)(𝒙^𝟐+ 𝟒))/(𝟑𝒙^𝟐+ 𝟒𝒙 + 𝟓)) (𝟏/((𝒙 − 𝟑) )+𝟐𝒙/(𝒙^𝟐+ 𝟒)−(𝟔𝒙 + 𝟒)/(𝟑𝒙^𝟐 + 𝟒𝒙 + 𝟓))
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo