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


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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 𝒍𝒐𝒈⁡𝒚 = 𝒍𝒐𝒈 √(((𝒙 − 𝟑) (𝒙𝟐 + 𝟒))/( 𝟑𝒙𝟐+ 𝟒𝒙 + 𝟓)) log⁡𝑦 = log⁡ (((𝑥 − 3) (𝑥2 + 4))/( 3𝑥2+ 4𝑥 + 5))^(1/2) 𝒍𝒐𝒈⁡𝒚 = 𝟏/𝟐 log ((𝑥 − 3) (𝑥2 + 4))/((3𝑥2+ 4𝑥 + 5) ) (𝑈𝑠𝑖𝑛𝑔 log⁡〖𝑎^𝑏 〗=𝑏 log⁡𝑎) 𝒍𝒐𝒈⁡𝒚 = 𝟏/𝟐 log ((𝑥 − 3) (𝑥2 + 4))/((3𝑥2+ 4𝑥 + 5) ) 𝑙𝑜𝑔⁡𝑦 = 1/2 (log⁡〖 (𝑥−3)〗+〖log 〗⁡(𝑥2 + 4)−log⁡〖 (3𝑥2 + 4𝑥 + 5)〗 ) Differentiating 𝑤.𝑟.𝑡.𝑥 (𝑑 (log⁡𝑦))/𝑑𝑥 = 1/2 ((𝑑(log⁡(𝑥−3)+〖log 〗⁡〖(𝑥^2+4)−log⁡〖 (3𝑥^2+4𝑥+5)〗 〗 ) )/𝑑𝑥) (𝑑 (log⁡𝑦))/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 𝑙𝑜𝑔⁡𝑎𝑏=𝑙𝑜𝑔⁡𝑎+𝑐𝑜𝑠⁡𝑏 &𝑙𝑜𝑔⁡〖𝑎/𝑏〗=𝑙𝑜𝑔⁡〖𝑎−𝑙𝑜𝑔⁡𝑏 〗 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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Davneet Singh

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.