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Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Example 39 Differentiate w.r.t. x, the following function: (i) √(3𝑥+2) + 1/√(2𝑥^2+ 4) Let y = √(3𝑥+2) + 1/√(2𝑥^2+ 4 ) Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2) " + " 1/√(2𝑥^2 + 4 ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2))/𝑑𝑥 + 𝑑(1/√(2𝑥^2 + 4 ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2))/𝑑𝑥 + (𝑑(2𝑥^2 + 4)^((−1)/2))/𝑑𝑥 Calculating 𝑑(√(3𝑥 + 2))/𝑑𝑥 & (𝑑(2𝑥^2 + 4)^((−1)/2))/𝑑𝑥 separately Calculating 𝐝(√(𝟑𝐱 + 𝟐))/𝒅𝒙 𝑑(√(3𝑥 + 2))/𝑑𝑥 = 1/(2√(3𝑥 + 2 )) × 𝑑(3𝑥 + 2)/𝑑𝑥 = 1/(2√(3𝑥 + 2 )) × (3+0) = 𝟑/(𝟐√(𝟑𝒙 + 𝟐 )) Calculating (𝒅(𝟐𝒙^𝟐 + 𝟒)^((−𝟏)/𝟐))/𝒅𝒙 (𝑑(2𝑥^2 + 4)^((−1)/2))/𝑑𝑥 = (−1)/2 〖(2𝑥^2+4)〗^((−1)/( 2) −1) . 𝑑(2𝑥^2+ 4)/𝑑𝑥 = (−1)/2 (2𝑥^2+ 4)^((−3)/( 2)) . (𝑑(2𝑥^2 )/𝑑𝑥 + 𝑑(4)/𝑑𝑥) = (−1)/2 (2𝑥^2+ 4)^((−3)/( 2)) . (4𝑥+0) = (−4𝑥)/2 (2𝑥^2+ 4)^((−3)/( 2)) = (−𝟐𝒙)/(𝟐𝒙^𝟐+ 𝟒)^(𝟑/𝟐) Hence, 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥+2))/𝑑𝑥 + 𝑑(1/√(2𝑥^2+ 4 ))/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝟑/(𝟐√(𝟑𝒙 + 𝟐 )) − 𝟐𝒙/(𝟐𝒙^𝟐+ 𝟒)^(𝟑/( 𝟐))