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Example 44 - Differentiate the function (i) root (3x+2) + 1/root(2x^2

Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Transcript

Example 44 Differentiate w.r.t. x, the following function: (i) √(3π‘₯+2) + 1/√(2π‘₯^2+ 4) Let y = √(3π‘₯+2) + 1/√(2π‘₯^2+ 4 ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(√(3π‘₯ + 2) " + " 1/√(2π‘₯^2 + 4 ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(√(3π‘₯ + 2))/𝑑π‘₯ + 𝑑(1/√(2π‘₯^2 + 4 ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(√(3π‘₯ + 2))/𝑑π‘₯ + (𝑑(2π‘₯^2 + 4)^((βˆ’1)/2))/𝑑π‘₯ Calculating 𝑑(√(3π‘₯ + 2))/𝑑π‘₯ & (𝑑(2π‘₯^2 + 4)^((βˆ’1)/2))/𝑑π‘₯ separately Calculating 𝐝(√(πŸ‘π± + 𝟐))/𝒅𝒙 𝑑(√(3π‘₯ + 2))/𝑑π‘₯ = 1/(2√(3π‘₯ + 2 )) Γ— 𝑑(3π‘₯ + 2)/𝑑π‘₯ = 1/(2√(3π‘₯ + 2 )) Γ— (3+0) = πŸ‘/(𝟐√(πŸ‘π’™ + 𝟐 )) Calculating (𝒅(πŸπ’™^𝟐 + πŸ’)^((βˆ’πŸ)/𝟐))/𝒅𝒙 (𝑑(2π‘₯^2 + 4)^((βˆ’1)/2))/𝑑π‘₯ = (βˆ’1)/2 γ€–(2π‘₯^2+4)γ€—^((βˆ’1)/( 2) βˆ’1) . 𝑑(2π‘₯^2+ 4)/𝑑π‘₯ = (βˆ’1)/2 (2π‘₯^2+ 4)^((βˆ’3)/( 2)) . (𝑑(2π‘₯^2 )/𝑑π‘₯ + 𝑑(4)/𝑑π‘₯) = (βˆ’1)/2 (2π‘₯^2+ 4)^((βˆ’3)/( 2)) . (4π‘₯+0) = (βˆ’4π‘₯)/2 (2π‘₯^2+ 4)^((βˆ’3)/( 2)) = (βˆ’πŸπ’™)/(πŸπ’™^𝟐+ πŸ’)^(πŸ‘/𝟐) Hence, 𝑑𝑦/𝑑π‘₯ = 𝑑(√(3π‘₯+2))/𝑑π‘₯ + 𝑑(1/√(2π‘₯^2+ 4 ))/𝑑π‘₯ π’…π’š/𝒅𝒙 = πŸ‘/(𝟐√(πŸ‘π’™ + 𝟐 )) βˆ’ πŸπ’™/(πŸπ’™^𝟐+ πŸ’)^(πŸ‘/( 𝟐))

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.