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Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 8
Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 9

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Transcript

Example 45 Differentiate the following w.r.t. x. (iii) sin^(βˆ’1) ((2^( π‘₯+1) )/( 1 +γ€– 4 γ€—^π‘₯ )) Let 𝑓(π‘₯) = sin^(βˆ’1) ((2^( π‘₯+1) )/( 1 +γ€– 4 γ€—^π‘₯ )) 𝑓(π‘₯) = sin^(βˆ’1) ((2^( π‘₯). 2)/( 1 + (2^π‘₯ )^2 )) Let 𝟐^𝒙 = tan ΞΈ 𝑓(π‘₯) = sin^(βˆ’1) ((tanβ‘γ€–πœƒ γ€—. 2)/( 1 + tan^2β‘πœƒ )) = sin^(βˆ’1) ((2 tanβ‘γ€–πœƒ γ€— )/( 1 +γ€– tan^2γ€—β‘πœƒ )) = sin^(βˆ’1) (sin 2πœƒ) = 2πœƒ (sin⁑2πœƒ "= " (2 tanβ‘πœƒ)/(1 +γ€– tan^2γ€—β‘πœƒ )) (As 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖(π‘ π‘–π‘›β‘πœƒ)γ€— =πœƒ) Since 2^π‘₯= tanβ‘πœƒ tan^(βˆ’1) (2^π‘₯ )=πœƒ ∴ 𝒇(𝒙) = 𝟐 (〖𝒕𝒂𝒏〗^(βˆ’πŸ) (𝟐^𝒙 )) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓’(π‘₯) = 2 (𝑑 (tan^(βˆ’1) 2^π‘₯ )" " )/𝑑π‘₯ 𝑓’(π‘₯) = 2 . 1/(1 + (2^π‘₯ )^2 ) . (𝒅 (𝟐^𝒙 )" " )/𝒅𝒙 𝑓’(π‘₯) = (2 )/(1 + (2^π‘₯ )^2 ) . 𝟐^𝒙 . π’π’π’ˆβ‘πŸ (𝐴𝑠 𝑑/𝑑π‘₯(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1))=1/(1+π‘₯^2 )) (𝐴𝑠 𝑑/𝑑π‘₯ (π‘Ž^π‘₯ )=π‘Ž^π‘₯. π‘™π‘œπ‘”β‘π‘₯ ) 𝑓’(π‘₯) = (γ€–2. 2γ€—^π‘₯.γ€– log〗⁑2)/(1 + (2^π‘₯ )^2 ) 𝑓’(π‘₯) = (2^(π‘₯ + 1).γ€– log〗⁑2)/(1 + (2^π‘₯ )^2 ) 𝑓’(π‘₯) = (2^(π‘₯ + 1).γ€– log〗⁑2)/(1 + (2^2 )^π‘₯ ) 𝒇’(𝒙) = (𝟐^(𝒙 + 𝟏).γ€– π’π’π’ˆγ€—β‘πŸ)/(𝟏 + πŸ’^𝒙 )

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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.