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


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Example 40 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

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.