Ex 5.4, 4 - Differentiate sin (tan-1 e^-x) - Chapter 5 Class 12

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Ex 5.4, 4 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.4, 4 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ in , sin⁑〖 (tan^(βˆ’1) 𝑒^(βˆ’π‘₯) )γ€—Let 𝑦 = sin⁑〖 (tan^(βˆ’1) 𝑒^(βˆ’π‘₯) )γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑦^β€² = (sin⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ) )^β€² = γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ) Γ— (tan^(βˆ’1) 𝑒^(βˆ’π‘₯) )^β€² = γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ) Γ— 1/(1 + (𝑒^(βˆ’π‘₯) )^2 ) Γ—(𝑒^(βˆ’π‘₯) )^β€² = γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ) Γ— 1/(1 + (𝑒^(βˆ’π‘₯) )^2 ) Γ— βˆ’π‘’^(βˆ’π‘₯) = (𝑒^(βˆ’π‘₯) γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ))/(1 + (𝑒^(βˆ’π‘₯) )^2 ) = (βˆ’π’†^(βˆ’π’™) γ€–πœπ¨π¬ 〗⁑(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒆^(βˆ’π’™) ))/(𝟏 + 𝒆^(βˆ’πŸπ’™) ) = (𝑒^(βˆ’π‘₯) γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ))/(1 + (𝑒^(βˆ’π‘₯) )^2 ) = (βˆ’π’†^(βˆ’π’™) γ€–πœπ¨π¬ 〗⁑(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒆^(βˆ’π’™) ))/(𝟏 + 𝒆^(βˆ’πŸπ’™) ) = (𝑒^(βˆ’π‘₯) γ€–cos 〗⁑(tan^(βˆ’1) 𝑒^(βˆ’π‘₯) ))/(1 + (𝑒^(βˆ’π‘₯) )^2 ) = (βˆ’π’†^(βˆ’π’™) γ€–πœπ¨π¬ 〗⁑(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒆^(βˆ’π’™) ))/(𝟏 + 𝒆^(βˆ’πŸπ’™) )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.