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

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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.