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Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

Example 44 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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Example 44 Differentiate w.r.t. x, the following function: (ii) 𝑒^(sec^2⁑π‘₯ ) + 3cos^(–1) π‘₯ Let y = 𝑒^(sec^2⁑π‘₯ ) + 3cos^(–1) π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒^(sec^2⁑π‘₯ )+ 3cos^(–1) π‘₯" " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒^(sec^2⁑π‘₯ ) )/𝑑π‘₯ + 𝒅(πŸ‘γ€–π’„π’π’”γ€—^(β€“πŸ) 𝒙)/𝒅𝒙 𝑑𝑦/𝑑π‘₯ = 𝑒^(sec^2⁑π‘₯ ) 𝑑(sec^2⁑π‘₯ )/𝑑π‘₯ + 3. ((βˆ’πŸ)/√(𝟏 βˆ’π’™^𝟐 )) 𝑑𝑦/𝑑π‘₯ = 𝑒^(sec^2⁑π‘₯ ). 2 sec π‘₯ . 𝒅(〖𝒔𝒆𝒄 〗⁑𝒙 )/𝒅𝒙 βˆ’ 3/√(1 βˆ’π‘₯^2 ) "As" 𝑑(𝑒^π‘₯ )/𝑑π‘₯=𝑒^π‘₯ & 𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1)⁑π‘₯ )/𝑑π‘₯=(βˆ’1)/√(1 βˆ’γ€– π‘₯γ€—^2 ) 𝑑𝑦/𝑑π‘₯ = 𝑒^(sec^2⁑π‘₯ ). 2 sec π‘₯ . 𝒔𝒆𝒄⁑𝒙 .𝒕𝒂𝒏⁑𝒙 βˆ’ 3/√(1 βˆ’ π‘₯^2 ) π’…π’š/𝒅𝒙 = γ€–πŸ 𝒆〗^(〖𝒔𝒆𝒄〗^πŸβ‘π’™ ). 〖𝒔𝒆𝒄〗^𝟐 𝒙 .𝒕𝒂𝒏⁑𝒙 βˆ’ πŸ‘/√(𝟏 βˆ’ 𝒙^𝟐 )

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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.