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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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.