Ex 5.5, 1 - Differentiate cos x . cos 2x . cos 3x - Class 12 - Ex 5.5

part 2 - Ex 5.5, 1 - Ex 5.5 - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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Ex 5.5, 1 Differentiate the functions in, cos⁑π‘₯ . cos⁑2π‘₯ . cos⁑3π‘₯ Let y = cos⁑π‘₯ . cos⁑2π‘₯ . cos⁑3π‘₯ Taking log both sides log⁑𝑦 = log (cos⁑π‘₯.cos⁑2π‘₯.cos⁑3π‘₯ ) π’π’π’ˆβ‘π’š = π’π’π’ˆ ⁑(𝒄𝒐𝒔⁑𝒙) + π’π’π’ˆ ⁑(𝒄𝒐𝒔 πŸπ’™) + π’π’π’ˆ ⁑(π’„π’π’”β‘πŸ‘π’™) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑(log ⁑(cos⁑π‘₯)" + " log ⁑(cos⁑2π‘₯) "+ " log ⁑(cos⁑3π‘₯))/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = (𝑑(log ⁑(cos⁑π‘₯)) )/𝑑π‘₯ + (𝑑(log ⁑(cos⁑2π‘₯)) )/𝑑π‘₯ + (𝑑(log ⁑(cos⁑3π‘₯)) )/𝑑π‘₯ 𝒅(π’π’π’ˆβ‘π’š )/π’…π’š (π’…π’š/𝒅𝒙) = 𝟏/πœπ¨π¬β‘π’™ . (𝒅 (πœπ¨π¬β‘π’™ ))/𝒅𝒙 + 𝟏/πœπ¨π¬β‘πŸπ’™ . (𝒅(πœπ¨π¬β‘πŸπ’™))/𝒅𝒙 + 𝟏/πœπ¨π¬β‘πŸ‘π’™ . 𝒅(πœπ¨π¬β‘πŸ‘π’™ )/𝒅𝒙 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 1/cos⁑π‘₯ .(βˆ’ sin⁑π‘₯) + 1/cos⁑2π‘₯ .(βˆ’ sin⁑2π‘₯).𝑑(2π‘₯)/𝑑π‘₯ + 1/cos⁑π‘₯ .(βˆ’ sin⁑3π‘₯).𝑑(3π‘₯)/𝑑π‘₯ 𝟏/π’š . π’…π’š/𝒅𝒙 = (βˆ’π¬π’π§β‘π’™)/πœπ¨π¬β‘π’™ βˆ’ π¬π’π§β‘πŸπ’™/πœπ¨π¬β‘π’™ . 𝟐 βˆ’ π¬π’π§β‘πŸ‘π’™/πœπ¨π¬β‘πŸ‘π’™ . πŸ‘ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’tan⁑π‘₯βˆ’tan⁑2π‘₯. 2 βˆ’tan⁑3π‘₯. 3 𝟏/π’š . π’…π’š/𝒅𝒙 = βˆ’ (𝒕𝒂𝒏⁑𝒙+𝟐 π’•π’‚π’β‘πŸπ’™+πŸ‘ π’•π’‚π’β‘πŸ‘π’™ ) 𝑑𝑦/𝑑π‘₯ = βˆ’π‘¦ (tan⁑π‘₯+2 tan⁑2π‘₯+3 tan⁑3π‘₯ ) π’…π’š/𝒅𝒙 = βˆ’ 𝒄𝒐𝒔⁑𝒙 . π’„π’π’”β‘πŸπ’™ . π’„π’π’”β‘πŸ‘π’™ (𝒕𝒂𝒏⁑𝒙+𝟐 π’•π’‚π’β‘πŸπ’™+πŸ‘ π’•π’‚π’β‘πŸ‘π’™ )

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