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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.