Check sibling questions

Slide61.JPG

Slide62.JPG

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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’™ )

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.