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Ex 5.5, 14 - Find dy/dx of (cos x)^y = (cos y)^x - Chapter 5 Class 12

Ex 5.5, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.5, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.5, 14 Find 𝑑𝑦/𝑑π‘₯ of the functions in, γ€–(cos⁑〖π‘₯ γ€—)γ€—^𝑦 = γ€–(cos⁑〖𝑦 γ€—)γ€—^π‘₯Given γ€–(cos⁑π‘₯)γ€—^𝑦 = γ€–(cos⁑𝑦)γ€—^π‘₯ Taking log both sides log γ€–(cos⁑π‘₯)γ€—^𝑦 = log γ€–(cos⁑𝑦)γ€—^π‘₯ 𝑦 . log (cos⁑π‘₯)=π‘₯.log⁑〖(cos⁑𝑦)γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(𝑦 . log (cos⁑π‘₯)))/𝑑π‘₯ = 𝑑(π‘₯.γ€– log〗⁑〖(cos⁑𝑦)γ€— )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏 )=𝑏 . π‘™π‘œπ‘”β‘π‘Ž) Finding (𝒅(π’š . π’π’π’ˆ (𝒄𝒐𝒔⁑𝒙)))/𝒅𝒙 (𝑑(𝑦 . π‘™π‘œπ‘” (π‘π‘œπ‘ β‘π‘₯)))/𝑑π‘₯ = (𝑑(𝑦))/𝑑π‘₯ . log co𝑠⁑π‘₯ + (𝑑(π‘™π‘œπ‘” (π‘π‘œπ‘ β‘π‘₯)))/𝑑π‘₯ . 𝑦 = 𝑑𝑦/𝑑π‘₯ . log co𝑠⁑π‘₯ + 1/π‘π‘œπ‘ β‘π‘₯ . 𝑑(π‘π‘œπ‘ β‘π‘₯ )/𝑑π‘₯ . 𝑦 = 𝑑𝑦/𝑑π‘₯ . log co𝑠⁑π‘₯ + 1/π‘π‘œπ‘ β‘π‘₯ . (βˆ’sin⁑π‘₯ ) . 𝑦 = 𝑑𝑦/𝑑π‘₯ . log co𝑠⁑π‘₯ + ((βˆ’sin⁑π‘₯ ))/π‘π‘œπ‘ β‘π‘₯ . 𝑦 = 𝑑𝑦/𝑑π‘₯ . log co𝑠⁑π‘₯βˆ’tan⁑π‘₯. 𝑦 Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Finding 𝒅(𝒙.γ€– π’π’π’ˆγ€—β‘γ€–(π’„π’π’”β‘π’š)γ€— )/𝒅𝒙 𝑑(π‘₯.γ€– π‘™π‘œπ‘”γ€—β‘γ€–(π‘π‘œπ‘ β‘π‘¦)γ€— )/𝑑π‘₯ = (𝑑(π‘₯))/𝑑π‘₯ . log co𝑠⁑𝑦 + (𝑑(π‘™π‘œπ‘” (π‘π‘œπ‘ β‘π‘¦)))/𝑑π‘₯ . π‘₯ = log co𝑠⁑𝑦 + 1/π‘π‘œπ‘ β‘π‘¦ . 𝑑(π‘π‘œπ‘ β‘π‘¦ )/𝑑π‘₯ . π‘₯ = log co𝑠⁑𝑦 + 1/π‘π‘œπ‘ β‘π‘¦ . 𝑑(π‘π‘œπ‘ β‘π‘¦ )/𝑑π‘₯ . 𝑑𝑦/𝑑𝑦 . π‘₯ = log co𝑠⁑𝑦 + 1/π‘π‘œπ‘ β‘π‘¦ . 𝑑(π‘π‘œπ‘ β‘π‘¦ )/𝑑𝑦 . 𝑑𝑦/𝑑π‘₯ . π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 = log co𝑠⁑𝑦 + 1/π‘π‘œπ‘ β‘π‘¦ . (βˆ’sin⁑𝑦) . 𝑑𝑦/𝑑π‘₯ . π‘₯ = log co𝑠⁑𝑦 + βˆ’tan⁑𝑦 . π‘₯ . 𝑑𝑦/𝑑π‘₯ Now , (𝑑(𝑦 . log (cos⁑π‘₯)))/𝑑π‘₯ = 𝑑(π‘₯.γ€– log〗⁑〖(cos⁑𝑦)γ€— )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ log co𝑠⁑π‘₯βˆ’tan⁑π‘₯. 𝑦 = log co𝑠⁑𝑦 βˆ’ tan⁑𝑦 . π‘₯ . 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ log co𝑠⁑π‘₯βˆ’π‘¦ . tan⁑π‘₯ = log co𝑠⁑𝑦 βˆ’ π‘₯ . tan⁑𝑦 . 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ log co𝑠⁑π‘₯+π‘₯ tan 𝑑𝑦/𝑑π‘₯ = log co𝑠⁑𝑦 + 𝑦 tan⁑π‘₯ 𝑑𝑦/𝑑π‘₯ (log co𝑠⁑π‘₯+π‘₯ tan 𝑦) = log co𝑠⁑𝑦 + 𝑦 tan⁑π‘₯ π’…π’š/𝒅𝒙 = (π₯𝐨𝐠 π’„π’π’”β‘π’š " + " π’š 𝒕𝒂𝒏⁑𝒙)/(π₯𝐨𝐠 𝒄𝒐𝒔⁑𝒙 + 𝒙 𝐭𝐚𝐧 π’š)

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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.