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Misc 3 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ the function, (5π‘₯)^(3cos⁑2π‘₯) Let 𝑦" = " (5π‘₯)^(3cos⁑2π‘₯) Taking log on both sides log⁑𝑦 = log (5π‘₯)^(3cos⁑2π‘₯) π’π’π’ˆβ‘π’š = πŸ‘ 𝐜𝐨𝐬 πŸπ’™ . π’π’π’ˆ β‘πŸ“π’™ Differentiating both sides 𝑀.π‘Ÿ.𝑑. x 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝟏/π’š . π’…π’š/𝒅𝒙 = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ )/𝑑π‘₯ . log ⁑5π‘₯ + 𝑑(log ⁑5π‘₯)/𝑑π‘₯ .3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 3 𝑑(cos 2π‘₯ )/𝑑π‘₯ . log ⁑5π‘₯ + 1/5π‘₯ . 𝑑(5π‘₯)/𝑑π‘₯ . 3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 3(βˆ’sin⁑2π‘₯ ) Γ— 2 Γ— log ⁑5π‘₯ + 1/5π‘₯ Γ— 5 Γ— 3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’6 sin 2π‘₯ . log⁑5π‘₯ + (3 cos⁑2π‘₯)/π‘₯ Using Product rule (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 where u = 3 π‘π‘œπ‘  2π‘₯ & 𝑣=π‘™π‘œπ‘” ⁑5π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑦 (βˆ’6 sin 2π‘₯ . log⁑5π‘₯ "+ " (3 cos⁑2π‘₯)/π‘₯) 𝑑𝑦/𝑑π‘₯ = (5π‘₯)^(3cos⁑2π‘₯) (βˆ’6 sin 2π‘₯ . log⁑5π‘₯ "+ " (3 cos⁑2π‘₯)/π‘₯) π’…π’š/𝒅𝒙 = (πŸ“π’™)^(πŸ‘π’„π’π’”β‘πŸπ’™) ((πŸ‘ π’„π’π’”β‘πŸπ’™)/π’™βˆ’πŸ” 𝐬𝐒𝐧 πŸπ’™ . π’π’π’ˆβ‘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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo