Misc 3 - Differentiate (5x)3cos2x - Chapter 5 Class 12 - Miscellaneous

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  1. Chapter 5 Class 12 Continuity and Differentiability
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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‘₯) log⁑𝑦 = 3 cos 2π‘₯ . log ⁑5π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ Using product rule 3 π‘π‘œπ‘  2π‘₯ . π‘™π‘œπ‘” ⁑5π‘₯ As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 where u = 3 π‘π‘œπ‘  2π‘₯ & 𝑣=π‘™π‘œπ‘” ⁑5π‘₯ 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‘₯ ) . 𝑑(2π‘₯)/𝑑π‘₯ . log ⁑5π‘₯ + 1/5π‘₯ Γ— 5 . 3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’3 sin 2π‘₯ . 2 . log⁑5π‘₯ + (3 cos⁑2π‘₯)/π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’6 sin 2π‘₯ . log⁑5π‘₯ + (3 cos⁑2π‘₯)/π‘₯ 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.