Check sibling questions

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

Misc 3 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc 3 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

This video is only available for Teachoo black users

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


Transcript

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’™ )

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.