

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Misc 2
Misc 3 You are here
Misc 4
Misc 5 Important
Misc 6 Important
Misc 7 Important
Misc 8
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12
Misc 13 Important
Misc 14 Important
Misc 15 Important
Misc 16 Important
Misc 17 Important
Misc 18
Misc 19
Misc 20
Misc 21
Misc 22 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Last updated at June 5, 2023 by Teachoo
Misc 3 Differentiate 𝑤.𝑟.𝑡. 𝑥 the function, (5𝑥)^(3cos2𝑥) Let 𝑦" = " (5𝑥)^(3cos2𝑥) Taking log on both sides log𝑦 = log (5𝑥)^(3cos2𝑥) 𝒍𝒐𝒈𝒚 = 𝟑 𝐜𝐨𝐬 𝟐𝒙 . 𝒍𝒐𝒈 𝟓𝒙 Differentiating both sides 𝑤.𝑟.𝑡. x 𝑑(log𝑦 )/𝑑𝑥 = 𝑑(3 cos 2𝑥 . log 5𝑥)/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑(3 cos 2𝑥 . log 5𝑥)/𝑑𝑥 (As 𝑙𝑜𝑔(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔𝑎) 𝑑(log𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑(3 cos 2𝑥 . log 5𝑥)/𝑑𝑥 𝟏/𝒚 . 𝒅𝒚/𝒅𝒙 = 𝑑(3 cos 2𝑥 . log 5𝑥)/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑(3 cos 2𝑥 )/𝑑𝑥 . log 5𝑥 + 𝑑(log 5𝑥)/𝑑𝑥 .3 cos 2𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 3 𝑑(cos 2𝑥 )/𝑑𝑥 . log 5𝑥 + 1/5𝑥 . 𝑑(5𝑥)/𝑑𝑥 . 3 cos 2𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 3(−sin2𝑥 ) × 2 × log 5𝑥 + 1/5𝑥 × 5 × 3 cos 2𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = −6 sin 2𝑥 . log5𝑥 + (3 cos2𝑥)/𝑥 Using Product rule (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = 3 𝑐𝑜𝑠 2𝑥 & 𝑣=𝑙𝑜𝑔 5𝑥 𝑑𝑦/𝑑𝑥 = 𝑦 (−6 sin 2𝑥 . log5𝑥 "+ " (3 cos2𝑥)/𝑥) 𝑑𝑦/𝑑𝑥 = (5𝑥)^(3cos2𝑥) (−6 sin 2𝑥 . log5𝑥 "+ " (3 cos2𝑥)/𝑥) 𝒅𝒚/𝒅𝒙 = (𝟓𝒙)^(𝟑𝒄𝒐𝒔𝟐𝒙) ((𝟑 𝒄𝒐𝒔𝟐𝒙)/𝒙−𝟔 𝐬𝐢𝐧 𝟐𝒙 . 𝒍𝒐𝒈𝟓𝒙 )