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Transcript

Misc 3 Differentiate 𝑤.𝑟.𝑡. 𝑥 the function, (5𝑥)^(3cos⁡2𝑥) Let 𝑦" = " (5𝑥)^(3cos⁡2𝑥) Taking log on both sides log⁡𝑦 = log (5𝑥)^(3cos⁡2𝑥) 𝒍𝒐𝒈⁡𝒚 = 𝟑 𝐜𝐨𝐬 𝟐𝒙 . 𝒍𝒐𝒈 ⁡𝟓𝒙 Differentiating both sides 𝑤.𝑟.𝑡. x 𝑑(log⁡𝑦 )/𝑑𝑥 = 𝑑(3 cos 2𝑥 . log ⁡5𝑥)/𝑑𝑥 𝑑(log⁡𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑(3 cos 2𝑥 . log ⁡5𝑥)/𝑑𝑥 (As 𝑙𝑜𝑔⁡(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔⁡𝑎) 𝑑(log⁡𝑦 )/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(−sin⁡2𝑥 ) × 2 × log ⁡5𝑥 + 1/5𝑥 × 5 × 3 cos 2𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = −6 sin 2𝑥 . log⁡5𝑥 + (3 cos⁡2𝑥)/𝑥 Using Product rule (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = 3 𝑐𝑜𝑠 2𝑥 & 𝑣=𝑙𝑜𝑔 ⁡5𝑥 𝑑𝑦/𝑑𝑥 = 𝑦 (−6 sin 2𝑥 . log⁡5𝑥 "+ " (3 cos⁡2𝑥)/𝑥) 𝑑𝑦/𝑑𝑥 = (5𝑥)^(3cos⁡2𝑥) (−6 sin 2𝑥 . log⁡5𝑥 "+ " (3 cos⁡2𝑥)/𝑥) 𝒅𝒚/𝒅𝒙 = (𝟓𝒙)^(𝟑𝒄𝒐𝒔⁡𝟐𝒙) ((𝟑 𝒄𝒐𝒔⁡𝟐𝒙)/𝒙−𝟔 𝐬𝐢𝐧 𝟐𝒙 . 𝒍𝒐𝒈⁡𝟓𝒙 )

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.