Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity   1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Miscellaneous

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 (As 𝑙𝑜𝑔⁡(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔⁡𝑎) 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 𝑑(log⁡𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑(3 cos 2𝑥 . log ⁡5𝑥)/𝑑𝑥 𝑑(log⁡𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑(3 cos 2𝑥 . log ⁡5𝑥)/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑(3 cos 2𝑥 . log ⁡5𝑥)/𝑑𝑥 Using product rule 3 𝑐𝑜𝑠 2𝑥 . 𝑙𝑜𝑔 ⁡5𝑥 As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = 3 𝑐𝑜𝑠 2𝑥 & 𝑣=𝑙𝑜𝑔 ⁡5𝑥 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 ) . 𝑑(2𝑥)/𝑑𝑥 . log ⁡5𝑥 + 1/5𝑥 × 5 . 3 cos 2𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = −3 sin 2𝑥 . 2 . log⁡5𝑥 + (3 cos⁡2𝑥)/𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = −6 sin 2𝑥 . log⁡5𝑥 + (3 cos⁡2𝑥)/𝑥 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 )

Miscellaneous 