# Misc 3 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Sept. 17, 2019 by Teachoo

Last updated at Sept. 17, 2019 by Teachoo

Transcript

Misc 3 Differentiate π€.π.π‘. π₯ the function, (5π₯)^(3cosβ‘2π₯) Let π¦" = " (5π₯)^(3cosβ‘2π₯) Taking log on both sides logβ‘π¦ = log (5π₯)^(3cosβ‘2π₯) logβ‘π¦ = 3 cos 2π₯ . log β‘5π₯ (As πππβ‘(π^π) = π πππβ‘π) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦ )/ππ₯ = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ 1/π¦ . ππ¦/ππ₯ = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ Using product rule 3 πππ 2π₯ . πππ β‘5π₯ As (π’π£)β = π’βπ£ + π£βπ’ where u = 3 πππ 2π₯ & π£=πππ β‘5π₯ 1/π¦ . ππ¦/ππ₯ = π(3 cos 2π₯ )/ππ₯ . log β‘5π₯ + π(log β‘5π₯)/ππ₯ .3 cos 2π₯ 1/π¦ . ππ¦/ππ₯ = 3 π(cos 2π₯ )/ππ₯ . log β‘5π₯ + 1/5π₯ . π(5π₯)/ππ₯ . 3 cos 2π₯ 1/π¦ . ππ¦/ππ₯ = 3(βsinβ‘2π₯ ) . π(2π₯)/ππ₯ . log β‘5π₯ + 1/5π₯ Γ 5 . 3 cos 2π₯ 1/π¦ . ππ¦/ππ₯ = β3 sin 2π₯ . 2 . logβ‘5π₯ + (3 cosβ‘2π₯)/π₯ 1/π¦ . ππ¦/ππ₯ = β6 sin 2π₯ . logβ‘5π₯ + (3 cosβ‘2π₯)/π₯ ππ¦/ππ₯ = π¦ (β6 sin 2π₯ . logβ‘5π₯ "+ " (3 cosβ‘2π₯)/π₯) ππ¦/ππ₯ = (5π₯)^(3cosβ‘2π₯) (β6 sin 2π₯ . logβ‘5π₯ "+ " (3 cosβ‘2π₯)/π₯) π π/π π = (ππ)^(ππππβ‘ππ) ((π πππβ‘ππ)/πβπ π¬π’π§ ππ . πππβ‘ππ )

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.