# Misc 3 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Nov. 19, 2019 by Teachoo

Last updated at Nov. 19, 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 )

Miscellaneous

Misc 1

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 Important

Misc 20

Misc 21

Misc 22

Misc 23 Important

About the Author

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.