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Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

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₯) πππβ‘π = π ππ¨π¬ ππ . πππ β‘ππ Differentiating both sides π€.π.π‘. x 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₯ (As πππβ‘(π^π) = π πππβ‘π) π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ π/π . π π/π π = π(3 cos 2π₯ . log β‘5π₯)/ππ₯ 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₯ ) Γ 2 Γ log β‘5π₯ + 1/5π₯ Γ 5 Γ 3 cos 2π₯ 1/π¦ . ππ¦/ππ₯ = β6 sin 2π₯ . logβ‘5π₯ + (3 cosβ‘2π₯)/π₯ Using Product rule (π’π£)β = π’βπ£ + π£βπ’ where u = 3 πππ 2π₯ & π£=πππ β‘5π₯ 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 )