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Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Misc 2 Differentiate π€.π.π‘. π₯ the function sin^3 π₯ + cos^6 π₯ Let π¦=sin^3 π₯ + cos^6 π₯ Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π((sin^3 π₯ )" + " (cos^6 π₯" " ))/ππ₯ = π((sin^3 π₯ ))/( ππ₯) + π((cos^6 π₯" " ))/ππ₯ = 3 sin^2 π₯ . π(sinβ‘π₯ )/ππ₯ + 6 cos^5 π₯" ". π(cosβ‘π₯ )/ππ₯ = 3 sin^2 π₯ . cosβ‘π₯ + 6 cos^5 π₯. (βsinβ‘π₯ ) = 3 sin^2 π₯ . cosβ‘π₯ β 6 sinβ‘π₯. cos^5 π₯ = 3 γπππγ^π π . πππβ‘π (πππβ‘πβπ γπππγ^πβ‘π )