Last updated at April 16, 2024 by Teachoo
Ex 5.2, 2 Differentiate the functions with respect to π₯ cos (sinβ‘π₯) Let π¦ = cos (sinβ‘π₯) We need to find derivative of π¦, π€.π.π‘.π₯ i.e. (ππ¦ )/ππ₯ = (π(cos (sinβ‘π₯ )))/ππ₯ = β sinβ‘(sinβ‘π₯) . (π(sinβ‘γπ₯)γ)/ππ₯ = β sinβ‘(sinβ‘π₯) . cosβ‘π₯ = β ππ¨π¬β‘π πππ β‘(π¬π’π§β‘γπ)γ