Finding derivative of a function by chain rule

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.2, 6 Differentiate the functions with respect to π₯ cosβ‘π₯3 . sin2 (π₯5)Let π¦ = cosβ‘π₯3 . sin2 (π₯5) Let π’ = cosβ‘π₯3 & π£=sin2 (π₯5) β΄ π¦ = ππ We need to find derivative of π¦ π€.π.π‘.π₯ π¦^β² = (π’π£)^β² = π’^β² π£+π£^β² π’ Finding πβ π’=cosβ‘π₯3 " " Differentiating π’^β² = (cosβ‘π₯3) = βsinβ‘π₯3 . (π₯^3 )^β² = βsinβ‘γπ₯^3 γ. 3π₯^(3 β1) = βsinβ‘γπ₯^3 γ. 3π₯^2 = β ππ^π . πππβ‘γπ^π γ Finding πβ π£=sin2 π₯5 π£=(sin π₯5)^2 Differentiating π£^β² = ((sin π₯5)^2 )^β² = 2(sin π₯5). (sin π₯^5 )^β² = 2 sin π₯^5 (cosβ‘γπ₯^5 γ ) (π₯^5 )^β² = 2 sin π₯^5 .cosβ‘γπ₯^5 γ . 5π₯^4 = 10π₯^4 . sin π₯^5 .cosβ‘γπ₯^5 γ Now π¦^β² = π’^β² π£+π£^β² π’ =(β 3π₯^2 . sinβ‘γπ₯^3 γ ) .(sin2 π₯5)+(10π₯^4 . sin π₯^5 .cosβ‘γπ₯^5 γ )(cosβ‘π₯3) =πππ^π . πππ π^π .πππβ‘γπ^π γ. πππβ‘γπ^π γβ ππ^π. πππβ‘γπ^π γ.ππππ ππ