Last updated at Dec. 16, 2024 by Teachoo
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¦ = ππ We need to find derivative of π¦ π€.π.π‘.π₯ π¦^β² = (π’π£)^β² = π’^β² π£+π£^β² π’ Finding πβ π’=cosβ‘π₯3 " " Differentiating π’^β² = (cosβ‘π₯3) = βsinβ‘π₯3 . (π₯^3 )^β² = βsinβ‘γπ₯^3 γ. 3π₯^(3 β1) = βsinβ‘γπ₯^3 γ. 3π₯^2 = β ππ^π . πππβ‘γπ^π γ Finding πβ π£=sin2 π₯5 π£=(sin π₯5)^2 Differentiating π£^β² = ((sin π₯5)^2 )^β² = 2(sin π₯5). (sin π₯^5 )^β² = 2 sin π₯^5 (cosβ‘γπ₯^5 γ ) (π₯^5 )^β² = 2 sin π₯^5 .cosβ‘γπ₯^5 γ . 5π₯^4 = 10π₯^4 . sin π₯^5 .cosβ‘γπ₯^5 γ Now π¦^β² = π’^β² π£+π£^β² π’ =(β 3π₯^2 . sinβ‘γπ₯^3 γ ) .(sin2 π₯5)+(10π₯^4 . sin π₯^5 .cosβ‘γπ₯^5 γ )(cosβ‘π₯3) =πππ^π . πππ π^π .πππβ‘γπ^π γ. πππβ‘γπ^π γβ ππ^π. πππβ‘γπ^π γ.ππππ ππ
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo