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Ex 5.2, 6 - Differentiate cos x3 sin2 (x5) - Chapter 5 CBSE

Ex 5.2, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.2, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.2, 6 Differentiate the functions with respect to π‘₯ cos⁑π‘₯3 . sin2 (π‘₯5)Let 𝑦 = cos⁑π‘₯3 . sin2 (π‘₯5) Let 𝑒 = cos⁑π‘₯3 & 𝑣=sin2 (π‘₯5) ∴ 𝑦 = 𝒖𝒗 We need to find derivative of 𝑦 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑦^β€² = (𝑒𝑣)^β€² = 𝑒^β€² 𝑣+𝑣^β€² 𝑒 Finding 𝒖’ 𝑒=cos⁑π‘₯3 " " Differentiating 𝑒^β€² = (cos⁑π‘₯3) = βˆ’sin⁑π‘₯3 . (π‘₯^3 )^β€² = βˆ’sin⁑〖π‘₯^3 γ€—. 3π‘₯^(3 βˆ’1) = βˆ’sin⁑〖π‘₯^3 γ€—. 3π‘₯^2 = βˆ’ πŸ‘π’™^𝟐 . π’”π’Šπ’β‘γ€–π’™^πŸ‘ γ€— Finding 𝒗’ 𝑣=sin2 π‘₯5 𝑣=(sin π‘₯5)^2 Differentiating 𝑣^β€² = ((sin π‘₯5)^2 )^β€² = 2(sin π‘₯5). (sin π‘₯^5 )^β€² = 2 sin π‘₯^5 (cos⁑〖π‘₯^5 γ€— ) (π‘₯^5 )^β€² = 2 sin π‘₯^5 .cos⁑〖π‘₯^5 γ€— . 5π‘₯^4 = 10π‘₯^4 . sin π‘₯^5 .cos⁑〖π‘₯^5 γ€— Now 𝑦^β€² = 𝑒^β€² 𝑣+𝑣^β€² 𝑒 =(βˆ’ 3π‘₯^2 . sin⁑〖π‘₯^3 γ€— ) .(sin2 π‘₯5)+(10π‘₯^4 . sin π‘₯^5 .cos⁑〖π‘₯^5 γ€— )(cos⁑π‘₯3) =πŸπŸŽπ’™^πŸ’ . π’”π’Šπ’ 𝒙^πŸ“ .𝒄𝒐𝒔⁑〖𝒙^πŸ“ γ€—. 𝒄𝒐𝒔⁑〖𝒙^πŸ‘ γ€—βˆ’ πŸ‘π’™^𝟐. π’”π’Šπ’β‘γ€–π’™^πŸ‘ γ€—.π’”π’Šπ’πŸ π’™πŸ“

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.