Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12



  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise


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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.