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

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.7, 3 Find the second order derivatives of the function π‘₯. cos⁑π‘₯ Let y = π‘₯. cos⁑π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ . 𝑑𝑦/𝑑π‘₯ = (𝑑(π‘₯". " cos⁑π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(π‘₯)/𝑑π‘₯ .cos⁑π‘₯ + (𝑑(cos⁑〖π‘₯)γ€—)/𝑑π‘₯ . π‘₯ 𝑑𝑦/𝑑π‘₯ = cos⁑π‘₯+(βˆ’ sin⁑π‘₯ ) . π‘₯ Using Product Rule As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 𝑑𝑦/𝑑π‘₯ = cos⁑π‘₯ βˆ’ π‘₯ sin⁑π‘₯ Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = (𝑑 (cos⁑π‘₯" βˆ’ " π‘₯ sin⁑π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑑(cos⁑〖π‘₯)γ€—)/𝑑π‘₯ βˆ’ (𝑑(γ€–x sin〗⁑〖π‘₯)γ€—)/𝑑π‘₯ Using product rule in π‘₯ 𝑠𝑖𝑛⁑π‘₯" " (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = "βˆ’ " sin⁑π‘₯ βˆ’ (𝑑(π‘₯)/𝑑π‘₯.sin⁑π‘₯+𝑑(sin⁑π‘₯ )/𝑑π‘₯.π‘₯) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = "βˆ’ " sin⁑π‘₯ βˆ’ (sin⁑〖π‘₯+cos⁑〖π‘₯ . π‘₯γ€— γ€— ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = "βˆ’ " sin⁑π‘₯ βˆ’ sin⁑〖π‘₯βˆ’γ€–x cos〗⁑〖π‘₯ γ€— γ€— (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = "βˆ’ " 𝒙 π’„π’π’”β‘π’™βˆ’πŸ π’”π’Šπ’β‘π’™

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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.