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Ex 5.7, 3 - Find second order derivatives of x cosx - Ex 5.7

Ex 5.7, 3 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.