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Ex 5.7, 6 Class 12 Maths - Find second order derivative of e^x sin 5x

Ex 5.7, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.7, 6 Find the second order derivatives of the function 𝑒^π‘₯ sin⁑5π‘₯ Let y = 𝑒^π‘₯ sin⁑5π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ . 𝑑𝑦/𝑑π‘₯ = (𝑑(𝑒^π‘₯ " " sin⁑5π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒^π‘₯ )/𝑑π‘₯ .sin⁑〖 5π‘₯γ€— + (𝑑(γ€–sin 5〗⁑π‘₯))/𝑑π‘₯ . 𝑒^π‘₯ 𝑑𝑦/𝑑π‘₯ =𝑒^π‘₯ .sin⁑〖 5π‘₯γ€— + cos⁑5π‘₯ (𝑑(5π‘₯))/𝑑π‘₯ . 𝑒^π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒^π‘₯. sin⁑〖 5π‘₯γ€— + 5.𝑒^π‘₯. cos⁑5π‘₯ using product rule in 𝑒^π‘₯ 𝑠𝑖𝑛⁑5π‘₯ As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯)= (𝑑 (𝑒^π‘₯. sin⁑〖 5π‘₯γ€— " + " 5.𝑒^π‘₯." " cos⁑5π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑑 (𝑒^π‘₯. sin⁑〖 5π‘₯γ€—))/𝑑π‘₯ + (𝑑 (5.𝑒^π‘₯." " cos⁑5π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑑 (𝑒^π‘₯ sin⁑5π‘₯))/𝑑π‘₯ + 5 (𝑑(𝑒^π‘₯." " cos⁑5π‘₯))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = ((𝑑 (𝑒^π‘₯))/𝑑π‘₯.sin⁑5π‘₯+(𝑑 (sin⁑5π‘₯))/𝑑π‘₯ .𝑒^π‘₯ ) + 5 ((𝑑(𝑒^π‘₯))/𝑑π‘₯ .cos⁑5π‘₯+(𝑑(cos⁑5π‘₯))/𝑑π‘₯ .𝑒^π‘₯ ) using product rule in 𝑒^π‘₯ 𝑠𝑖𝑛⁑5π‘₯ & 𝑒^π‘₯." " π‘π‘œπ‘ β‘5π‘₯ As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑒^π‘₯.sin⁑5π‘₯+(cos⁑5π‘₯ ) .(𝑑(⁑5π‘₯))/𝑑π‘₯.𝑒^π‘₯ ) + 5 (𝑒^π‘₯.cos⁑5π‘₯+(βˆ’sin⁑〖5 π‘₯)γ€—.(𝑑(⁑5π‘₯))/𝑑π‘₯.𝑒^π‘₯ ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑒^π‘₯ sin⁑5π‘₯+cos⁑5π‘₯.5.𝑒^π‘₯ ) + 5 (𝑒^π‘₯ cos⁑5π‘₯βˆ’sin⁑〖5 π‘₯γ€—.5.𝑒^π‘₯ ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑒^π‘₯ sin⁑5π‘₯+5𝑒^π‘₯ cos⁑5π‘₯+5𝑒^π‘₯ cos⁑5π‘₯βˆ’25𝑒^π‘₯ sin⁑〖5 π‘₯γ€— (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 10 𝑒^π‘₯ cos⁑5π‘₯ βˆ’ 24 𝑒^π‘₯ sin⁑〖5 π‘₯γ€— (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = 2𝒆^𝒙 (πŸ“π’„π’π’”β‘πŸ“π’™ βˆ’ 1𝟐 π’”π’Šπ’β‘γ€–πŸ“ 𝒙〗)

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Davneet Singh

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.