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Ex 5.7, 16 - Ex 5.7

Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 8
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 9
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 10

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Transcript

Ex 5.7, 16 (Method 1) If 𝑒^𝑦 (x+1)= 1, show that 𝑑2𝑦/𝑑π‘₯2 = (𝑑𝑦/𝑑π‘₯)^2 We need to show that 𝑑2𝑦/𝑑π‘₯2 = (𝑑𝑦/𝑑π‘₯)^2 𝑒^𝑦 (x+1)= 1 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(𝑒^𝑦 (x+1))/𝑑π‘₯ = (𝑑(1))/𝑑π‘₯ 𝑑(𝑒^𝑦 (x + 1))/𝑑π‘₯ = 0 Using product rule in ey(x + 1) As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 where u = ey & v = x + 1 (𝑑(𝑒^𝑦))/𝑑π‘₯ . (x+1) + (𝑑 (x + 1))/𝑑π‘₯ . 𝑒^𝑦 = 0 (𝑑(𝑒^𝑦))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 (x+1) + ((𝑑(π‘₯))/𝑑π‘₯ + (𝑑(1))/𝑑π‘₯) . 𝑒^𝑦 = 0 (𝑑(𝑒^𝑦))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ (x+1) + (1+0) . 𝑒^𝑦 = 0 𝑒^𝑦 Γ— 𝑑𝑦/𝑑π‘₯ (x+1) + 𝑒^𝑦 = 0 𝑒^𝑦 (𝑑𝑦/𝑑π‘₯) (x+1) = βˆ’ 𝑒^𝑦 𝑑𝑦/𝑑π‘₯ = ("βˆ’ " 𝑒^𝑦)/(𝑒^𝑦 (π‘₯ + 1)) 𝑑𝑦/𝑑π‘₯ = ("βˆ’ " 1)/((π‘₯ + 1)) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑/𝑑π‘₯ (("βˆ’ " 1)/((π‘₯+1) )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’[((𝑑(1))/𝑑π‘₯ . (π‘₯ + 1) βˆ’ 𝑑(π‘₯ + 1)/𝑑π‘₯ . 1)/γ€–(π‘₯ + 1)γ€—^2 ] using Quotient Rule As, (𝑒/𝑣)^β€²= (𝑒’𝑣 βˆ’ 𝑣’𝑒)/𝑣^2 where U = 1 & V = x + 1 = βˆ’[(0 . (π‘₯+1) βˆ’ 𝑑(π‘₯+1)/𝑑π‘₯ . 1)/γ€–(π‘₯ + 1)γ€—^2 ] = βˆ’[(0 βˆ’ (1 + 0) . 1)/γ€–(π‘₯ + 1)γ€—^2 ] = βˆ’[(βˆ’1)/γ€–(π‘₯ + 1)γ€—^2 ] = 1/γ€–(π‘₯ + 1)γ€—^2 Hence (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 1/γ€–(π‘₯ + 1)γ€—^2 = ((βˆ’1)/( π‘₯ + 1))^2 = (𝑑𝑦/𝑑π‘₯)^2 Hence proved Ex 5.7, 16 (Method 2) If 𝑦= 𝑒^𝑦 (x+1)= 1, show that 𝑑2𝑦/𝑑π‘₯2 = (𝑑𝑦/𝑑π‘₯)^2 If 𝑦= 𝑒^𝑦 (x+1)= 1 We need to show that 𝑑2𝑦/𝑑π‘₯2 = (𝑑𝑦/𝑑π‘₯)^2 𝑒^𝑦 (π‘₯+1)= 1 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(𝑒^𝑦 (x + 1))/𝑑π‘₯ = (𝑑(1))/𝑑π‘₯ 𝑑(𝑒^𝑦 (x + 1))/𝑑π‘₯ = 0 Using product rule in ey(x + 1) As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 where u = ey & v = x + 1 (𝑑(𝑒^𝑦))/𝑑π‘₯ . (x+1) + (𝑑 (x + 1))/𝑑π‘₯ . 𝑒^𝑦 = 0 (𝑑(𝑒^𝑦))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 (x+1) + ((𝑑(π‘₯))/𝑑π‘₯ + (𝑑(1))/𝑑π‘₯) . 𝑒^𝑦 = 0 (𝑑(𝑒^𝑦))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ (x+1) + (1+0) . 𝑒^𝑦 = 0 𝑒^𝑦 Γ— 𝑑𝑦/𝑑π‘₯ (x+1) + 𝑒^𝑦 = 0 𝑒^𝑦 (𝑑𝑦/𝑑π‘₯) (x+1) = βˆ’ 𝑒^𝑦 𝑑𝑦/𝑑π‘₯ = ("βˆ’ " 𝑒^𝑦)/(𝑒^𝑦 (π‘₯ + 1)) 𝑑𝑦/𝑑π‘₯ = ("βˆ’ " 1)/((π‘₯ + 1)) Given, 𝑒^𝑦 (x + 1) = 1 𝒆^π’š = 𝟏/(𝒙 + 𝟏) Putting (2) in (1) 𝑑𝑦/𝑑π‘₯ = ("βˆ’ " 1)/((π‘₯ + 1)) 𝑑𝑦/𝑑π‘₯ = βˆ’ 𝑒^𝑦 …(1) …(2) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = (𝑑("βˆ’" 𝑒^𝑦))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (𝑑(𝑒^𝑦))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (𝑑(𝑒^𝑦))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (𝑑(𝑒^𝑦))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = "βˆ’ " 𝒆^π’šΓ— 𝑑𝑦/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = π’…π’š/𝒅𝒙 Γ— 𝑑𝑦/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑𝑦/𝑑π‘₯ Γ— 𝑑𝑦/𝑑π‘₯ (From (1) "βˆ’ " 𝑒^𝑦 " = " 𝑑𝑦/𝑑π‘₯) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (𝑑𝑦/𝑑π‘₯)^2 Hence proved

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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.