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Ex 5.7, 16 - If ey (x + 1)=1, show d2y/dx2 = (dy/dx)2 - Finding second order derivatives- Implicit form

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.7, 16 (Method 1) If 𝑒﷮𝑦﷯ (x+1)= 1, show that 𝑑2𝑦﷮𝑑𝑥2﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ We need to show that 𝑑2𝑦﷮𝑑𝑥2﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ 𝑒﷮𝑦﷯ (x+1)= 1 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑 𝑒﷮𝑦﷯ x+1﷯﷯﷮𝑑𝑥﷯ = 𝑑(1)﷮𝑑𝑥﷯ 𝑑 𝑒﷮𝑦﷯ x + 1﷯﷯﷮𝑑𝑥﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ . (x+1) + 𝑑 (x + 1)﷮𝑑𝑥﷯ . 𝑒﷮𝑦﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ (x+1) + 𝑑(𝑥)﷮𝑑𝑥﷯ + 𝑑(1)﷮𝑑𝑥﷯﷯ . 𝑒﷮𝑦﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ (x+1) + 1+0﷯ . 𝑒﷮𝑦﷯ = 0 𝑒﷮𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ (x+1) + 𝑒﷮𝑦﷯ = 0 𝑒﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ (x+1) = − 𝑒﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑒﷮𝑦﷯﷮ 𝑒﷮𝑦﷯ (𝑥 + 1)﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 1﷮(𝑥 + 1)﷯ Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ − 1﷮ 𝑥+1﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑(1)﷮𝑑𝑥﷯ . 𝑥+1﷯ − 𝑑 𝑥+1﷯﷮𝑑𝑥﷯ . 1﷮ (𝑥+1)﷮2﷯﷯﷯ = − 0 . 𝑥+1﷯ − 𝑑 𝑥+1﷯﷮𝑑𝑥﷯ . 1﷮ (𝑥+1)﷮2﷯﷯﷯ = − 0 − 1 + 0﷯ . 1﷮ (𝑥+1)﷮2﷯﷯﷯ = − −1﷮ (𝑥+1)﷮2﷯﷯﷯ = 1﷮ (𝑥+1)﷮2﷯﷯ Hence 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 1﷮ (𝑥+1)﷮2﷯﷯ = −1﷮ 𝑥+1﷯﷯﷮2﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ Hence proved Ex 5.7, 16 (Method 2) If 𝑦= 𝑒﷮𝑦﷯ (x+1)= 1, show that 𝑑2𝑦﷮𝑑𝑥2﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ If 𝑦= 𝑒﷮𝑦﷯ (x+1)= 1 We need to show that 𝑑2𝑦﷮𝑑𝑥2﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ 𝑒﷮𝑦﷯ (𝑥+1)= 1 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑 𝑒﷮𝑦﷯ x+1﷯﷯﷮𝑑𝑥﷯ = 𝑑(1)﷮𝑑𝑥﷯ 𝑑 𝑒﷮𝑦﷯ x + 1﷯﷯﷮𝑑𝑥﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ . (x+1) + 𝑑 (x + 1)﷮𝑑𝑥﷯ . 𝑒﷮𝑦﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ (x+1) + 𝑑(𝑥)﷮𝑑𝑥﷯ + 𝑑(1)﷮𝑑𝑥﷯﷯ . 𝑒﷮𝑦﷯ = 0 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ (x+1) + 1+0﷯ . 𝑒﷮𝑦﷯ = 0 𝑒﷮𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ (x+1) + 𝑒﷮𝑦﷯ = 0 𝑒﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ (x+1) = − 𝑒﷮𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑒﷮𝑦﷯﷮ 𝑒﷮𝑦﷯ (𝑥 + 1)﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 1﷮(𝑥 + 1)﷯ Given, 𝑒﷮𝑦﷯ x + 1﷯ = 1 𝒆﷮𝒚﷯ = 𝟏﷮𝒙 + 𝟏﷯ Putting (2) in (1) 𝑑𝑦﷮𝑑𝑥﷯ = − 1﷮(𝑥 + 1)﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 𝑒﷮𝑦﷯ Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑(− 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑦﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑( 𝑒﷮𝑦﷯)﷮𝑑𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝒆﷮𝒚﷯× 𝑑𝑦﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝒅𝒚﷮𝒅𝒙﷯ × 𝑑𝑦﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯ × 𝑑𝑦﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯ Hence proved .

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