# Ex 5.7, 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.7, 16 (Method 1) If 𝑒𝑦 (x+1)= 1, show that 𝑑2𝑦𝑑𝑥2 = 𝑑𝑦𝑑𝑥2 We need to show that 𝑑2𝑦𝑑𝑥2 = 𝑑𝑦𝑑𝑥2 𝑒𝑦 (x+1)= 1 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑 𝑒𝑦 x+1𝑑𝑥 = 𝑑(1)𝑑𝑥 𝑑 𝑒𝑦 x + 1𝑑𝑥 = 0 𝑑( 𝑒𝑦)𝑑𝑥 . (x+1) + 𝑑 (x + 1)𝑑𝑥 . 𝑒𝑦 = 0 𝑑( 𝑒𝑦)𝑑𝑥 × 𝑑𝑦𝑑𝑦 (x+1) + 𝑑(𝑥)𝑑𝑥 + 𝑑(1)𝑑𝑥 . 𝑒𝑦 = 0 𝑑( 𝑒𝑦)𝑑𝑦 × 𝑑𝑦𝑑𝑥 (x+1) + 1+0 . 𝑒𝑦 = 0 𝑒𝑦 × 𝑑𝑦𝑑𝑥 (x+1) + 𝑒𝑦 = 0 𝑒𝑦 𝑑𝑦𝑑𝑥 (x+1) = − 𝑒𝑦 𝑑𝑦𝑑𝑥 = − 𝑒𝑦 𝑒𝑦 (𝑥 + 1) 𝑑𝑦𝑑𝑥 = − 1(𝑥 + 1) Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 − 1 𝑥+1 𝑑2𝑦𝑑 𝑥2 = − 𝑑(1)𝑑𝑥 . 𝑥+1 − 𝑑 𝑥+1𝑑𝑥 . 1 (𝑥+1)2 = − 0 . 𝑥+1 − 𝑑 𝑥+1𝑑𝑥 . 1 (𝑥+1)2 = − 0 − 1 + 0 . 1 (𝑥+1)2 = − −1 (𝑥+1)2 = 1 (𝑥+1)2 Hence 𝑑2𝑦𝑑 𝑥2 = 1 (𝑥+1)2 = −1 𝑥+12 = 𝑑𝑦𝑑𝑥2 Hence proved Ex 5.7, 16 (Method 2) If 𝑦= 𝑒𝑦 (x+1)= 1, show that 𝑑2𝑦𝑑𝑥2 = 𝑑𝑦𝑑𝑥2 If 𝑦= 𝑒𝑦 (x+1)= 1 We need to show that 𝑑2𝑦𝑑𝑥2 = 𝑑𝑦𝑑𝑥2 𝑒𝑦 (𝑥+1)= 1 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑 𝑒𝑦 x+1𝑑𝑥 = 𝑑(1)𝑑𝑥 𝑑 𝑒𝑦 x + 1𝑑𝑥 = 0 𝑑( 𝑒𝑦)𝑑𝑥 . (x+1) + 𝑑 (x + 1)𝑑𝑥 . 𝑒𝑦 = 0 𝑑( 𝑒𝑦)𝑑𝑥 × 𝑑𝑦𝑑𝑦 (x+1) + 𝑑(𝑥)𝑑𝑥 + 𝑑(1)𝑑𝑥 . 𝑒𝑦 = 0 𝑑( 𝑒𝑦)𝑑𝑦 × 𝑑𝑦𝑑𝑥 (x+1) + 1+0 . 𝑒𝑦 = 0 𝑒𝑦 × 𝑑𝑦𝑑𝑥 (x+1) + 𝑒𝑦 = 0 𝑒𝑦 𝑑𝑦𝑑𝑥 (x+1) = − 𝑒𝑦 𝑑𝑦𝑑𝑥 = − 𝑒𝑦 𝑒𝑦 (𝑥 + 1) 𝑑𝑦𝑑𝑥 = − 1(𝑥 + 1) Given, 𝑒𝑦 x + 1 = 1 𝒆𝒚 = 𝟏𝒙 + 𝟏 Putting (2) in (1) 𝑑𝑦𝑑𝑥 = − 1(𝑥 + 1) 𝑑𝑦𝑑𝑥 = − 𝑒𝑦 Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑(− 𝑒𝑦)𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = − 𝑑( 𝑒𝑦)𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = − 𝑑( 𝑒𝑦)𝑑𝑥 × 𝑑𝑦𝑑𝑦 𝑑2𝑦𝑑 𝑥2 = − 𝑑( 𝑒𝑦)𝑑𝑦 × 𝑑𝑦𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = − 𝒆𝒚× 𝑑𝑦𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = 𝒅𝒚𝒅𝒙 × 𝑑𝑦𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = 𝑑𝑦𝑑𝑥 × 𝑑𝑦𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = 𝑑𝑦𝑑𝑥2 Hence proved .

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.