# Ex 5.4, 7 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.4, 7 (Method 1) Differentiate w.r.t. x in, 𝑒 𝑥, x > 0 Let 𝑦 = 𝑒 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑦𝑑𝑥 = 𝑑 𝑒 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 12 𝑒 𝑥 × 𝑑 𝑒 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 12 𝑒 𝑥 × 𝑒 𝑥 . 𝑑 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑒 𝑥4 𝑥 . 𝑒 𝑥 𝒅𝒚𝒅𝒙 = 𝒆 𝒙𝟒 𝒙 . 𝒆 𝒙 Ex 5.4, 7 (Method 2) Differentiate 𝑤.𝑟.𝑡. 𝑥 in, 𝑒 𝑥, x > 0 Let 𝑦 = 𝑒 𝑥 Let 𝑒 𝑥=𝑡 𝑦 = 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑 𝑦𝑑𝑥 = 𝑑 𝑡𝑑𝑥 We need 𝑑𝑡 in denominator, so multiplying & Dividing by 𝑑𝑡 . 𝑑𝑦𝑑𝑥 = 𝑑 𝑡𝑑𝑥 × 𝑑𝑡𝑑𝑡 𝑑𝑦𝑑𝑥 = 𝑑 𝑡𝑑𝑡 × 𝑑𝑡𝑑𝑥 𝑑𝑦𝑑𝑥 = 12 𝑡 × 𝑑𝑡𝑑𝑥 Putting value of t = 𝑒 𝑥 𝑑𝑦𝑑𝑥 = 12 𝑒 𝑥 × 𝑑 𝑒 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 12 𝑒 𝑥 × 𝑒 𝑥 . 𝑑 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑒 𝑥4 𝑥 . 𝑒 𝑥 𝒅𝒚𝒅𝒙 = 𝒆 𝒙𝟒 𝒙 . 𝒆 𝒙

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.