Ex 5.4, 7 - Differentiate root e root x - Chapter 5 NCERT - Ex 5.4

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  1. Chapter 5 Class 12 Continuity and Differentiability
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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 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2 ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ × 𝑑 𝑒﷮ ﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2 ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ × 𝑒﷮ ﷮𝑥﷯﷯ . 𝑑 ﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ ﷮𝑥﷯﷯﷮4 ﷮𝑥﷯ . ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒆﷮ ﷮𝒙﷯﷯﷮𝟒 ﷮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 𝑦﷯﷮𝑑𝑥﷯ = 𝑑 ﷮𝑡﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑡 in denominator, so multiplying & Dividing by 𝑑𝑡 . 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮𝑡﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮𝑡﷯﷯﷮𝑑𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2 ﷮𝑡﷯﷯ × 𝑑𝑡﷮𝑑𝑥﷯ Putting value of t = 𝑒﷮ ﷮𝑥﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2 ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ × 𝑑 𝑒﷮ ﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2 ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ × 𝑒﷮ ﷮𝑥﷯﷯ . 𝑑 ﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ ﷮𝑥﷯﷯﷮4 ﷮𝑥﷯ . ﷮ 𝑒﷮ ﷮𝑥﷯﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒆﷮ ﷮𝒙﷯﷯﷮𝟒 ﷮𝒙 . 𝒆﷮ ﷮𝒙﷯﷯﷯﷯

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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.