Ex 5.3, 5 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.3, 5 Find ๐๐ฆ/๐๐ฅ in, ๐ฅ2 + ๐ฅ๐ฆ + ๐ฆ2 = 100 ๐ฅ2 + ๐ฅ๐ฆ + ๐ฆ2 = 100 Differentiating both sides ๐ค.๐.๐ก.๐ฅ . ๐(๐ฅ2 + ๐ฅ๐ฆ + ๐ฆ2)/๐๐ฅ = (๐ (100))/๐๐ฅ ๐(๐ฅ2)/๐๐ฅ + ๐(๐ฅ๐ฆ)/๐๐ฅ + (๐(๐ฆ2))/๐๐ฅ = (๐ (100))/๐๐ฅ 2๐ฅ^(2โ1) + (๐(๐ฅ๐ฆ))/๐๐ฅ + (๐(๐ฆ2))/๐๐ฅ ร ๐๐ฆ/๐๐ฆ = 0 2๐ฅ + (๐(๐ฅ๐ฆ))/๐๐ฅ + (๐(๐ฆ2))/๐๐ฆ ร ๐๐ฆ/๐๐ฅ = 0 As (๐ฅ^๐ )^โฒ=๐๐ฅ^(๐โ1) & derivative of a constant is zero 2๐ฅ + (๐(๐ฅ๐ฆ))/๐๐ฅ + 2๐ฆ^(2โ1) . ๐๐ฆ/๐๐ฅ = 0 2๐ฅ + (๐(๐ฅ๐ฆ))/๐๐ฅ + 2๐ฆ . ๐๐ฆ/๐๐ฅ = 0 Using product rule in ๐ฅ๐ฆ = ๐ฅ^โฒ ๐ฆ+๐ฆ^โฒ ๐ฅ 2๐ฅ + ((๐(๐ฅ))/๐๐ฅ .๐ฆ+๐(๐ฆ)/๐๐ฅ.๐ฅ) + 2๐ฆ . ๐๐ฆ/๐๐ฅ = 0 2๐ฅ + (1 .๐ฆ+๐ฅ.๐(๐ฆ)/๐๐ฅ) + 2๐ฆ . ๐๐ฆ/๐๐ฅ = 0 2๐ฅ + ๐ฆ + ๐ฅ . ๐๐ฆ/๐๐ฅ + 2๐ฆ . ๐๐ฆ/๐๐ฅ = 0 (2๐ฅ + ๐ฆ) + ๐๐ฆ/๐๐ฅ . (๐ฅ + 2๐ฆ) = 0 ๐๐ฆ/๐๐ฅ . (๐ฅ + 2๐ฆ) = โ (2๐ฅ + ๐ฆ) ๐ ๐/๐ ๐ = (โ( ๐๐ + ๐ ))/( ( ๐ + ๐๐ ))
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo