Last updated at Dec. 16, 2024 by Teachoo
Misc 23 If ๐ฆ=๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ) , โ 1 โค ๐ฅ โค 1, show that (1โ๐ฅ^2 ) (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 โ๐ฅ ๐๐ฆ/๐๐ฅ โ ๐2 ๐ฆ =0 . ๐ฆ=๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ) Differentiating ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ฅ = ๐(๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ" " ) )/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ" " ) ร ๐(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ" " ) ร ๐ ((โ1)/โ(1 โ ๐ฅ^2 )) ๐๐ฆ/๐๐ฅ = (โ๐ ๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ" " ))/โ(1 โ ๐ฅ^2 ) โ(1 โ ๐ฅ^2 ) ๐๐ฆ/๐๐ฅ = โ๐๐^(ใ๐ ๐๐๐ ใ^(โ1) ๐ฅ" " ) โ(1 โ ๐ฅ^2 ) ๐๐ฆ/๐๐ฅ = โ๐๐ฆ Since we need to prove (1โ๐ฅ^2 ) (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 โ ๐ฅ ๐๐ฆ/๐๐ฅ โ๐2 ๐ฆ =0 Squaring (1) both sides (โ(1 โ ๐ฅ^2 ) ๐๐ฆ/๐๐ฅ)^2 = (โ๐๐ฆ)^2 (1โ๐ฅ^2 ) (๐ฆ^โฒ )^2 = ๐^2 ๐ฆ^2 Differentiating again w.r.t x ๐((1 โ ๐ฅ^2 ) (๐ฆ^โฒ )^2 )/๐๐ฅ = (d(๐^2 ๐ฆ^2))/๐๐ฅ ๐((1 โ ๐ฅ^2 ) (๐ฆ^โฒ )^2 )/๐๐ฅ = ๐^2 (๐(๐ฆ^2))/๐๐ฅ ๐((1 โ ๐ฅ^2 ) (๐ฆ^โฒ )^2 )/๐๐ฅ = ๐^2 ร 2๐ฆ ร๐๐ฆ/๐๐ฅ ๐(1 โ ๐ฅ^2 )/๐๐ฅ (๐ฆ^โฒ )^2+(1 โ ๐ฅ^2 ) ๐ ((๐^โฒ )^๐ )/๐ ๐ = ๐^2 ร 2๐ฆ๐ฆ^โฒ (โ2๐ฅ)(๐ฆ^โฒ )^2+(1 โ ๐ฅ^2 )(๐๐^โฒ ร ๐ (๐^โฒ )/๐ ๐) = ๐^2 ร 2๐ฆ๐ฆ^โฒ (โ2๐ฅ)(๐ฆ^โฒ )^2+(1 โ ๐ฅ^2 )(๐๐^โฒ ร ๐^โฒโฒ ) = ๐^2 ร 2๐ฆ๐ฆ^โฒ Dividing both sides by ๐๐^โฒ โ๐ฅ๐ฆ^โฒ+(1 โ ๐ฅ^2 ) ๐ฆ^โฒโฒ = ๐^2 ร ๐ฆ โ๐ฅ๐ฆ^โฒ+(1 โ ๐ฅ^2 ) ๐ฆ^โฒโฒ = ๐^2 ๐ฆ (๐ โ ๐^๐ ) ๐^โฒโฒโ๐๐^โฒโ๐^๐ ๐=๐ Hence proved
Miscellaneous
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About the Author
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