Ex 5.4, 2 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.4, 2 (Method 1) Differentiate ๐ค.๐.๐ก. x in , ๐^(sin^(โ1) ๐ฅ)Let ๐ฆ = ๐^(sin^(โ1) ๐ฅ) Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐(๐ฆ)/๐๐ฅ = ๐(๐^(sin^(โ1) ๐ฅ) )/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐^(sin^(โ1) ๐ฅ) . ๐(sin^(โ1) ๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐^(sin^(โ1) ๐ฅ) . (1/โ(1 โ ๐ฅ^2 )) ๐ (๐)/๐ ๐ = ๐^(ใ๐๐๐ใ^(โ๐) ๐)/โ(๐โ๐^๐ ) (๐(๐^๐ฅ )/๐๐ฅ " = " ๐^๐ฅ " " ) Ex 5.4, 2 (Method 2) Differentiate ๐ค.๐.๐ก. x in , ๐^(sin^(โ1) ๐ฅ)Let ๐ฆ = ๐^(sin^(โ1) ๐ฅ) Let sin^(โ1) ๐ฅ=๐ก ๐ฆ = ๐^๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐(๐ฆ)/๐๐ฅ = ๐(๐^๐ก )/๐๐ฅ We need ๐๐ก in denominator, so multiplying & Dividing by ๐๐ก . ๐๐ฆ/๐๐ฅ= ๐(๐^๐ก )/๐๐ฅ ร ๐๐ก/๐๐ก ๐๐ฆ/๐๐ฅ= ๐(๐^๐ก )/๐๐ฅ ร ๐๐ก/๐๐ก ๐๐ฆ/๐๐ฅ= ๐(๐^๐ก )/๐๐ก ร ๐๐ก/๐๐ฅ ๐๐ฆ/๐๐ฅ= ๐^๐ก ร ๐๐ก/๐๐ฅ Putting value of ๐ก ๐๐ฆ/๐๐ฅ= ๐^(sin^(โ1) ๐ฅ) ร ๐(sin^(โ1) ๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ= ๐^(sin^(โ1) ๐ฅ) ร 1/โ(1 โ ๐ฅ^2 ) ๐ ๐/๐ ๐ = ๐^(ใ๐๐๐ใ^(โ๐) ๐)/โ(๐ โ ๐^๐ )
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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