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Ex 5.4, 2 (Method 1) Differentiate ๐‘ค.๐‘Ÿ.๐‘ก. x in , ๐‘’^(sin^(โˆ’1) ๐‘ฅ)Let ๐‘ฆ = ๐‘’^(sin^(โˆ’1) ๐‘ฅ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘ฆ)/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^(sin^(โˆ’1) ๐‘ฅ) )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(sin^(โˆ’1) ๐‘ฅ) . ๐‘‘(sin^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(sin^(โˆ’1) ๐‘ฅ) . (1/โˆš(1 โˆ’ ๐‘ฅ^2 )) ๐’…(๐’š)/๐’…๐’™ = ๐’†^(ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ๐’™)/โˆš(๐Ÿโˆ’๐’™^๐Ÿ ) (๐‘‘(๐‘’^๐‘ฅ )/๐‘‘๐‘ฅ " = " ๐‘’^๐‘ฅ " " ) Ex 5.4, 2 (Method 2) Differentiate ๐‘ค.๐‘Ÿ.๐‘ก. x in , ๐‘’^(sin^(โˆ’1) ๐‘ฅ)Let ๐‘ฆ = ๐‘’^(sin^(โˆ’1) ๐‘ฅ) Let sin^(โˆ’1) ๐‘ฅ=๐‘ก ๐‘ฆ = ๐‘’^๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘ฆ)/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ฅ We need ๐‘‘๐‘ก in denominator, so multiplying & Dividing by ๐‘‘๐‘ก . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ก/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ก/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘’^๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ Putting value of ๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘’^(sin^(โˆ’1) ๐‘ฅ) ร— ๐‘‘(sin^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘’^(sin^(โˆ’1) ๐‘ฅ) ร— 1/โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐’…๐’š/๐’…๐’™ = ๐’†^(ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ๐’™)/โˆš(๐Ÿ โˆ’ ๐’™^๐Ÿ )

  1. Chapter 5 Class 12 Continuity and Differentiability
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About the Author

Davneet Singh

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