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Ex 5.4, 2 - Differentiate e^sin-1 x - Teachoo - Ex 5.4

Ex 5.4, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.4, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Transcript

Ex 5.4, 2 (Method 1) Differentiate 𝑤.𝑟.𝑡. x in , 𝑒^(sin^(−1) 𝑥)Let 𝑦 = 𝑒^(sin^(−1) 𝑥) Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑦)/𝑑𝑥 = 𝑑(𝑒^(sin^(−1) 𝑥) )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(sin^(−1) 𝑥) . 𝑑(sin^(−1) 𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(sin^(−1) 𝑥) . (1/√(1 − 𝑥^2 )) 𝒅(𝒚)/𝒅𝒙 = 𝒆^(〖𝒔𝒊𝒏〗^(−𝟏) 𝒙)/√(𝟏−𝒙^𝟐 ) (𝑑(𝑒^𝑥 )/𝑑𝑥 " = " 𝑒^𝑥 " " ) Ex 5.4, 2 (Method 2) Differentiate 𝑤.𝑟.𝑡. x in , 𝑒^(sin^(−1) 𝑥)Let 𝑦 = 𝑒^(sin^(−1) 𝑥) Let sin^(−1) 𝑥=𝑡 𝑦 = 𝑒^𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑(𝑦)/𝑑𝑥 = 𝑑(𝑒^𝑡 )/𝑑𝑥 We need 𝑑𝑡 in denominator, so multiplying & Dividing by 𝑑𝑡 . 𝑑𝑦/𝑑𝑥= 𝑑(𝑒^𝑡 )/𝑑𝑥 × 𝑑𝑡/𝑑𝑡 𝑑𝑦/𝑑𝑥= 𝑑(𝑒^𝑡 )/𝑑𝑥 × 𝑑𝑡/𝑑𝑡 𝑑𝑦/𝑑𝑥= 𝑑(𝑒^𝑡 )/𝑑𝑡 × 𝑑𝑡/𝑑𝑥 𝑑𝑦/𝑑𝑥= 𝑒^𝑡 × 𝑑𝑡/𝑑𝑥 Putting value of 𝑡 𝑑𝑦/𝑑𝑥= 𝑒^(sin^(−1) 𝑥) × 𝑑(sin^(−1) 𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥= 𝑒^(sin^(−1) 𝑥) × 1/√(1 − 𝑥^2 ) 𝒅𝒚/𝒅𝒙 = 𝒆^(〖𝒔𝒊𝒏〗^(−𝟏) 𝒙)/√(𝟏 − 𝒙^𝟐 )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.