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

Go Ad-free

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 )

Davneet Singh's photo - Co-founder, Teachoo

Made by

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