Ex 5.7, 16 - Ex 5.7

Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 8
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 9
Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 10

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.7, 16 (Method 1) If ๐‘’^๐‘ฆ (x+1)= 1, show that ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 We need to show that ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 ๐‘’^๐‘ฆ (x+1)= 1 Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘’^๐‘ฆ (x+1))/๐‘‘๐‘ฅ = (๐‘‘(1))/๐‘‘๐‘ฅ ๐‘‘(๐‘’^๐‘ฆ (x + 1))/๐‘‘๐‘ฅ = 0 Using product rule in ey(x + 1) As (๐‘ข๐‘ฃ)โ€™= ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข where u = ey & v = x + 1 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ . (x+1) + (๐‘‘ (x + 1))/๐‘‘๐‘ฅ . ๐‘’^๐‘ฆ = 0 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ (x+1) + ((๐‘‘(๐‘ฅ))/๐‘‘๐‘ฅ + (๐‘‘(1))/๐‘‘๐‘ฅ) . ๐‘’^๐‘ฆ = 0 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (x+1) + (1+0) . ๐‘’^๐‘ฆ = 0 ๐‘’^๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (x+1) + ๐‘’^๐‘ฆ = 0 ๐‘’^๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) (x+1) = โˆ’ ๐‘’^๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ("โˆ’ " ๐‘’^๐‘ฆ)/(๐‘’^๐‘ฆ (๐‘ฅ + 1)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ("โˆ’ " 1)/((๐‘ฅ + 1)) Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ (("โˆ’ " 1)/((๐‘ฅ+1) )) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’[((๐‘‘(1))/๐‘‘๐‘ฅ . (๐‘ฅ + 1) โˆ’ ๐‘‘(๐‘ฅ + 1)/๐‘‘๐‘ฅ . 1)/ใ€–(๐‘ฅ + 1)ใ€—^2 ] using Quotient Rule As, (๐‘ข/๐‘ฃ)^โ€ฒ= (๐‘ขโ€™๐‘ฃ โˆ’ ๐‘ฃโ€™๐‘ข)/๐‘ฃ^2 where U = 1 & V = x + 1 = โˆ’[(0 . (๐‘ฅ+1) โˆ’ ๐‘‘(๐‘ฅ+1)/๐‘‘๐‘ฅ . 1)/ใ€–(๐‘ฅ + 1)ใ€—^2 ] = โˆ’[(0 โˆ’ (1 + 0) . 1)/ใ€–(๐‘ฅ + 1)ใ€—^2 ] = โˆ’[(โˆ’1)/ใ€–(๐‘ฅ + 1)ใ€—^2 ] = 1/ใ€–(๐‘ฅ + 1)ใ€—^2 Hence (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = 1/ใ€–(๐‘ฅ + 1)ใ€—^2 = ((โˆ’1)/( ๐‘ฅ + 1))^2 = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 Hence proved Ex 5.7, 16 (Method 2) If ๐‘ฆ= ๐‘’^๐‘ฆ (x+1)= 1, show that ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 If ๐‘ฆ= ๐‘’^๐‘ฆ (x+1)= 1 We need to show that ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 ๐‘’^๐‘ฆ (๐‘ฅ+1)= 1 Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘’^๐‘ฆ (x + 1))/๐‘‘๐‘ฅ = (๐‘‘(1))/๐‘‘๐‘ฅ ๐‘‘(๐‘’^๐‘ฆ (x + 1))/๐‘‘๐‘ฅ = 0 Using product rule in ey(x + 1) As (๐‘ข๐‘ฃ)โ€™= ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข where u = ey & v = x + 1 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ . (x+1) + (๐‘‘ (x + 1))/๐‘‘๐‘ฅ . ๐‘’^๐‘ฆ = 0 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ (x+1) + ((๐‘‘(๐‘ฅ))/๐‘‘๐‘ฅ + (๐‘‘(1))/๐‘‘๐‘ฅ) . ๐‘’^๐‘ฆ = 0 (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (x+1) + (1+0) . ๐‘’^๐‘ฆ = 0 ๐‘’^๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (x+1) + ๐‘’^๐‘ฆ = 0 ๐‘’^๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) (x+1) = โˆ’ ๐‘’^๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ("โˆ’ " ๐‘’^๐‘ฆ)/(๐‘’^๐‘ฆ (๐‘ฅ + 1)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ("โˆ’ " 1)/((๐‘ฅ + 1)) Given, ๐‘’^๐‘ฆ (x + 1) = 1 ๐’†^๐’š = ๐Ÿ/(๐’™ + ๐Ÿ) Putting (2) in (1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ("โˆ’ " 1)/((๐‘ฅ + 1)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’ ๐‘’^๐‘ฆ โ€ฆ(1) โ€ฆ(2) Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = (๐‘‘("โˆ’" ๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (๐‘‘(๐‘’^๐‘ฆ))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = "โˆ’ " ๐’†^๐’šร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = ๐’…๐’š/๐’…๐’™ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (From (1) "โˆ’ " ๐‘’^๐‘ฆ " = " ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 Hence proved

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.