Ex 6.3, 2 - Find slope of tangent y = x-1/x-2, at x = 10

Ex 6.3,2 - Chapter 6 Class 12 Application of Derivatives - Part 2

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.3, 2 Find the slope of the tangent to the curve ๐‘ฆ=(๐‘ฅ โˆ’ 1)/(๐‘ฅ โˆ’ 2) , ๐‘ฅโ‰ 2 at ๐‘ฅ=10 ๐‘ฆ=(๐‘ฅ โˆ’ 1)/(๐‘ฅ โˆ’ 2) ๐‘ฅโ‰ 2 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘((๐‘ฅ โˆ’ 1)/(๐‘ฅ โˆ’ 2))" " /๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=((๐‘ฅ โˆ’ 1)^โ€ฒ (๐‘ฅ โˆ’ 2) โˆ’ (๐‘ฅ โˆ’ 2)^โ€ฒ (๐‘ฅ โˆ’ 1))/(๐‘ฅ โˆ’ 2)^2 Using Quotient Rule As (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃ โˆ’ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(1 (๐‘ฅ โˆ’ 2) โˆ’1 (๐‘ฅ โˆ’ 1))/(๐‘ฅ โˆ’ 2)^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘ฅ โˆ’ 2 โˆ’ ๐‘ฅ + 1)/(๐‘ฅ โˆ’ 2)^2 So, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’ 1)/(๐‘ฅ โˆ’ 2)^2 Putting ๐‘ฅ=10 (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)_(๐‘ฅ = 10)=(โˆ’ 1)/(10 โˆ’ 2)^2 =(โˆ’ 1)/ใ€– 8ใ€—^2 =(โˆ’ 1)/64 Hence Slope of a tangent at ๐‘ฅ = 10 is (โˆ’๐Ÿ)/๐Ÿ”๐Ÿ’

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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.