1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Tangents and Normals (using Differentiation)
Check sibling questions

Question 2 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

Transcript

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

    3. Tangents and Normals (using Differentiation)

    Approximations (using Differentiation) →
Approximations (using Differentiation) →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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