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

Transcript

Ex 6.3, 3 Find the slope of the tangent to curve ๐‘ฆ=๐‘ฅ^3โˆ’๐‘ฅ+1 at the point whose ๐‘ฅโˆ’๐‘๐‘œ๐‘œ๐‘Ÿ๐‘‘๐‘–๐‘›๐‘Ž๐‘ก๐‘’ is 2. ๐‘ฆ=๐‘ฅ^3โˆ’๐‘ฅ+1 We know that slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3 โˆ’ ๐‘ฅ + 1)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2โˆ’1+0 We need to find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ at the point whose ๐‘ฅโˆ’๐‘๐‘œ๐‘œ๐‘Ÿ๐‘‘๐‘–๐‘›๐‘Ž๐‘ก๐‘’ is 2 Putting ๐‘ฅ=2 in ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_(๐‘ฅ = 2)=3(2)^2โˆ’1 =3 ร—4โˆ’1 =12โˆ’1 =11 Hence slope of a tangent is 11

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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.