Ex 6.3, 3 - Find slope of tangent y = x3 - x + 1 at x = 2 - Ex 6.3

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.3,3 Find the slope of the tangent to curve ๐‘ฆ= ๐‘ฅ๏ทฎ3๏ทฏโˆ’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๏ทฎ3๏ทฏโˆ’๐‘ฅ+1 We know that slope of tangent is ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ= ๐‘‘ ๐‘ฅ๏ทฎ3๏ทฏ โˆ’ ๐‘ฅ + 1๏ทฏ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ=3 ๐‘ฅ๏ทฎ2๏ทฏโˆ’1+0 We need to find ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ 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๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏโ”‚๏ทฎ๐‘ฅ = 2๏ทฏ=3 2๏ทฏ๏ทฎ2๏ทฏโˆ’1 =3 ร—4โˆ’1 =12โˆ’1 =11 Hence slope of a tangent is 11

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