# Ex 6.3,3 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

Last updated at Aug. 19, 2021 by Teachoo

Transcript

Ex 6.3, 3 Find the slope of the tangent to curve ๐ฆ=๐ฅ^3โ๐ฅ+1 at the point whose ๐ฅโ๐๐๐๐๐๐๐๐๐ก๐ is 2. ๐ฆ=๐ฅ^3โ๐ฅ+1 We know that slope of tangent is ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=๐(๐ฅ^3 โ ๐ฅ + 1)/๐๐ฅ ๐๐ฆ/๐๐ฅ=3๐ฅ^2โ1+0 We need to find ๐๐ฆ/๐๐ฅ at the point whose ๐ฅโ๐๐๐๐๐๐๐๐๐ก๐ is 2 Putting ๐ฅ=2 in ๐๐ฆ/๐๐ฅ ใ๐๐ฆ/๐๐ฅโใ_(๐ฅ = 2)=3(2)^2โ1 =3 ร4โ1 =12โ1 =11 Hence slope of a tangent is 11

Ex 6.3

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