Question 3 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at July 14, 2026 by

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

Ex 6.3,3 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Question 3 Find the slope of the tangent to curve š‘¦=š‘„^3āˆ’š‘„+1 at the point whose š‘„āˆ’š‘š‘œš‘œš‘Ÿš‘‘š‘–š‘›š‘Žš‘”š‘’ is 2. š‘¦=š‘„^3āˆ’š‘„+1 We know that slope of tangent is š‘‘š‘¦/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=š‘‘(š‘„^3 āˆ’ š‘„ + 1)/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=3š‘„^2āˆ’1+0 We need to find š‘‘š‘¦/š‘‘š‘„ at the point whose š‘„āˆ’š‘š‘œš‘œš‘Ÿš‘‘š‘–š‘›š‘Žš‘”š‘’ is 2 Putting š‘„=2 in š‘‘š‘¦/š‘‘š‘„ 怖š‘‘š‘¦/š‘‘š‘„ā”‚怗_(š‘„ = 2)=3(2)^2āˆ’1 =3 Ɨ4āˆ’1 =12āˆ’1 =11 Hence slope of a tangent is 11

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