Question 4 - Finding slope of tangent/normal - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 4 - Find slope of tangent at y = x3 - 3x + 2 at x = 3

Ex 6.3,4 - Chapter 6 Class 12 Application of Derivatives - Part 2

Transcript

Question 4 Find the slope of the tangent to the curve 𝑦=𝑥^3−3𝑥+2 at the point whose 𝑥−coordinate is 3 𝑦=𝑥^3−3𝑥+2 We know that slope of tangent =𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2−3 Since 𝑥−coordinate is 3 Putting 𝑥=3 in (1) 〖𝑑𝑦/𝑑𝑥│〗_(𝑥 = 3)=3(3)^2−3 =3 ×9−3 =27−3 =24 Hence slope of tangent is 24

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