Ex 6.3, 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

Article by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Popular