Ex 6.3, 3 Find the slope of the tangent to curve 𝑦=𝑥^3−𝑥+1 at the point whose 𝑥−𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 is 2. 𝑦=𝑥^3−𝑥+1 We know that slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^3 − 𝑥 + 1)/𝑑𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2−1+0 We need to find 𝑑𝑦/𝑑𝑥 at the point whose 𝑥−𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 is 2 Putting 𝑥=2 in 𝑑𝑦/𝑑𝑥 〖𝑑𝑦/𝑑𝑥│〗_(𝑥 = 2)=3(2)^2−1 =3 ×4−1 =12−1 =11 Hence slope of a tangent is 11

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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