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Ex 6.3
Ex 6.3,2
Ex 6.3,3 Important You are here
Ex 6.3,4
Ex 6.3, 5 Important
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12
Ex 6.3,13
Ex 6.3, 14 (i)
Ex 6.3, 14 (ii) Important
Ex 6.3, 14 (iii)
Ex 6.3, 14 (iv) Important
Ex 6.3, 14 (v)
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19
Ex 6.3,20
Ex 6.3,21 Important
Ex 6.3,22
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25
Ex 6.3,26 (MCQ) Important
Ex 6.3,27 (MCQ)
Last updated at Aug. 19, 2021 by Teachoo
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