Question 2 Find the slope of the tangent to the curve 𝑦=(𝑥 − 1)/(𝑥 − 2) , 𝑥≠2 at 𝑥=10 𝑦=(𝑥 − 1)/(𝑥 − 2) 𝑥≠2 We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥= 𝑑((𝑥 − 1)/(𝑥 − 2))" " /𝑑𝑥 𝑑𝑦/𝑑𝑥=((𝑥 − 1)^′ (𝑥 − 2) − (𝑥 − 2)^′ (𝑥 − 1))/(𝑥 − 2)^2 Using Quotient Rule As (𝑢/𝑣)^′=(𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 𝑑𝑦/𝑑𝑥=(1 (𝑥 − 2) −1 (𝑥 − 1))/(𝑥 − 2)^2 𝑑𝑦/𝑑𝑥=(𝑥 − 2 − 𝑥 + 1)/(𝑥 − 2)^2 So, 𝑑𝑦/𝑑𝑥=(− 1)/(𝑥 − 2)^2 Putting 𝑥=10 (𝑑𝑦/𝑑𝑥)_(𝑥 = 10)=(− 1)/(10 − 2)^2 =(− 1)/〖 8〗^2 =(− 1)/64 Hence Slope of a tangent at 𝑥 = 10 is (−𝟏)/𝟔𝟒

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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