Question 8 Find the approximate change in the value of 1/𝑥^2 , when x changes from x = 2 to x = 2.002 Let y = 1/𝑥^2 We need to find change in value of y when x changes from 2 to 0.002 ∴ ∆x = 2.002 – 2 = 0.002 We need to find ∆y Now, ∆y = 𝑑𝑦/𝑑𝑥 ∆x ∆y = 𝑑(1/𝑥^2 )/𝑑𝑥 ∆x ∆y = 𝑑(𝑥^(−2) )/𝑑𝑥 × 0.002 ∆y = –2 × x–3 × 0.002 Since we go from x = 2 to x = 2.002 We put x = 2 in ∆y ∆y = –2 × 2–3 × 0.002 ∆y = –2 × 1/2^3 × 0.002 ∆y = –2 × 1/2^3 × 2/1000 ∆y = (−1)/2000 ∆y = – 0.0005 ∴ y decreases by 0.0005

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

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