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

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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