Question

Find the approximate change in the value of $\frac{1}{x^2}$, when x changes from x = 2 to x = 2.002.

Answer 1

Let $y=\frac{1}{x^2}.$ Then $\frac{dy}{dx}=-\frac{2}{x^3}.$
$\therefore dy=(\frac{dy}{dx}{)}_{atx=2}\times dx=(-\frac{2}{2^3})\times 0.002=-\frac{0.002}{4}=-\mathrm{0.0005.}$
Hence is decreased by 0.0005.