Question

Using differential, find the approximate value of the \[\frac{1}{(2 . 002 )^2}\] ?

Answer 1

\[\text { Consider the function y } = f\left( x \right) = \frac{1}{x^2} . \]

\[\text { Let }: \]

\[ x = 2 \]

\[x + ∆ x = 2 . 002\]

\[\text { Then }, \]

\[ ∆ x = - 0 . 002\]

\[\text { For } x = 2 , \]

\[ y = \frac{1}{2^2} = \frac{1}{4}\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 002\]

\[\text { Now,} y = \frac{1}{x^2}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2}{x^3}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 2} = \frac{1}{4}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{4} \times - 0 . 002 = - 0 . 0005\]

\[ \Rightarrow ∆ y = - 0 . 0005\]

\[ \therefore \frac{1}{\left( 2 . 002 \right)^2} = y + ∆ y = 0 . 2495\]