Question

Using differential, find the approximate value of the \[\frac{1}{\sqrt{25 . 1}}\] ?

Answer 1

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

\[\text { Let }: \]

\[ x = 25 \]

\[x + ∆ x = 25 . 1\]

\[\text { Then }, \]

\[ ∆ x = 0 . 1\]

\[\text { For }  x = , \]

\[ y = \frac{1}{\sqrt{25}} = 0 . 2\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 1\]

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

\[ \Rightarrow \frac{dy}{dx} = \frac{- 1}{2 \left( x \right)^\frac{3}{2}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 25} = - 0 . 004\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = - 0 . 004 \times 0 . 1 = - 0 . 0004\]

\[ \Rightarrow ∆ y = - 0 . 0004\]

\[ \therefore \frac{1}{\sqrt{25 . 1}} = y + ∆ y = 0 . 1996\]