Question

Using differential, find the approximate value of the  \[\sqrt{0 . 082}\] ?

Answer 1

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

\[\text { Let }: \]

\[x = 0 . 0841\]

\[x + ∆ x = 0 . 082\]

\[\text { Then }, \]

\[ ∆ x = - 0 . 0021\]

\[\text { For } x = 0 . 0841, \]

\[ y = \sqrt{0 . 0841} = 0 . 29\]

\[\text { Let }: \]

\[ dx = ∆ x = - 0 . 0021\]

\[\text { Now,} y = \left( x \right)^\frac{1}{2} \]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 0 . 0841} = \frac{1}{0 . 58} = \frac{50}{29}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{50}{29} \times \left( - 0 . 0021 \right) = - 0 . 0036\]

\[ \Rightarrow ∆ y = - 0 . 0036\]

\[ \therefore \sqrt{0 . 082} = y + ∆ y = 0 . 2864\]