Question

1 Using differential, find the approximate value of the following:

\[\sqrt{25 . 02}\]

Answer 1

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

\[\text { Let }: \]

\[ x = 25 \]

\[ x + ∆ x = 25 . 02\]

\[\text { Then, } \]

\[ ∆ x = 0 . 02\]

\[\text { For}   x = 25, \]

\[ y = \sqrt{25} = 5\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 02\]

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

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 25} = \frac{1}{10}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002\]

\[ \Rightarrow ∆ y = 0 . 002\]

\[ \therefore \sqrt{25 . 02} = y + ∆ y = 5 . 002\]