Question

Using differential, find the approximate value of the \[\sqrt{26}\] ?

Answer 1

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

\[\text { Let }: \]

\[ x = 25\]

\[x + ∆ x = 26\]

\[\text { Then }, \]

\[ ∆ x = 1\]

\[\text { For } x = 25, \]

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

\[\text { Let }: \]

\[ dx = ∆ x = 1\]

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

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

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

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

\[ \Rightarrow ∆ y = 0 . 1\]

\[ \therefore \sqrt{26} = y + ∆ y = 5 . 1\]