Question

Using differential, find the approximate value of the  \[\sqrt{37}\] ?

Answer 1

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

\[\text { Let }: \]

\[ x = 36 \]

\[x + ∆ x = 37\]

\[\text { Then }, \]

\[ ∆ x = 1\]

\[\text { For } x = 36, \]

\[ y = \sqrt{36} = 6\]

\[\text { Let }: \]

\[ dx = ∆ x = 1\]

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

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 36} = \frac{1}{12}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{12} \times 1 = 0 . 0833\]

\[ \Rightarrow ∆ y = 0 . 0833\]

\[ \therefore \sqrt{37} = y + ∆ y = 6 . 0833\]