Question

Using differential, find the approximate value of the \[\sqrt{49 . 5}\] ?

Answer 1

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

\[\text { Let }: \]

\[ x = 49\]

\[x + ∆ x = 49 . 5\]

\[\text { Then }, \]

\[ ∆ x = 0 . 5\]

\[\text { For } x = 49, \]

\[ y = \sqrt{49} = 7\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 5\]

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

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 49} = \frac{1}{14}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{14} \times 0 . 5 = 0 . 0357\]

\[ \Rightarrow ∆ y = 0 . 0357\]

\[ \therefore \sqrt{49 . 5} = y + ∆ y = 7 . 0357\]