Question

Using differential, find the approximate value of the \[\sqrt{36 . 6}\] ?

Answer 1

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

\[\text { Let }: \]

\[ x = 36\]

\[x + ∆ x = 36 . 6\]

\[\text { Then}, \]

\[ ∆ x = 0 . 6\]

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

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

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 6\]

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

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

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

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{12} \times 0 . 6 = 0 . 05\]

\[ \Rightarrow ∆ y = 0 . 05\]

\[ \therefore \sqrt{36 . 6} = y + ∆ y = 6 . 05\]