Question

Using differential, find the approximate value of the \[\left( \frac{17}{81} \right)^\frac{1}{4}\] ?

Answer 1

\[\text { Consider the function } y = f\left( x \right) = \left( x \right)^\frac{1}{4} . \]

\[\text { Let }: \]

\[ x = \frac{16}{81} \]

\[ x + ∆ x = \frac{17}{81}\]

\[\text { Then }, \]

\[ ∆ x = \frac{1}{81}\]

\[\text { For } x = \frac{16}{81}, \]

\[ y = \left( \frac{16}{81} \right)^\frac{1}{4} = \frac{2}{3}\]

\[\text { Let }: \]

\[ dx = ∆ x = \frac{1}{81}\]

\[\text { Now }, y = \left( x \right)^\frac{1}{4} \]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{4 \left( x \right)^\frac{3}{4}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = \frac{16}{81}} = \frac{27}{32}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{27}{32} \times \frac{1}{81} = \frac{1}{96} = 0 . 01042\]

\[ \Rightarrow ∆ y = 0 . 01042\]

\[ \therefore \left( \frac{17}{81} \right)^\frac{1}{4} = y + ∆ y = 0 . 6771\]