Question

Using differential, find the approximate value of the \[\left( 15 \right)^\frac{1}{4}\] ?

Answer 1

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

\[\text{ Let }: \]

\[ x = 16 \]

\[x + ∆ x = 15\]

\[\text { Then }, \]

\[ ∆ x = - 1\]

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

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

\[\text { Let }: \]

\[ dx = ∆ x = - 1\]

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

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 16} = \frac{1}{32}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{32} \times \left( - 1 \right) = \frac{- 1}{32}\]

\[ \Rightarrow ∆ y = \frac{- 1}{32} = - 0 . 03125\]

\[ \therefore \left( 15 \right)^\frac{1}{4} = y + ∆ y = 1 . 96875\]