Using differentials, find the approximate values of the following:
(v) $(15)^{\frac{1}{4}}$

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Solution

Let $y = x^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 x^{\frac{3}{4}}}$
And x = 16 and $Δ x = - 1$
Then,
$Δ y = (x + Δ x)^{\frac{1}{4}} - x^{\frac{1}{4}}$
$\Rightarrow Δ y = (16 - 1)^{\frac{1}{4}} - (16)^{\frac{1}{4}}$
$\Rightarrow Δ y = (15)^{\frac{1}{4}} - 2$
$\Rightarrow (15)^{\frac{1}{4}} = 2 + Δ y$ ...(1)
Now, approximate change in value of y is given by,
$∵ Δ y \approx (\frac{d y}{d x}) \times Δ x$
$\Rightarrow Δ y \approx \frac{1}{4 x^{\frac{3}{4}}} \times Δ x$
$\Rightarrow Δ y \approx \frac{1}{(16)^{\frac{3}{4}}} \times (- 1)$
$\Rightarrow Δ y \approx - \frac{1}{4 \times 8} \approx - 0.03125$
From equation(1)
$(15)^{\frac{1}{4}} \approx 2 - 0.03125$
$∴ (15)^{\frac{1}{4}} \approx 1.96875$

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