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

Answer

consider the function $\text{y}=\text{f}(\text{x})=\text{x}^{\frac{1}{4}}$
Let:
x = 16
$\text{x}+\triangle\text{x}=15$
Then,
$\triangle\text{x}=-1$
For x = 16
$\text{y}=(16)^{\frac{1}{4}}=2$
Let:
$\text{dx}=\triangle\text{x}=-1$
Now, $\text{y}=(\text{x})^{\frac{1}{4}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{4(\text{x})^{\frac{3}{4}}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=16}=\frac{1}{32}$
$\therefore\triangle\text{y}=\text{dy}=\Big(\frac{\text{dy}}{\text{dx}}\Big)\text{dx}=\frac{1}{32}\times(-1)=\frac{-1}{32}$
$\Rightarrow\triangle\text{y}=\frac{-1}{32}=-0.03125$
$\therefore(15)^{\frac{1}{4}}=\text{y}+\triangle\text{y}=1.96875$

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