Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(15{)}^{\frac{1}{4}}$

Answer 1

$(15{)}^{\frac{1}{4}}$
Consider $f\left(x\right)=x^{\frac{1}{4}}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}{x}^{-\frac{3}{4}}$
Let $x=16and\Delta x=-1$
Now, $f(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)$
$\Rightarrow (x+\Delta x{)}^{\frac{1}{4}}\simeq x^{\frac{1}{4}}+\frac{\Delta x}{4x^{\frac{3}{4}}}\Rightarrow (16-1{)}^{\frac{1}{4}}\simeq (16{)}^{\frac{1}{4}}+\frac{(-1)}{4(16{)}^{\frac{3}{4}}}=2-\frac{1}{4\times 2^3}=2-0.031=1.969$
$\therefore (15{)}^{\frac{1}{4}}\simeq 1.969$