Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(255{)}^{\frac{1}{4}}$

Answer 1

$\left(255{)}^{\frac{1}{4}}Considerf\right(x)=x^{\frac{1}{4}}\Rightarrow f^{\prime }(x)=\frac{1}{4}{x}^{\frac{-3}{4}}Now,f(x+\Delta x)\simeq f(x)+\Delta x{f}^{\prime }(x)\Rightarrow (x+\Delta x{)}^{\frac{1}{4}}+\frac{\Delta x}{4x^{\frac{3}{4}}}\Rightarrow (256-1{)}^{\frac{1}{4}}\simeq (256{)}^{\frac{1}{4}}+\frac{(-1)}{4(256{)}^{\frac{3}{4}}}=4+\frac{-1}{4\left\{\right(256{)}^{\frac{1}{4}}{\}}^3}=4-\frac{1}{4\times 4^3}=4-\frac{1}{256}=4-0.004=3.996\Rightarrow (255{)}^{\frac{1}{4}}\simeq 3.996$