Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(3.968{)}^{\frac{3}{2}}$

Answer 1

$\left(3.968{)}^{\frac{3}{2}}Considerf\right(x)=x^{\frac{3}{2}}\Rightarrow f^{\prime }(x)=\frac{3}{2}{x}^{\frac{1}{2}}=\frac{3}{2}\sqrt{x}Letx=4and\Delta x=-0.032Now,f(x+\Delta x{)}^{\frac{3}{2}}\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)\Rightarrow (x+\Delta x{)}^{\frac{3}{2}}\simeq x^{\frac{3}{2}}+\Delta x(\frac{3}{2}\sqrt{x})\Rightarrow (4-0.032{)}^{\frac{3}{2}}\simeq (4{)}^{\frac{3}{2}}+\frac{3(-0.032)}{2}\sqrt{4}=8-0.096=7.904\Rightarrow (3.968{)}^{\frac{3}{2}}\simeq 7.904$