Question

Using differentials, find the approximate value of each of the following:

$(\frac{17}{81}{)}^{\frac{1}{4}}$

$(33{)}^{\frac{-1}{5}}$

Answer 1

Consider $f\left(x\right)=x^{\frac{1}{4}}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}{x}^{\frac{-3}{4}}$
Let $x=\frac{16}{81}$ and $\Delta x=\frac{1}{81},$
Now, $f^{\prime }(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)$

$\Rightarrow (\frac{16}{81}+\frac{1}{81}{)}^{\frac{1}{4}}\simeq (\frac{16}{81}{)}^{\frac{1}{4}}+\frac{\frac{1}{81}}{4(\frac{16}{81}{)}^{\frac{3}{4}}}$
$\Rightarrow (\frac{17}{81}{)}^{\frac{1}{4}}\simeq \frac{2}{3}+\frac{1}{81\times 4\times (\frac{2}{3}{)}^3}=\frac{2}{3}+\frac{1}{96}=\frac{65}{96}$
Hence, $(\frac{17}{81}{)}^{\frac{1}{4}}\simeq \mathrm{0.677.}$

Consider $f\left(x\right)=x^{\frac{-1}{5}}\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{5}{x}^{\frac{-6}{5}}$
Let $x=32and\Delta x=1,$
Now, $f(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)$
$\Rightarrow (x+\Delta x{)}^{\frac{-1}{5}}\simeq x^{\frac{-1}{5}}-\frac{1}{5x^{\frac{6}{5}}}\Delta x\Rightarrow (33{)}^{\frac{-1}{5}}\simeq (32{)}^{\frac{-1}{5}}-\frac{1}{5(32{)}^{\frac{6}{5}}}=\frac{1}{2}-\frac{1}{5\times 2^6}=0.5-\frac{1}{320}=0.5-0.003\Rightarrow (33{)}^{\frac{-1}{5}}\simeq 0.497$
Hence, the approximate value of $(33{)}^{\frac{-1}{5}}is0.497$