Question

Using differentials, find the approximate value of the following:
$(33{)}^{\frac{-1}{5}}$

Answer 1

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 $(33{)}^{\frac{-1}{5}}is0.497$