Question

Show that the percentage error in the $n^{th}$ root of a number is approximately $\frac{1}{n}$ times the percentage error in the number.

Answer 1

Let $x$ be the number
Let $y=f\left(x\right)=x^{1/n}$
Then $\log y=\frac{1}{n}\log x$
Taking differential on the both sides, we have
$\frac{1}{y}dy=\frac{1}{n}.\frac{1}{x}dx$
$\frac{\Delta y}{y}\approx \frac{1}{y}dy=\frac{1}{n}.\frac{1}{x}dx$
$\frac{\Delta y}{y}\times 100\approx \frac{1}{n}(\frac{dx}{x}\times 100)$
$=\frac{1}{n}$ times the percentage error in the number