Discussion
If \( y = x^{3} \) and \( x = 2 \) is measured with an error of 1 per cent, the resulting percentage error in \( y \) is about
Answer: A
Take logarithms and differentiate, which converts a product into a sum of relative errors:
\[ \ln y = 3 \ln x \quad \Longrightarrow \quad \frac{dy}{y} \;=\; 3\,\frac{dx}{x}. \]
So a one per cent error in \( x \) becomes three per cent in \( y \). The exponent multiplies the relative error, and the value \( x = 2 \) does not enter at all
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