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Question
Show that the percentage error in the nth root of a number is approximately `1/"n"` times the percentage error in the number
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Solution
Let x be the number
Let y = `x^(1/"n")`
log y = `1/"n" log x`
Taking differentiate on both sides, we have
`1/y "d"y = 1/"n" xx 1/x "d"x`
`(Deltay)/y = 1/y "d"y`
= `1/"n" xx 1/x "d"x`
`(Deltay)/y xx 100 = 1/"n"(("d"x)/x xx 100)`
`1/"n"` times the percentage error in the number.
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