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### Answers (1)

Answer: $n:1$

Hint:  Here we use the concept of relative error

$\frac{d y}{y}=\left(\frac{d y}{d x}\right) \Delta x$

Given: $y=x^{n}$

Solution:

$y=x^{n}$

Let’s differentiate

$\frac{d y}{d x}=n x^{n-1}$

Approximate error in $y$ is

\begin{aligned} d y &=\left(\frac{d y}{d x}\right) \Delta x \\\\ &=n x^{n-1} \times \Delta x \end{aligned}

Relative error in $y$ is  $\frac{d y}{y}=\frac{n}{x} \Delta x$

Approximate error  $x$ is $dx$

\begin{aligned} &=\left(\frac{d x}{d y}\right) \Delta y \\\\ &=\frac{1}{n x^{n-1}} \Delta y \end{aligned}

Relative error in  $x$  is $\frac{d x}{x}=\frac{1}{n x^{n}} \Delta y$

Required solution,

$\frac{\frac{n}{x} \times \Delta x}{\frac{1}{n x^{n}} \times \Delta y}=n^{2} x^{n-1} \frac{\Delta x}{\Delta y}$

\begin{aligned} &=\frac{x}{y} \times \frac{n \times x^{n-1} \times \Delta x}{\Delta x} \\\\ &=\frac{n \times x^{n}}{x^{n}} \\\\ &=\frac{n}{1} \end{aligned}

So, ratio is $n:1$

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