If `y=x^n ,` then the ratio of relative errors in `ya n dx` is `1:1` (b) 2:1 (c) 1:n (d) n:1

Explanation

`{y}={x}^{{n}} `


`\Rightarrow \frac{{dy}}{{dx}}={nx} {n}^{{n}-1}`



Approximate error in `y` is `d y=(\frac{d y}{d x}) \Delta x`


`={nx}^{{n}-1} \Delta {x}`
...