This is related to derivatives.

`y = x^5 ----(1)`

Let us say for for marginal change in `deltax` the change in y is `deltay `

`y+deltay = (x+deltax)^5 ----(2)`

(2)-(1)

`deltay = (x+deltax)^5-x^5`

Rate of change means `(deltay)/(deltax)`

`(deltay)/(deltax) = [(x+deltax)^5-x^5]/(deltax)`

When `(deltay)/(deltax)`...

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This is related to derivatives.

`y = x^5 ----(1)`

Let us say for for marginal change in `deltax` the change in y is `deltay `

`y+deltay = (x+deltax)^5 ----(2)`

(2)-(1)

`deltay = (x+deltax)^5-x^5`

Rate of change means `(deltay)/(deltax)`

`(deltay)/(deltax) = [(x+deltax)^5-x^5]/(deltax)`

When `(deltay)/(deltax)` be a small marginal change;

`lim_(xrarr0)(deltay)/(deltax) ` referred as the derivative of a function.

Therefore rate of marginal change can be expressed as the derivative.

Rate of change at x = 2 and y = 32;

`((dy)/dx)_(x=2) = 5*2^4 = 80`

**So the statement is correct and we should agree with that.**