# We are first asked to estimate the instantaneous rate of change for the function `y = x^2` at the x-values, `x=1 , x=2 , x=3` , and our estimations are 2 (for x=1), 4 (for x=2), and 6 (for x=3)....

We are first asked to estimate the instantaneous rate of change for the function `y = x^2` at the x-values, `x=1 , x=2 , x=3` , and our estimations are 2 (for x=1), 4 (for x=2), and 6 (for x=3). We are then asked to estimate the instantaneous rate of change for the function `y = x^3` at the x-values, `x = 1, x = 2, x = 3` , and our estimations are 3 (for x=1), 12 (for x=2), and 27 (for x=3). Our next task is to create two functions which can be developed from our `y = x^2 , and x^3` which will give us the instaneous rate of change - without reducing the interval. Is there a non-calulus way of approaching this? Note, our following task after creating the functions will be to use the reasoning we found to find the instantaneous rate of change for: `y = x^4` at `x = 3` , and `y = 3x^3` at `x=2.`

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The instantaneous rate of change of `y = x^2` at x = 1 is 2 at x = 2 it is 4 and at x = 3 it is 6. The instantaneous rate of change of `y = x^3` at x = 1 is 3, at x = 2 it is 12 and at x = 3 it is 27.

The values given follows from the fact that the derivative of `y = x^2` is `dy/dx = 2x` and for `y = x^3` the derivative is `dy/dx = 3x^2`

To determine the rate of change of `y = x^4` , differentiate this with respect to x. The result is `dy/dx = 4x^3` . The instantaneous rate of change at x = 3 is 4*3^3 = 108. This can also be derived by multiplying 3 with the instantaneous rate of change for `y = x^3` at x = 3 which has been determined to be 27. Similarly, for `y = 3x^3` , `dy/dx = 9x^2` . At x = 2, the instantaneous rate of change is 36; this is similar to the earlier case as 36 is equal to the product of 9 and the instantaneous rate of change determined to be 4 for y = x^2 at x = 2.

Note, calculus solutions are welcome as well. I was just curious if there was a non-calculus approach.