Question

How to find instantaneous rate of change for `y=4x^3+2x-3` at x=2?

Answer 1

`50`

Explanation:

We need to differentiate the expression, `4x^3+2x-3` to find the slope of the tangent, `d/dx[4x^3+2x-3]` = `12x^2+2`, [ by the general power rule for differentiation, i.e, if `y=ax^n`, `dy/dx=anx^[n-1]`]
and so when `x=2`, `dy/dx` =`12[2]^2+2` = `50`.

This is the rate of change of `y` with respect to `x` at the point where `x=2`, and means `y` is changing fifty times faster than `x` at this point. Hope this was helpful.

Answer 2

`50`

Explanation:

"Instantaneous rate of change" is just a fancy way of saying "derivative". We need to differentiate this business and plug in `2` at the end.

We can find the derivative using the power rule. Here, we multiply the constant times the exponent, and the power gets decremented. Doing this, we get

`y'=12x^2+2`

NOTE: Recall that the derivative of an `x` term is just its coefficient, and the derivative of a constant is `0`.

Now, we can plug `2` in for `x` to get

`=12(2)^2+2`

`=48+2`

`=50`

The instantaneous rate of change at `x=2` is `50`.

Hope this helps!