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

May 10, 2018

$50$

#### Explanation:

We need to differentiate the expression, $4 {x}^{3} + 2 x - 3$ to find the slope of the tangent, $\frac{d}{\mathrm{dx}} \left[4 {x}^{3} + 2 x - 3\right]$ = $12 {x}^{2} + 2$, [ by the general power rule for differentiation, i.e, if $y = a {x}^{n}$, $\frac{\mathrm{dy}}{\mathrm{dx}} = a n {x}^{n - 1}$]
and so when $x = 2$, $\frac{\mathrm{dy}}{\mathrm{dx}}$ =$12 {\left[2\right]}^{2} + 2$ = $50$.

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.

May 10, 2018

$50$

#### 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

$y ' = 12 {x}^{2} + 2$

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

$= 12 {\left(2\right)}^{2} + 2$

$= 48 + 2$

$= 50$

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

Hope this helps!