12.4 Derivatives  (Page 5/18)

If $f\left(x\right)=x^2-100x$ describes the cost of producing $x$ computers, $f^{\prime }\left(x\right)$ will describe the marginal cost. We need to find the derivative. For purposes of calculating the derivative, we can use the following functions:

$\begin{array}{ll}{f}^{\prime }(x)=\frac{f(a+h)-f(a)}{h}\hfill & \text{Formula for a derivative}\hfill \\ =\frac{{(x+h)}^2-100(x+h)-(x^2-100x)}{h}\hfill & \text{Substitute }f(a+h)\text{ and }f(a).\hfill \\ =\frac{x^2+2xh+h^2-100x-100h-x^2+100x}{h}\hfill & \text{Multiply polynomials, distribute}.\hfill \\ =\frac{2xh+h^2-100h}{h}\hfill & \text{Collect like terms}.\hfill \\ =\frac{\cancel{h}(2x+h-100)}{\cancel{h}}\hfill & \text{Factor and cancel like terms}.\hfill \\ =2x+h-100\hfill & \text{Simplify}.\hfill \\ =2x-100\hfill & \text{Evaluate when }h=0.\hfill \\ f^{\prime }(x)=2x-100\hfill & \text{Formula for marginal cost}\hfill \\ f^{\prime }(200)=2(200)-100=300\hfill & \text{Evaluate for 200 units}.\hfill \end{array}$

The marginal cost of producing the 201 ^st unit will be approximately $300.

First we need to evaluate the function $f(t)$ and the derivative of the function $f^{\prime }(t),$ and distinguish between the two. When we evaluate the function $f(t),$ we are finding the distance the car has traveled in $t$ hours. When we evaluate the derivative $f^{\prime }(t),$ we are finding the speed of the car after $t$ hours.

1. $f(0)=0$ means that in zero hours, the car has traveled zero miles.
2. $f^{\prime }(1)=60$ means that one hour into the trip, the car is traveling 60 miles per hour.
3. $f(1)=70$ means that one hour into the trip, the car has traveled 70 miles. At some point during the first hour, then, the car must have been traveling faster than it was at the 1-hour mark.
4. $f(2.5)=150$ means that two hours and thirty minutes into the trip, the car has traveled 150 miles.