# Difference between revisions of "022 Exam 2 Sample A, Problem 8"

Use differentials to approximate the change in profit given $x=10$ units and $dx=0.2$ units, where profit is given by $P(x)=-4x^{2}+90x-128$ .

Foundations:
A differential is a method of linearly approximating the change of a function. We use the derivative of the function at an initial point $x_{0}$ as the slope of a line, and use the standard relation
$m\,=\,{\frac {\Delta y}{\Delta x}},$ where $\Delta y$ represents the change in $y$ values, and $\Delta x$ represents the change in $x$ values. Due to the use of the derivative $f'\left(x_{0}\right)$ as the slope, we usually rewrite this using $dy$ and $dx$ to indicate the relative changes. Thus,
$f'(x_{0})\,=\,m\,=\,{\frac {dy}{dx}}.$ We can then rearrange this to find $dy=f'(x_{0})\cdot dx.$ Solution:

Step 1:
First, we must find the derivative. We have $P'(x)=-8x+90$ .
Step 2:
We need the derivative at our initial point, or $x_{0}=10$ . This is
$P'(x_{0})\,=\,P'(10)\,=\,-8(10)+90\,=\,10.$ Step 3:
Finally, we plug in the values to find
$dy\,=\,P'(x_{0})\cdot dx\,=\,10\cdot 0.2\,=\,2.$ Note that if a teacher gives you units (thousands of dollars, dollars, cubits...), you should include them in your answer.
$dy\,=\,2.$ 