# Difference between revisions of "022 Sample Final A, Problem 13"

Use differentials to find ${\displaystyle dy}$ given ${\displaystyle y=x^{2}-6x,~x=4,~dx=-0.5.}$

Foundations:
When we use differentials, we are approximating a value for a function by using the slope of the derivative. The idea is that given a distance ${\displaystyle dx}$ from a point ${\displaystyle x}$, we can use ${\displaystyle f'(x)}$, the slope of the tangent line, to find the rise, ${\displaystyle dy}$. Recalling that we can write
${\displaystyle f'(x)\,=\,{\frac {dy}{dx}},}$
the relation is
${\displaystyle dy\,=\,f'(x)\cdot dx,}$
where we use the given specific ${\displaystyle x}$-value to evaluate ${\displaystyle f'(x)}$.

Solution:

Step 1:
By the power rule, we have
${\displaystyle f'(x)\,=\,2x-6.}$
We need to evaluate this at the given value ${\displaystyle x=4}$, so
${\displaystyle f'(4)\,=\,2(4)-6\,=\,2.}$
Step 2:
We use the values given and the result from step 1 to find
${\displaystyle dy\,=\,f'(x)\cdot dx\,=\,2(-0.5)\,=\,-1.}$
${\displaystyle dy\,=\,-1.}$