# 009A Sample Final 1, Problem 8

Let

${\displaystyle y=x^{3}.}$

(a) Find the differential  ${\displaystyle dy}$  of  ${\displaystyle y=x^{3}}$  at  ${\displaystyle x=2}$.

(b) Use differentials to find an approximate value for  ${\displaystyle 1.9^{3}}$.

Foundations:
What is the differential  ${\displaystyle dy}$  of  ${\displaystyle y=x^{2}}$  at  ${\displaystyle x=1?}$

Since  ${\displaystyle x=1,}$  the differential is  ${\displaystyle dy=2xdx=2dx.}$

Solution:

(a)

Step 1:
First, we find the differential  ${\displaystyle dy.}$
Since  ${\displaystyle y=x^{3},}$  we have

${\displaystyle dy\,=\,3x^{2}\,dx.}$

Step 2:
Now, we plug  ${\displaystyle x=2}$  into the differential from Step 1.
So, we get

${\displaystyle dy\,=\,3(2)^{2}\,dx\,=\,12\,dx.}$

(b)

Step 1:
First, we find  ${\displaystyle dx.}$  We have
${\displaystyle dx=1.9-2=-0.1.}$
Then, we plug this into the differential from part (a).
So, we have

${\displaystyle dy\,=\,12(-0.1)\,=\,-1.2.}$

Step 2:
Now, we add the value for  ${\displaystyle dy}$  to  ${\displaystyle 2^{3}}$  to get an
approximate value of  ${\displaystyle 1.9^{3}.}$
Hence, we have

${\displaystyle 1.9^{3}\,\approx \,2^{3}+-1.2\,=\,6.8.}$

(a)     ${\displaystyle dy=12\,dx}$
(b)     ${\displaystyle 6.8}$