# 009A Sample Final 3, Problem 10

Let  ${\displaystyle y=\tan(x).}$

(a) Find the differential  ${\displaystyle dy}$  of  ${\displaystyle y=\tan(x)}$  at  ${\displaystyle x={\frac {\pi }{4}}.}$

(b) Use differentials to find an approximate value for  ${\displaystyle \tan(0.885).}$  Hint:  ${\displaystyle {\frac {\pi }{4}}\approx 0.785.}$

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  ${\displaystyle y=\tan x,}$  we have

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

Step 2:
Now, we plug  ${\displaystyle x={\frac {\pi }{4}}}$  into the differential from Step 1.
So, we get

${\displaystyle dy\,=\,{\bigg (}\sec {\bigg (}{\frac {\pi }{4}}{\bigg )}{\bigg )}^{2}\,dx\,=\,2\,dx.}$

(b)

Step 1:
First, we find  ${\displaystyle dx.}$  We have
${\displaystyle dx=0.885-{\frac {\pi }{4}}\approx 0.885-0.785=0.1.}$
Then, we plug this into the differential from part (a).
So, we have

${\displaystyle dy\,=\,2(0.1)\,=\,0.2.}$

Step 2:
Now, we add the value for  ${\displaystyle dy}$  to  ${\displaystyle \tan {\bigg (}{\frac {\pi }{4}}{\bigg )}}$  to get an
approximate value of  ${\displaystyle \tan(0.885).}$
Hence, we have

${\displaystyle \tan(0.885)\,\approx \,\tan {\bigg (}{\frac {\pi }{4}}{\bigg )}+0.2\,=\,1.2.}$

(a)     ${\displaystyle dy=2\,dx}$
(b)     ${\displaystyle 1.2}$