# 009A Sample Final 1, Problem 4

If  ${\displaystyle y=\cos ^{-1}(2x)}$ compute  ${\displaystyle {\frac {dy}{dx}}}$  and find the equation for the tangent line at  ${\displaystyle x_{0}={\frac {\sqrt {3}}{4}}.}$

Foundations:
1. Chain Rule
${\displaystyle {\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)}$
2. Recall
${\displaystyle {\frac {d}{dx}}(\cos ^{-1}(x))={\frac {-1}{\sqrt {1-x^{2}}}}}$
3. The equation of the tangent line to  ${\displaystyle f(x)}$  at the point  ${\displaystyle (a,b)}$  is
${\displaystyle y=m(x-a)+b}$  where  ${\displaystyle m=f'(a).}$

Solution:

Step 1:
First, we compute  ${\displaystyle {\frac {dy}{dx}}.}$
Using the Chain Rule, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {{\frac {-1}{\sqrt {1-(2x)^{2}}}}(2x)'}\\&&\\&=&\displaystyle {{\frac {-2}{\sqrt {1-4x^{2}}}}.}\end{array}}}$

Step 2:
To find the equation of the tangent line, we first find the slope of the line.
Using  ${\displaystyle x_{0}={\frac {\sqrt {3}}{4}}}$  in the formula for  ${\displaystyle {\frac {dy}{dx}}}$  from Step 1, we get

${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-2}{\sqrt {1-4({\frac {\sqrt {3}}{4}})^{2}}}}\\&&\\&=&\displaystyle {\frac {-2}{\sqrt {\frac {1}{4}}}}\\&&\\&=&\displaystyle {-4.}\end{array}}}$

Step 3:
To get a point on the line, we plug in  ${\displaystyle x_{0}={\frac {\sqrt {3}}{4}}}$  into the equation given.
So, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {y_{0}}&=&\displaystyle {\cos ^{-1}{\bigg (}2{\frac {\sqrt {3}}{4}}{\bigg )}}\\&&\\&=&\displaystyle {\cos ^{-1}{\bigg (}{\frac {\sqrt {3}}{2}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}.}\end{array}}}$

Thus, the equation of the tangent line is   ${\displaystyle y=-4{\bigg (}x-{\frac {\sqrt {3}}{4}}{\bigg )}+{\frac {\pi }{6}}.}$

${\displaystyle {\frac {dy}{dx}}={\frac {-2}{\sqrt {1-4x^{2}}}}}$
${\displaystyle y=-4{\bigg (}x-{\frac {\sqrt {3}}{4}}{\bigg )}+{\frac {\pi }{6}}}$