009A Sample Final 1, Problem 4

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

You may leave your answers in point-slope form.

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

Solution:

Step 1:
First, we compute  ${\frac {dy}{dx}}.$
Using the Chain Rule, we get
${\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {{\frac {-1}{\sqrt {1-(2x)^{2}}}}(2x)'}\\&&\\&=&\displaystyle {{\frac {-2}{\sqrt {1-4x^{2}}}}.}\end{array}}$

Step 2:
To find the equation of the tangent line, we first find the slope of the line.
Using  $x_{0}={\frac {\sqrt {3}}{4}}$  in the formula for  ${\frac {dy}{dx}}$  from Step 1, we get
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-2}{\sqrt {1-4({\frac {\sqrt {3}}{4}})^{2}}}}\\&&\\&=&\displaystyle {\frac {-2}{\sqrt {\frac {1}{4}}}}\\&&\\&=&\displaystyle {-4.}\end{array}}$

Step 3:
To get a point on the line, we plug in  $x_{0}={\frac {\sqrt {3}}{4}}$  into the equation given.
So, we have
${\begin{array}{rcl}\displaystyle {y_{0}}&=&\displaystyle {\cos ^{-1}{\bigg (}2{\frac {\sqrt {3}}{4}}{\bigg )}}\\&&\\&=&\displaystyle {\cos ^{-1}{\bigg (}{\frac {\sqrt {3}}{2}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{6}}.}\end{array}}$
Thus, the equation of the tangent line is   $y=-4{\bigg (}x-{\frac {\sqrt {3}}{4}}{\bigg )}+{\frac {\pi }{6}}.$

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

Return to Sample Exam