Find y’ if x = 2 arccos 2t and y = 4 arcsin 2t

$dy = 4 \left[ \dfrac{2 \, dt}{\sqrt{1 - (2t)^2}} \right]$

$dy = \dfrac{8 \, dt}{\sqrt{1 - 4t^2}}$
 

$x = 2\arccos 2t$

$dx = 2 \left[ \dfrac{-2 \, dt}{\sqrt{1 - (2t)^2}} \right]$

$dx = \dfrac{-4 \, dt}{\sqrt{1 - 4t^2}}$
 

$y' = \dfrac{dy}{dx}$

$y' = \dfrac{\dfrac{8 \, dt}{\sqrt{1 - 4t^2}}}{\dfrac{-4 \, dt}{\sqrt{1 - 4t^2}}}$

$y' = \dfrac{8}{-4}$

$y' = -2$   ←   answer
 

Recommended Solution
Notice the the choices are free from any variable, which means that the answer is the same regardless of the value of t. You can therefore use your calculator directly by assigning a limit to its d/dx function. Set the angle of your calculator into RAD.
$y' = \dfrac{\dfrac{d}{dx}\big[ 4\arcsin 2x \big]_{x = 0.325}}{\dfrac{d}{dx}\big[ 2\arccos 2x \big]_{x = 0.325}}$

$y' = -2$   ←   answer