$y = 4\arcsin 2t$

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

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

$x = 2\arccos 2t$

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

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

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

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

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

$y' = -2$ ← *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*

0.325 is a randomly chosen number between -1 and 1. Recall that sin θ and cos θ cannot be greater than 1 and cannot be less than -1 for any value of θ