(N/A) Let $y = e^{\sec ^{2} x} + 3 \cos ^{-1} x$.
This function is defined for all $x \in [-1, 1]$.
To find the derivative,we use the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{\sec ^{2} x}) + \frac{d}{dx}(3 \cos ^{-1} x)$
$= e^{\sec ^{2} x} \cdot \frac{d}{dx}(\sec ^{2} x) + 3 \cdot \left( -\frac{1}{\sqrt{1 - x^{2}}} \right)$
$= e^{\sec ^{2} x} \cdot (2 \sec x \cdot \frac{d}{dx}(\sec x)) - \frac{3}{\sqrt{1 - x^{2}}}$
$= e^{\sec ^{2} x} \cdot (2 \sec x \cdot \sec x \tan x) - \frac{3}{\sqrt{1 - x^{2}}}$
$= 2 \sec ^{2} x \tan x e^{\sec ^{2} x} - \frac{3}{\sqrt{1 - x^{2}}}$
Note that the derivative is valid for $x \in (-1, 1)$ because the derivative of $\cos ^{-1} x$ is only defined in the open interval $(-1, 1)$.