$$ y = \sinh ^{-1} (\tan{x})$$ $$ y' = \frac{1}{\sqrt{1 + \tan^2x}} \cdot \frac{d}{dx} \tan{x}$$
$$ = \frac{\sec^2x}{\sqrt{1+\tan^2x}}$$ $$ = \sec{x}$$
Is that right?
$$y = \cosh^{-1}\sqrt{x}$$
$$y' = \frac{1}{\sqrt{x-1}} \frac{d}{dx} \sqrt{x}$$ $$ = \frac{1}{2 \cdot \sqrt{x-1} \sqrt{x}}$$
Anything else I can do here?