Differentiate y = (1)/(2) + sqrt(x)

First, rewrite the function using exponent notation.

$$y = \frac{1}{2} + x^{1/2}$$

Now, differentiate each term. The derivative of a constant is 0, and for $x^n$, the derivative is $nx^{n-1}$.

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\right) + \frac{d}{dx}\left(x^{1/2}\right)$$ $$\frac{dy}{dx} = 0 + \frac{1}{2}x^{(1/2)-1}$$ $$\frac{dy}{dx} = \frac{1}{2}x^{-1/2}$$ $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$

The derivative is $\boxed{\frac{1}{2\sqrt{x}}}$.

The standard derivative of $\csc x$ is $-\csc x \cot x$.

$$\frac{dy}{dx} = -\csc x \cot x$$

The derivative is $\boxed{-\csc x \cot x}$.

This requires the product rule, $\frac{d}{dx}(uv) = u'v + uv'$, and the chain rule for $\cos x^2$.
Let $u = 2x$ and $v = \cos x^2$.
First, find the derivatives of $u$ and $v$:

$$u' = \frac{d}{dx}(2x) = 2$$

For $v = \cos x^2$, let $w = x^2$. Then $\frac{dw}{dx} = 2x$.

$$\frac{dv}{dx} = \frac{d}{dw}(\cos w) \cdot \frac{dw}{dx} = (-\sin w) \cdot (2x) = -2x \sin x^2$$

Now apply the product rule:

$$\frac{dy}{dx} = u'v + uv'$$ $$\frac{dy}{dx} = (2)(\cos x^2) + (2x)(-2x \sin x^2)$$ $$\frac{dy}{dx} = 2\cos x^2 - 4x^2 \sin x^2$$

The derivative is $\boxed{2\cos x^2 - 4x^2 \sin x^2}$.