Given y = sqrt(x2 − 5x + 8).

Let $u = x^2 - 5x + 8$. Then $y = u^{1/2}$.
The derivative of $y$ with respect to $u$ is $\frac{dy}{du} = \frac{1}{2}u^{-1/2}$.
The derivative of $u$ with respect to $x$ is $\frac{du}{dx} = 2x - 5$.
Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

The derivative is $\boxed{\frac{2x - 5}{2\sqrt{x^2 - 5x + 8}}}$.

.6:
Given $y = \frac{1}{\sqrt{x^3 + 1}}$.

Let $u = x^3 + 1$. Then $y = u^{-1/2}$.
The derivative of $y$ with respect to $u$ is $\frac{dy}{du} = -\frac{1}{2}u^{-3/2}$.
The derivative of $u$ with respect to $x$ is $\frac{du}{dx} = 3x^2$.
Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

The derivative is $\boxed{-\frac{3x^2}{2\sqrt{(x^3 + 1)^3}}}$.

.14:
Given $f(x) = \ln(\sin 3x)$.

The derivative of $\ln(g(x))$ is $\frac{g'(x)}{g(x)}$.
Here, $g(x) = \sin 3x$. So, $f'(x) = \frac{1}{\sin 3x} \cdot \frac{d}{dx}(\sin 3x)$.

Let $h(x) = 3x$. Then $\sin 3x = \sin(h(x))$.
The derivative of $\sin(h(x))$ is $\cos(h(x)) \cdot h'(x)$.
So, $\frac{d}{dx}(\sin 3x) = \cos(3x) \cdot \frac{d}{dx}(3x)$.
$\frac{d}{dx}(3x) = 3$.
Therefore, $\frac{d}{dx}(\sin 3x) = 3 \cos 3x$.

Since $\frac{\cos \theta}{\sin \theta} = \cot \theta$:

The derivative is $\boxed{3 \cot 3x}$.