Differentiate the following with respect to x.

The derivative of $\sin(u)$ is $\cos(u) \cdot u'$.
Here, $u = 6x$, so $u' = 6$.

The correct answer is $\boxed{\mathbf{6 \cos 6x}}$.

2) $y = 3 \tan x$
The student's answer is correct.

The derivative of $\tan x$ is $\sec^2 x$.

The correct answer is $\boxed{\mathbf{3 \sec^2 x}}$.

3) $y = \sin \frac{1}{2}x$

The derivative of $\sin(u)$ is $\cos(u) \cdot u'$.
Here, $u = \frac{1}{2}x$, so $u' = \frac{1}{2}$.

The answer is $\boxed{\mathbf{\frac{1}{2} \cos\left(\frac{1}{2}x\right)}}$.

4) $y = \cos(2x + 5)$

The derivative of $\cos(u)$ is $-\sin(u) \cdot u'$.
Here, $u = 2x + 5$, so $u' = 2$.

The answer is $\boxed{\mathbf{-2 \sin(2x + 5)}}$.

5) $y = \tan(3x - 1)$

The derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$.
Here, $u = 3x - 1$, so $u' = 3$.

The answer is $\boxed{\mathbf{3 \sec^2(3x - 1)}}$.

6) $y = 4 \tan 2x$

The derivative of $\tan(u)$ is $\sec^2(u) \cdot u'$.
Here, $u = 2x$, so $u' = 2$.

The answer is $\boxed{\mathbf{8 \sec^2(2x)}}$.

7) $y = \sin x - \cos x$

The derivative of $\sin x$ is $\cos x$.
The derivative of $\cos x$ is $-\sin x$.

The answer is $\boxed{\mathbf{\cos x + \sin x}}$.

8) $y = 2 \cos x + 3 \tan x$

The derivative of $2 \cos x$ is $2(-\sin x) = -2 \sin x$.
The derivative of $3 \tan x$ is $3(\sec^2 x) = 3 \sec^2 x$.

The answer is $\boxed{\mathbf{-2 \sin x + 3 \sec^2 x}}$.

9) $y = \cos 2x - \sin 2x$

For $\cos 2x$: $u = 2x$, $u' = 2$. Derivative is $-\sin(2x) \cdot 2 = -2 \sin 2x$.
For $\sin 2x$: $u = 2x$, $u' = 2$. Derivative is $\cos(2x) \cdot 2 = 2 \cos 2x$.

The answer is $\boxed{\mathbf{-2(\sin 2x + \cos 2x)}}$.

10) $y = \tan 2x + \sin 4x$

For $\tan 2x$: $u = 2x$, $u' = 2$. Derivative is $\sec^2(2x) \cdot 2 = 2 \sec^2 2x$.
For $\sin 4x$: $u = 4x$, $u' = 4$. Derivative is $\cos(4x) \cdot 4 = 4 \cos 4x$.

The answer is $\boxed{\mathbf{2 \sec^2 2x + 4 \cos 4x}}$.

11) $f(x) = 2x \sin \frac{x}{2}$

$(uv)' = u'v + uv'$.
Let $u = 2x$ and $v = \sin \frac{x}{2}$.

$u' = \frac{d}{dx}(2x) = 2$.
$v' = \frac{d}{dx}\left(\sin \frac{x}{2}\right) = \cos\left(\frac{x}{2}\right) \cdot \frac{1}{2} = \frac{1}{2} \cos\left(\frac{x}{2}\right)$.

The answer is $\boxed{\mathbf{2 \sin \frac{x}{2} + x \cos \frac{x}{2}}}$.

12) $f(x) = \frac{\sin 2x}{\cos 3x}$

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.
Let $u = \sin 2x$ and $v = \cos 3x$.

$u' = \frac{d}{dx}(\sin 2x) = \cos(2x) \cdot 2 = 2 \cos 2x$.
$v' = \frac{d}{dx}(\cos 3x) = -\sin(3x) \cdot 3 = -3 \sin 3x$.

The answer is $\boxed{\mathbf{\frac{2 \cos 2x \cos 3x + 3 \sin 2x \sin 3x}{\cos^2 3x}}}$.

13) $f(x) = \sin^3 x$

Let $u = \sin x$, so $u' = \cos x$.
The derivative of $u^3$ is $3u^2 \cdot u'$.

The answer is $\boxed{\mathbf{3 \sin^2 x \cos x}}$.

14) $y = \sqrt{\cos 2x}$

Let $u = \cos 2x$. Then $u' = -\sin(2x) \cdot 2 = -2 \sin 2x$.
The derivative of $u^{1/2}$ is $\frac{1}{2}u^{-1/2} \cdot u'$.

The answer is $\boxed{\mathbf{\frac{-\sin 2x}{\sqrt{\cos 2x}}}}$.

15) $y = \frac{1}{\sin x}$

Let $u = \sin x$. Then $u' = \cos x$.
The derivative of $u^{-1}$ is $-1u^{-2} \cdot u'$.

This can also be written as $-\cot x \csc x$.
The answer is $\boxed{\mathbf{-\frac{\cos x}{\sin^2 x}}}$.

16) $f(x) = \frac{1}{\cos x}$

Let $u = \cos x$. Then $u' = -\sin x$.
The derivative of $u^{-1}$ is $-1u^{-2} \cdot u'$.

This can also be written as $\tan x \sec x$.
The answer is $\boxed{\mathbf{\frac{\sin x}{\cos^2 x}}}$.