The problem asks us to find the derivatives of six different functions.

1. Problem Description

The problem asks us to find the derivatives of six different functions.

2. Solution Steps

We will differentiate each function separately.

1. $y = x^3 \sin(2x)$

We use the product rule: $(uv)' = u'v + uv'$.

Let $u = x^3$ and $v = \sin(2x)$.

Then $u' = 3x^2$ and $v' = 2\cos(2x)$.

$y' = (x^3)'\sin(2x) + x^3(\sin(2x))'$

$y' = 3x^2\sin(2x) + x^3(2\cos(2x))$

$y' = 3x^2\sin(2x) + 2x^3\cos(2x)$

2. $y = 3x^3 - x\cos(3x)$

We use the difference rule and product rule.

$y' = (3x^3)' - (x\cos(3x))'$

$y' = 9x^2 - [(x)'\cos(3x) + x(\cos(3x))']$

$y' = 9x^2 - [\cos(3x) + x(-3\sin(3x))]$

$y' = 9x^2 - \cos(3x) + 3x\sin(3x)$

3. $y = -x\cos(4-3x^2)$

We use the product rule.

$y' = -(x)'\cos(4-3x^2) - x[\cos(4-3x^2)]'$

$y' = -\cos(4-3x^2) - x[-\sin(4-3x^2)(-6x)]$

$y' = -\cos(4-3x^2) - 6x^2\sin(4-3x^2)$

4. $f(x) = \frac{\sin^2(3x)}{x^2}$

We use the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

Let $u = \sin^2(3x)$ and $v = x^2$.

Then $u' = 2\sin(3x)\cos(3x)(3) = 6\sin(3x)\cos(3x) = 3\sin(6x)$ and $v' = 2x$.

$f'(x) = \frac{(3\sin(6x))(x^2) - (\sin^2(3x))(2x)}{(x^2)^2}$

$f'(x) = \frac{3x^2\sin(6x) - 2x\sin^2(3x)}{x^4}$

$f'(x) = \frac{3x\sin(6x) - 2\sin^2(3x)}{x^3}$

5. $f(x) = \frac{\tan(2x)}{x^2-1}$

We use the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

Let $u = \tan(2x)$ and $v = x^2 - 1$.

Then $u' = 2\sec^2(2x)$ and $v' = 2x$.

$f'(x) = \frac{(2\sec^2(2x))(x^2-1) - (\tan(2x))(2x)}{(x^2-1)^2}$

$f'(x) = \frac{2(x^2-1)\sec^2(2x) - 2x\tan(2x)}{(x^2-1)^2}$

6. $f(x) = \cos(\sin^2(3x))$

We use the chain rule.

$f'(x) = -\sin(\sin^2(3x)) \cdot (2\sin(3x) \cdot \cos(3x) \cdot 3)$

$f'(x) = -6\sin(\sin^2(3x))\sin(3x)\cos(3x)$

$f'(x) = -3\sin(\sin^2(3x))\sin(6x)$

3. Final Answer

1. $y' = 3x^2\sin(2x) + 2x^3\cos(2x)$

2. $y' = 9x^2 - \cos(3x) + 3x\sin(3x)$

3. $y' = -\cos(4-3x^2) - 6x^2\sin(4-3x^2)$

4. $f'(x) = \frac{3x\sin(6x) - 2\sin^2(3x)}{x^3}$

5. $f'(x) = \frac{2(x^2-1)\sec^2(2x) - 2x\tan(2x)}{(x^2-1)^2}$

6. $f'(x) = -3\sin(\sin^2(3x))\sin(6x)$