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The image contains a mathematics question asking to find the derivatives of the following functions:

1. \( y = \frac{(2x^3 + 4x)^5}{(3x^2)} \)
2. \( y = 2x^{3x} \)
3. \( y = \frac{\sin(4x)}{x^2} \)
4. \( y = \sqrt{2x^3 + \sqrt{2x + 5}} \)

Let's solve each derivative step by step:

### 1. \( y = \frac{(2x^3 + 4x)^5}{3x^2} \)

To differentiate this, we can use the quotient rule. The quotient rule is given by:

\[
\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

Where \( u = (2x^3 + 4x)^5 \) and \( v = 3x^2 \).

First, differentiate \( u = (2x^3 + 4x)^5 \) using the chain rule:

\[
u' = 5(2x^3 + 4x)^4 \cdot (6x^2 + 4)
\]

Now, differentiate \( v = 3x^2 \):

\[
v' = 6x
\]

Applying the quotient rule:

\[
\frac{dy}{dx} = \frac{3x^2 \cdot [5(2x^3 + 4x)^4 \cdot (6x^2 + 4)] - (2x^3 + 4x)^5 \cdot 6x}{(3x^2)^2}
\]

Simplify the expression.

### 2. \( y = 2x^{3x} \)

To differentiate this, it's easier to first take the natural logarithm of both sides to deal with the exponent:

\[
\ln y = \ln(2x^{3x}) = \ln 2 + \ln(x^{3x}) = \ln 2 + 3x \ln x
\]

Now differentiate both sides with respect to \( x \):

\[
\frac{1}{y} \cdot \frac{dy}{dx} = 3 \ln x + 3
\]

Thus:

\[
\frac{dy}{dx} = y \cdot (3 \ln x + 3) = 2x^{3x} \cdot (3 \ln x + 3)
\]

### 3. \( y = \frac{\sin(4x)}{x^2} \)

This is another quotient rule problem where \( u = \sin(4x) \) and \( v = x^2 \).

Differentiate \( u \) and \( v \):

\[
u' = 4\cos(4x)
\]
\[
v' = 2x
\]

Applying the quotient rule:

\[
\frac{dy}{dx} = \frac{x^2 \cdot 4\cos(4x) - \sin(4x) \cdot 2x}{x^4}
\]

Simplify the expression.

### 4. \( y = \sqrt{2x^3 + \sqrt{2x + 5}} \)

Use the chain rule to differentiate this function.

Let \( u = 2x^3 + \sqrt{2x + 5} \), then \( y = u^{1/2} \).

First, differentiate \( u \):

\[
u' = 6x^2 + \frac{1}{2\sqrt{2x + 5}} \cdot 2
\]

Now, differentiate \( y \):

\[
\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot u'
\]

Substitute \( u \) back in:

\[
\frac{dy}{dx} = \frac{1}{2\sqrt{2x^3 + \sqrt{2x + 5}}} \cdot \left(6x^2 + \frac{1}{\sqrt{2x + 5}}\right)
\]

These are the steps to find the derivatives for each given function. Would you like any further details or explanations?

Here are some related questions:

1. How does the chain rule work for composite functions?
2. What is the importance of the quotient rule in differentiation?
3. Can the natural logarithm method be used for other exponential functions?
4. How do trigonometric derivatives interact with the chain rule?
5. Why do we simplify derivatives, and what are the benefits?

**Tip:** When solving derivative problems, always start by identifying which rules apply (product, quotient, chain, etc.), which makes the process more straightforward.

Explore detailed solutions for finding derivatives of complex functions including quotient rule applications, chain rule complexities, and handling trigonometric and exponential functions. Suitable for advanced high school students and calculus enthusiasts.

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