https://file.aiquickdraw.com/mathbot/file/1728866065703nayocjqu.jpg

The function provided in the image is:

\[
y = \sqrt{\frac{2x^3 + 2}{2 - 3x}}
\]

We are asked to find the derivative \(\frac{dy}{dx}\). Here's the step-by-step process:

### Step 1: Rewrite the function
We can rewrite the function in a form that's easier to differentiate by expressing the square root as an exponent:

\[
y = \left( \frac{2x^3 + 2}{2 - 3x} \right)^{1/2}
\]

### Step 2: Use the Chain Rule
To differentiate, we apply the chain rule. First, we treat the function inside the parentheses as \( u(x) = \frac{2x^3 + 2}{2 - 3x} \), so that:

\[
y = u^{1/2}
\]

The chain rule gives us:

\[
\frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx}
\]

So we need to find \( \frac{du}{dx} \).

### Step 3: Differentiate \( u(x) \)
The function \( u(x) = \frac{2x^3 + 2}{2 - 3x} \) is a quotient, so we apply the quotient rule:

\[
\frac{du}{dx} = \frac{(2 - 3x) \cdot \frac{d}{dx}(2x^3 + 2) - (2x^3 + 2) \cdot \frac{d}{dx}(2 - 3x)}{(2 - 3x)^2}
\]

#### Step 3.1: Differentiate the numerator and denominator
- The derivative of the numerator \( 2x^3 + 2 \) is \( 6x^2 \).
- The derivative of the denominator \( 2 - 3x \) is \( -3 \).

Now, substitute these derivatives into the quotient rule:

\[
\frac{du}{dx} = \frac{(2 - 3x)(6x^2) - (2x^3 + 2)(-3)}{(2 - 3x)^2}
\]

### Step 4: Simplify \( \frac{du}{dx} \)

\[
\frac{du}{dx} = \frac{(12x^2 - 18x^3) + 3(2x^3 + 2)}{(2 - 3x)^2}
\]
\[
= \frac{12x^2 - 18x^3 + 6x^3 + 6}{(2 - 3x)^2}
\]
\[
= \frac{-12x^3 + 12x^2 + 6}{(2 - 3x)^2}
\]

### Step 5: Substitute into the chain rule

Now substitute \( \frac{du}{dx} \) into the chain rule formula:

\[
\frac{dy}{dx} = \frac{1}{2} \left( \frac{2x^3 + 2}{2 - 3x} \right)^{-1/2} \cdot \frac{-12x^3 + 12x^2 + 6}{(2 - 3x)^2}
\]

### Final Answer:
Thus, the derivative \(\frac{dy}{dx}\) is:

\[
\frac{dy}{dx} = \frac{-12x^3 + 12x^2 + 6}{2(2 - 3x)^2 \cdot \sqrt{\frac{2x^3 + 2}{2 - 3x}}}
\]

Let me know if you would like more detailed steps or further clarifications! Here are some follow-up questions related to the concepts used:

1. What is the chain rule in calculus, and how is it applied?
2. Can you explain the quotient rule in detail and its importance in calculus?
3. How would you simplify complex fractions within a derivative?
4. What are some applications of derivatives in real-world problems?
5. How do square roots affect the differentiation process?

**Tip:** When dealing with fractions and derivatives, always apply the quotient rule carefully and simplify where possible to avoid calculation errors.

Given the function y = sqrt((2x^3 + 2) / (2 - 3x)), find dy/dx in any form.

Find the derivative dy/dx of the function y = sqrt((2x^3 + 2) / (2 - 3x)) using chain rule and quotient rule. This problem involves advanced calculus concepts suitable for university-level students.

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