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The problem appears to involve finding the derivative \( y'(x) \) of an implicitly defined function. Based on the text in the image, it can be translated as:

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Find the derivative \( y'(x) \) of the implicitly defined function given by:

\[
3x + 2y^3 + x^2 e^y = 4.
\]

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Let me solve this for you step-by-step.

### Step 1: Differentiate both sides with respect to \( x \)
Using implicit differentiation:
\[
\frac{d}{dx}\big(3x + 2y^3 + x^2e^y\big) = \frac{d}{dx}(4).
\]

On the left:
1. For \( 3x \): \( \frac{d}{dx}(3x) = 3 \),
2. For \( 2y^3 \): Apply the chain rule: \( \frac{d}{dx}(2y^3) = 6y^2 \frac{dy}{dx} = 6y^2 y' \),
3. For \( x^2 e^y \): Use the product rule:
   \[
   \frac{d}{dx}(x^2 e^y) = \frac{d}{dx}(x^2) \cdot e^y + x^2 \cdot \frac{d}{dx}(e^y) = 2x e^y + x^2 e^y \frac{dy}{dx}.
   \]

On the right:
\[
\frac{d}{dx}(4) = 0.
\]

Thus, the differentiated equation becomes:
\[
3 + 6y^2 y' + 2x e^y + x^2 e^y y' = 0.
\]

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### Step 2: Solve for \( y' \)
Group the terms involving \( y' \):
\[
6y^2 y' + x^2 e^y y' = - (3 + 2x e^y).
\]

Factor out \( y' \) on the left:
\[
y'(6y^2 + x^2 e^y) = - (3 + 2x e^y).
\]

Solve for \( y' \):
\[
y' = \frac{- (3 + 2x e^y)}{6y^2 + x^2 e^y}.
\]

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### Final Answer:
The derivative is:
\[
y' = \frac{- (3 + 2x e^y)}{6y^2 + x^2 e^y}.
\]

Would you like me to clarify any step or explore this further?

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#### Related Questions:
1. What is the geometric meaning of \( y'(x) \) in this context?
2. How do you interpret the role of implicit differentiation in solving such problems?
3. What happens when \( 6y^2 + x^2 e^y = 0 \)? Can \( y' \) still be defined?
4. Can this equation be solved explicitly for \( y \)?
5. How does the exponential term \( e^y \) impact the derivative?

#### Tip:
Always isolate \( y' \) systematically by grouping terms when performing implicit differentiation.

Calculate the derivative y'(x) of the implicitly defined function given by 3x + 2y^3 + x^2 e^y = 4.

Explore the step-by-step process of finding the derivative y'(x) for the implicitly defined equation 3x + 2y^3 + x^2 e^y = 4 using implicit differentiation. This problem covers key calculus concepts such as the product rule and chain rule, suitable for high school students in Grades 11-12.

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