The equation provided is:

$2xe^{-x^2} + 3y^2 e^y = 5x$

We will use implicit differentiation to find $\frac{dy}{dx}$. Since the equation involves both $x$ and $y$, and $y$ is a function of $x$, we will differentiate both sides of the equation with respect to $x$, remembering to apply the chain rule where necessary for terms involving $y$.

Steps:

1. Differentiate the left-hand side $2xe^{-x^2} + 3y^2 e^y$ with respect to $x$.
2. Differentiate the right-hand side $5x$ with respect to $x$.
3. Solve for $\frac{dy}{dx}$.

Step 1: Differentiating the left-hand side

• For $2xe^{-x^2}$, use the product rule: $\frac{d}{dx} \left( 2xe^{-x^2} \right) = 2e^{-x^2} + 2x \cdot \frac{d}{dx} \left( e^{-x^2} \right)$ Using the chain rule for $\frac{d}{dx} \left( e^{-x^2} \right)$: $\frac{d}{dx} \left( e^{-x^2} \right) = e^{-x^2} \cdot (-2x)$ So, the derivative of $2xe^{-x^2}$ is: $2e^{-x^2} - 4x^2 e^{-x^2}$

• For $3y^2 e^y$, again use the product rule and the chain rule: $\frac{d}{dx} \left( 3y^2 e^y \right) = \frac{d}{dx} \left( 3y^2 \right) \cdot e^y + 3y^2 \cdot \frac{d}{dx} \left( e^y \right)$ Since $y$ is a function of $x$, we apply the chain rule: $\frac{d}{dx} \left( 3y^2 \right) = 6y \cdot \frac{dy}{dx}, \quad \frac{d}{dx} \left( e^y \right) = e^y \cdot \frac{dy}{dx}$ Thus, the derivative of $3y^2 e^y$ is: $6y e^y \frac{dy}{dx} + 3y^2 e^y \frac{dy}{dx}$ Factoring out $\frac{dy}{dx}$, we get: $\frac{dy}{dx} \left( 6y e^y + 3y^2 e^y \right)$

Step 2: Differentiating the right-hand side

The derivative of $5x$ with respect to $x$ is simply $5$.

Step 3: Putting everything together

Now, we combine all the differentiated terms:

$2e^{-x^2} - 4x^2 e^{-x^2} + \frac{dy}{dx} \left( 6y e^y + 3y^2 e^y \right) = 5$

Solving for $\frac{dy}{dx}$:

$\frac{dy}{dx} \left( 6y e^y + 3y^2 e^y \right) = 5 - 2e^{-x^2} + 4x^2 e^{-x^2}$

Finally, we have:

$\frac{dy}{dx} = \frac{5 - 2e^{-x^2} + 4x^2 e^{-x^2}}{6y e^y + 3y^2 e^y}$

This is the derivative $\frac{dy}{dx}$ using implicit differentiation.

Would you like more details or clarifications?

Here are 5 questions to expand on this topic:

1. How would the process change if the equation included higher powers of $y$?
2. Can implicit differentiation be used for equations involving trigonometric functions?
3. What is the significance of the chain rule in implicit differentiation?
4. How do you handle cases where both sides of the equation involve products of $x$ and $y$?
5. How does implicit differentiation differ from explicit differentiation?

Tip: Always apply the chain rule when differentiating terms involving a function of $x$, like $y(x)$.