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The image shows a question asking to solve an equation for the variable \( y \). The equation appears to be:

\[ \frac{d}{dx} (7xy) + \frac{d}{dx} (5y^2) = 0 \]

Let me solve it step by step.

### 1. Differentiate the terms:

- First term: \( \frac{d}{dx} (7xy) \)

This requires the product rule:
\[ \frac{d}{dx} (7xy) = 7 \left( x \frac{dy}{dx} + y \right) \]

- Second term: \( \frac{d}{dx} (5y^2) \)

This is just the derivative of \( y^2 \) with respect to \( x \), assuming \( y = y(x) \), so we apply the chain rule:
\[ \frac{d}{dx} (5y^2) = 10y \frac{dy}{dx} \]

### 2. Substituting into the equation:

Substitute the derivatives back into the original equation:
\[
7 \left( x \frac{dy}{dx} + y \right) + 10y \frac{dy}{dx} = 0
\]

### 3. Group the \( \frac{dy}{dx} \) terms:

\[
7x \frac{dy}{dx} + 7y + 10y \frac{dy}{dx} = 0
\]
\[
(7x + 10y) \frac{dy}{dx} = -7y
\]

### 4. Solve for \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{-7y}{7x + 10y}
\]

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

\[
\boxed{\frac{dy}{dx} = \frac{-7y}{7x + 10y}}
\]

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Would you like further clarification on any of these steps, or should I explain in more detail?

Here are 5 follow-up questions related to this problem:

1. How is the product rule applied in differentiation?
2. Can you explain the chain rule for derivatives?
3. What is the physical interpretation of a derivative in this context?
4. How does implicit differentiation work in solving this equation?
5. Why is it important to group terms before solving for \( \frac{dy}{dx} \)?

**Tip:** Always keep track of where the product or chain rule is necessary when differentiating terms involving multiple variables.

Solve the equation using differentiation: (7xy) + (5y^2) = 0

This solution walks you through solving a differentiation problem using the product and chain rules. The equation is differentiated implicitly to find dy/dx for the expression 7xy + 5y^2 = 0. Suitable for students in Grades 10-12.

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