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We are given the equation \( y = 5x \), and the rate of change of \( x \) with respect to time, \( \frac{dx}{dt} = 9 \). We are asked to find \( \frac{dy}{dt} \).

### Step-by-Step Solution:

1. **Differentiate \( y = 5x \) with respect to time \( t \):**

   Since both \( y \) and \( x \) are functions of time \( t \), we apply implicit differentiation. Differentiating both sides with respect to \( t \):

   \[
   \frac{d}{dt}(y) = \frac{d}{dt}(5x)
   \]

2. **Differentiate the right-hand side:**

   The derivative of \( y \) with respect to \( t \) is \( \frac{dy}{dt} \), and the derivative of \( 5x \) with respect to \( t \) is:

   \[
   \frac{dy}{dt} = 5 \cdot \frac{dx}{dt}
   \]

3. **Substitute \( \frac{dx}{dt} = 9 \):**

   \[
   \frac{dy}{dt} = 5 \cdot 9 = 45
   \]

Thus, the rate of change of \( y \) with respect to \( t \), or \( \frac{dy}{dt} \), is **45**.

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Would you like further clarification on any part of this problem?

Here are 5 related questions you might consider:
1. What if \( \frac{dx}{dt} \) was a function of time? How would that change the solution?
2. What is the geometric interpretation of \( \frac{dy}{dt} = 45 \)?
3. Can we use this method if the relation between \( y \) and \( x \) was nonlinear, say \( y = 5x^2 \)?
4. How do we interpret \( \frac{dy}{dt} \) when both \( y \) and \( x \) depend on time?
5. What other techniques could be used if the equation had multiple variables or was more complex?

**Tip:** Always make sure to differentiate every term involving a variable with respect to the given independent variable, in this case, time \( t \).

Assume that y = 5x and dx/dt = 9. Find dy/dt.

This problem involves finding the rate of change of y with respect to time given that y = 5x and dx/dt = 9. The solution uses implicit differentiation and the chain rule, making it suitable for students in Grades 10-12 studying calculus.

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