Y= root X find dx/dt at X = 1 given that y increases at 2 units per secodn

We are given the relationship \( y = \sqrt{x} \) and asked to find \( \frac{dx}{dt} \) at \( x = 1 \), given that \( \frac{dy}{dt} = 2 \) units per second.

### Steps:

1. Differentiate both sides of the equation \( y = \sqrt{x} \) with respect to \( t \):
   \[
   \frac{d}{dt}(y) = \frac{d}{dt}(\sqrt{x})
   \]
   Using the chain rule:
   \[
   \frac{dy}{dt} = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt}
   \]

2. Substitute the known values:
   - \( \frac{dy}{dt} = 2 \)
   - \( x = 1 \)

   At \( x = 1 \), \( \sqrt{x} = 1 \), so:
   \[
   2 = \frac{1}{2(1)} \cdot \frac{dx}{dt}
   \]
   Simplifying:
   \[
   2 = \frac{1}{2} \cdot \frac{dx}{dt}
   \]

3. Solve for \( \frac{dx}{dt} \):
   \[
   \frac{dx}{dt} = 4
   \]

Thus, the rate of change of \( x \) with respect to \( t \) at \( x = 1 \) is \( \boxed{4} \) units per second.

Would you like more details on any of these steps, or have any questions?

### Related Questions:
1. How do you use the chain rule for more complex functions?
2. What happens if the value of \( x \) changes? How does \( \frac{dx}{dt} \) change?
3. Can this method apply to higher powers of \( x \) like \( y = x^{n/2} \)?
4. What are the physical interpretations of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) in real-world problems?
5. How does implicit differentiation help in related rates problems?

### Tip:
Always check if you're applying the chain rule correctly when differentiating with respect to time, especially when the variables are functions of \( t \).

Y= root X find dx/dt at X = 1 given that y increases at 2 units per second

Learn how to find the rate of change of x (dx/dt) for the equation y = sqrt(x) at x = 1, given that y increases at 2 units per second. This problem uses related rates and applies the chain rule of differentiation. Ideal for students in Grades 11-12.

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