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 $t$.