91Ó°ÊÓ

Implicit differentiation is a useful technique when dealing with relationships between variables that are not explicitly solved for one in terms of another. In this exercise, we have the function given as \( y = \sqrt{x} \), but we need to find the rate of change of \( y \) with respect to time \( t \), denoted as \( \frac{dy}{dt} \). Since \( y \) is dependent on \( x \), which in turn is a function of \( t \), we cannot directly differentiate \( y \) with respect to \( t \). Instead, we use implicit differentiation.
By differentiating both sides of the equation \( y = \sqrt{x} \) with respect to \( t \), and using the chain rule (a method used to differentiate compositions of functions), we can find how \( y \) changes as \( x \) changes.

The chain rule is an essential calculus tool when dealing with functions that are composed of other functions. In our exercise, both \( y \) and \( x \) are functions of time \( t \). To differentiate \( y = \sqrt{x} \) with respect to \( t \), we apply the chain rule.
This rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Mathematically, if \( y = f(u) \) and \( u = g(t) \), then \( \frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt} \).
In our case, the outer function is \( \sqrt{u} \) and the inner function is the variable \( x \). Thus, we differentiate \( \sqrt{x} \) to get \( \frac{1}{2\sqrt{x}} \) and multiply by \( \frac{dx}{dt} \). This gives \( \frac{dy}{dt} = \frac{1}{2\sqrt{x}} \times \frac{dx}{dt} \).

Differentiable functions are those that have a well-defined derivative at each point in their domain. In this problem, both \( x \) and \( y \) are differentiated with respect to time, which means they are smooth and consistent in how they change.
Differentiability is crucial in ensuring that the rates of change, \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), exist and can be calculated. For the function \( y = \sqrt{x} \), we are assuming that \( x \) is always positive, ensuring that the square root is a real number and thus \( y \) is differentiable for \( x > 0 \).
The differentiability allows us to apply calculus techniques like the chain rule to solve problems related to the changes in \( x \) and \( y \) over time.

Calculus problem solving often involves breaking down complex relationships into simpler, manageable parts using various rules and techniques. Related rates problems, like the one in our exercise, are common in calculus, where we relate the rates of change of different quantities using implicit differentiation and the chain rule.
The process includes:

• Understanding the given relationship between variables
• Identifying which rates are known and which need to be calculated
• Applying the chain rule and implicit differentiation to connect the rates

In this exercise, we started with a known relation, \( y = \sqrt{x} \), differentiated it with respect to time to find a relation between \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), and then used given values to find specific rates for different scenarios.
By systematically breaking down the problem, we can solve it step by step, which is the essence of calculus problem solving.