91Ó°ÊÓ

Implicit differentiation often uses a powerful tool called the **Chain Rule**. Let's see how it works in our problem. For the given equation \( u^2 + v^3 = 12 \), both \( u \) and \( v \) are functions of \( t \).
The Chain Rule helps us differentiate complex functions by breaking them down into more manageable parts. Essentially, it allows us to take the derivative of a composite function: a function within another function. In this scenario, since both \( u \) and \( v \) depend on \( t \), when differentiating their terms with respect to \( t \), we multiply by the derivative of these inner functions.
Specifically, for the term \( u^2 \), its derivative is calculated using the Chain Rule as follows:

• Differentiate the outer function: \( 2u \).
• Multiply by the derivative of the inner function: \( \frac{du}{dt} \).

This gives us \( 2u \frac{du}{dt} \). Similarly, for \( v^3 \):

• Differentiate the outer function: \( 3v^2 \).
• Multiply by the derivative of the inner function: \( \frac{dv}{dt} \).

This results in \( 3v^2 \frac{dv}{dt} \). Applying the Chain Rule effectively allows us to find rates of change within interconnected systems.

Differentiability is a crucial property for functions involved in calculus. When we say a function is **differentiable**, it means that at every point in its domain, it has a well-defined derivative. In other words, the function's graph is smooth and not broken or kinked.
In our exercise, both \( u \) and \( v \) are specified to be differentiable functions of \( t \). This ensures that when we differentiate \( u^2 + v^3 = 12 \) with respect to \( t \), using implicit differentiation, we have valid derivatives \( \frac{du}{dt} \) and \( \frac{dv}{dt} \).
Smooth, continuous functionalities:

• The existence of \( \frac{dv}{dt} = 2 \) highlights that \( v \) changes smoothly with respect to \( t \).
• The process helps us to fathom \( \frac{du}{dt} \) as well, as there are no sharp turns or discontinuities impacting \( u \).

Being differentiable is essential for applying most calculus techniques, as discontinuities or abrupt changes would invalidate many derivative calculations.

Understanding the **Rate of Change** is fundamental in interpreting derivatives.
Derivatives in essence tell us how one quantity changes with respect to another. In this context, \( \frac{du}{dt} \) and \( \frac{dv}{dt} \) represent the rates at which \( u \) and \( v \) change over time.
In the exercise, we needed to find \( \frac{du}{dt} \) given specific values to \( v \) and its rate of change. Since \( \frac{dv}{dt} = 2 \), it indicates \( v \) is increasing at a consistent pace. Our goal was to determine how this affects \( u \) — specifically, what \( \frac{du}{dt} \) is when \( v = 2 \).Steps to nail down the rate:

• Using derivatives, we engage with the relationship between variables \( u \) and \( v \) as entwined by the equation \( u^2 + v^3 = 12 \).
• After applying the Chain Rule and simplifying, we derived \( \frac{du}{dt} = \frac{-24}{2u} \).
• Replacing \( u \) with its corresponding value of 2 results in a readily computable rate \( \frac{du}{dt} = -6 \), exemplifying a decrease in \( u \) as \( t \) progresses.

Understanding these rates helps us predict how variables interact and change over time, which is vital for dynamic systems modeling.