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The Chain Rule is a fundamental concept in calculus used for finding the derivative of composite functions. It's like a tool that helps us handle more complex functions by breaking them down into simpler parts. If we have a function composed of another function, like our example where \[ y = \sqrt{3x - 2} \]we break it into parts: - Let \( u = 3x - 2 \), turning our function into \( y = \sqrt{u} \).

• This step simplifies finding the derivative because we can focus on simpler components.

The Chain Rule formula:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]In words, to find the derivative of \( y = \sqrt{3x - 2} \):1. Find the derivative of the outer function \( y = \sqrt{u} \), which is \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \).2. Find the derivative of the inner function \( u = 3x - 2 \), which gives \( \frac{du}{dx} = 3 \).By multiplying these derivatives, we obtain the full derivative of the original function, illustrating how the chain rule works.

Once we have a derivative, the next step is to evaluate it at a specific value of \( x \) to understand the rate of change at that point.
For our function, \[ \frac{dy}{dx} = \frac{3}{2\sqrt{3x - 2}} \]we substitute \( x = 2 \) to find the slope or the rate of change of \( y \) when \( x \) is exactly 2.

• Substituting gives us \[ \frac{dy}{dx} = \frac{3}{2\sqrt{4}} = \frac{3}{4} \].

This means that at \( x = 2 \), the function's rate of change is \( \frac{3}{4} \). Evaluating at a point is crucial because it gives us the exact value of the slope at that particular moment, which is key for understanding how small changes in \( x \) will affect \( y \). It tells us precisely how steep the function is at \( x = 2 \), helping to approximate changes in the function's value.

Differential calculation allows us to estimate changes in a function's output based on small changes in the input. Recognizing differentials helps us to model small increments very accurately.
Let's calculate the differential \( dy \) in our example:

• First, we find \( dx \), which is the change in \( x \), calculated as \( 2.03 - 2 = 0.03 \).
• The differential \( dy \) is given by \( dy = \frac{dy}{dx} \cdot dx \).

So, with our earlier work giving \( \frac{dy}{dx} = \frac{3}{4} \),

• We compute \( dy = \frac{3}{4} \times 0.03 = 0.0225 \).

This calculation tells us that for a tiny change in \( x \), from 2 to 2.03, the change in \( y \) is approximately \( 0.0225 \). It highlights the predictability of dynamic systems over small intervals and allows us to effectively approximate changes in function values using the linear properties of derivatives.