91Ó°ÊÓ

The concept of a derivative is fundamental in calculus. Essentially, it measures the rate at which a function changes at any given point. If you think about a car driving down a road, the derivative would be similar to the speedometer reading, showing your speed at a specific moment.
To find this instantaneous rate of change for a function \( f(x) \), we use the definition of the derivative, which is formally written as:

• \( f'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h} \)

This formula looks at a small change around the point \( a \), and calculates how much the function \( f(x) \) changes relative to that small change. The \( \lim_{{h \to 0}} \) part means that we want to find the limit as \( h \), an arbitrarily tiny number, approaches 0.
In practical terms for our exercise, this definition helps us calculate the derivative of the function \( f(x) = 4 - x^2 \) step by step, showing how changes in \( x \) affect the output of the function.

Differentiation is the process of applying rules and formulas to determine the derivative of a function. For the function \( f(x) = 4 - x^2 \), we want to find its derivative \( f'(x) \).
We use the steps outlined in the definition of the derivative:

• Start by substituting \( f(x) \) into the definition: \( \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \)
• For \( f(x) = 4 - x^2 \), this becomes \( \lim_{{h \to 0}} \frac{(4 - (x+h)^2) - (4 - x^2)}{h} \)

By expanding and simplifying the expression \( (x+h)^2 \) as \( x^2 + 2xh + h^2 \), and substituting back, we further simplify:

• \( \lim_{{h \to 0}} \frac{-2xh - h^2}{h} \)
• After factoring out \( h \), it becomes \( \lim_{{h \to 0}} (-2x - h) \)

As \( h \) approaches 0, the expression simplifies to \( -2x \). This process shows us how the differentiation rules neatly lead us to conclude that the derivative, or the function's rate of change, is \( f'(x) = -2x \).

Now that we have the general derivative of \( f(x) = 4 - x^2 \), which is \( f'(x) = -2x \), we can evaluate this at specific points. This means substituting different values of \( x \) into our derived function to find the rate of change at those points.
Let's take three examples:

• For \( x = -3 \): Substitute into the derivative function to get \( f'(-3) = -2(-3) = 6 \). This tells us that at \( x = -3 \), the function is increasing at a rate of 6.
• For \( x = 0 \): Substitute to receive \( f'(0) = -2(0) = 0 \). The derivative at this point is 0, indicating no change or a flat slope.
• For \( x = 1 \): Substituting gives us \( f'(1) = -2(1) = -2 \). Here, the function is decreasing at a rate of 2.

These evaluations provide insight into how the function behaves at these specific points, whether it's increasing, decreasing, or remaining constant.