91Ó°ÊÓ

Imagine driving along a winding road. Your speed at any given moment reflects a kind of change. In calculus, the first derivative does something similar by describing the rate of change of a function at each point. Turning to our exercise, we have a function, a mathematical rule, expressing how one quantity changes concerning another. Here, the function is represented as \(w = \frac{1}{z} + \frac{8}{3} - z\).

To find the first derivative, we calculate how quickly \(w\) changes with respect to \(z\). This is done by examining each term separately:

• \(\frac{1}{z}\) becomes \(-\frac{1}{z^2}\) after differentiation. This captures the idea that as \(z\) increases, the reciprocal \(\frac{1}{z}\) diminishes, and hence, the rate change is captured by a power of \(z\).
• The constant \(\frac{8}{3}\) disappears since it doesn't affect the change - constants have zero rates of change.
• The term \(-z\) changes at a constant rate of \(-1\).

Therefore, combining these pieces, the first derivative is given by \(w' = -\frac{1}{z^2} - 1\). This derivative tells us how steeply the function \(w\) changes for small changes in \(z\). It acts like a speedometer showing our speed (rate of change) at every value of \(z\).

Now, think of acceleration—how fast your speed is changing. The second derivative holds a similar meaning in mathematics by measuring how the rate of change itself is changing. Continuing with our example, we started with the first derivative \(w' = -\frac{1}{z^2} - 1\).

The second derivative is derived by differentiating this expression further with respect to \(z\). Let's break it down:

• \(-\frac{1}{z^2}\) turns into \(\frac{2}{z^3}\) upon differentiation. The negative sign disappears because squaring \(z\) results in positive differentiation; we similarly get another power of \(z\) indicating three-level change rate.
• The \(-1\) disappears here because its rate of change is zero.

Thus, the second derivative \(w'' = \frac{2}{z^3}\) expresses how the slope from the first derivative changes, depicting the curvature of our original function \(w\). It's like detailing how your speed increases or decreases on the road based on how you accelerate or decelerate.

Simplification is the process of transforming a complex expression into a simpler one, making it easier to work with. When we first encountered the function, it was structured as \(w = \left(\frac{1+3z}{3z}\right)(3-z)\). Multiplication across the terms is the first move to simplify.

Upon expanding, the function initially becomes \(w = \frac{3 + 8z - 3z^2}{3z}\). The principal aim is to perform division for each term in the numerator separately:

• \(\frac{3}{3z}\) simplifies to \(\frac{1}{z}\), reducing complexity by canceling out common factors.
• \(\frac{8z}{3z}\) simplifies to \(\frac{8}{3}\), showing that \(z\) cancels out here.
• \(\frac{3z^2}{3z}\) simplifies to \(z\), with \(z\) reducing the power by one.

By performing these algebraic operations, the expression boils down to \(w = \frac{1}{z} + \frac{8}{3} - z\). This simplified form aids in straightforward differentiation later, illustrating the beauty of turning a complex function into something manageable and easy to interpret.