91Ó°ÊÓ

Let's begin by understanding the derivative of a function. A derivative measures how one quantity changes as another quantity changes. Think of it as the rate of change or the slope of a function. Mathematically, if we have a function \(f(t)\), its derivative is represented as \(f'(t)\).

In the given exercise, we start with the function \(f(t) = t^{10}\). To find the derivative, we apply the power rule. This rule states that if \(f(t) = t^n\), then \(f'(t) = nt^{n-1}\). Here, \(n\) is \(10\), so the derivative becomes:

$$f'(t) = 10t^9.$$

This means the rate of change of the function \(t^{10}\) with respect to \(t\) is given by \(10t^9\).

Next, let's dive into the logarithmic derivative. The logarithmic derivative of a function combines ideas from both differentiation and logarithms. It is especially useful for simplifying the differentiation of products and quotients of functions. The logarithmic derivative of a function \(f(t)\) is given by the ratio of its derivative \(f'(t)\) to the function itself \(f(t)\).

Mathematically, it is expressed as:

$$\frac{f'(t)}{f(t)}.$$

Using our example \(f(t) = t^{10}\) and its derivative \(f'(t) = 10t^{9}\), we substitute these into the formula:

$$\frac{f'(t)}{f(t)} = \frac{10t^9}{t^{10}} = \frac{10}{t}.$$

This result, \(\frac{10}{t}\), is the logarithmic derivative of \(f(t) = t^{10}\). Notice how it simplifies to a much easier form. Understanding this concept is crucial for tackling more complex functions in calculus.

Finally, let's explore the concept of the percentage rate of change. The percentage rate of change tells us how much a quantity increases or decreases, expressed as a percentage. For a more intuitive understanding, think of it as converting our usual rate of change into a percentage to better appreciate its magnitude.

We use the logarithmic derivative to find the percentage rate of change. We multiply the logarithmic derivative by \(100\text{\%}\).

In our given exercise, for \(t=10\):

$$\text{Percentage rate of change} = \frac{10}{10} \times 100\text{\%} = 100\text{\%}.$$

This means at \(t=10\), the function \(t^{10}\) is increasing at a rate of \(100\text{\%}\) per unit.

For \(t=50\):

$$\text{Percentage rate of change} = \frac{10}{50} \times 100\text{\%} = 20\text{\%}.$$

This indicates that at \(t=50\), the function \(t^{10}\) is increasing at a rate of \(20\text{\%}\) per unit. Understanding the percentage rate of change helps in interpreting how rapidly a function grows or shrinks at specific points.