Problem 20 - Thomas Calculus

Question

In Exercises \(19-22,\) find the values of the derivatives. $$ \left.\frac{d y}{d x}\right|_{x=\sqrt{3}} \text { if } \quad y=1-\frac{1}{x} $$

Step-by-Step Solution

Answer

The derivative at \( x = \sqrt{3} \) is \( \frac{1}{3} \).

Step 1: Find the Derivative of the Function

The function given is \( y = 1 - \frac{1}{x} \). The derivative of \( y \) with respect to \( x \) is calculated using the power rule. The derivative of \( \frac{1}{x} \) can be rewritten as \( x^{-1} \), whose derivative is \( -x^{-2} \). Therefore, \( \frac{dy}{dx} = 0 - (-x^{-2}) = \frac{1}{x^2} \).

Step 2: Evaluate the Derivative at Specific Point

We need to find \( \frac{dy}{dx} \) at \( x = \sqrt{3} \). Substitute \( x = \sqrt{3} \) into the derivative we found. Thus, \( \frac{dy}{dx}\bigg|_{x=\sqrt{3}} = \frac{1}{(\sqrt{3})^2} = \frac{1}{3} \).

Key Concepts

The Power Rule in Calculus

The power rule is one of the simplest and most frequently used techniques in calculus for finding derivatives. This rule applies specifically to power functions, which are functions of the form \( x^n \), where \( n \) is a real number.

To find the derivative of such a function using the power rule, you multiply the coefficient by the current power of \( x \), and then subtract one from the power. This can be expressed in the formula:

• If \( y = x^n \), then \( \frac{dy}{dx} = n \cdot x^{n-1} \).

This rule is extremely useful because it allows us to quickly compute derivatives without needing complicated algebra. Remember that each term in a function is treated separately. For any term that is a constant coefficient, such as \( 1 \), the derivative is zero.

In the original exercise, you saw this in action with the term \( x^{-1} \), which became \(-x^{-2}\) when differentiated. Using the power rule simplifies the process of finding derivatives, making calculus more accessible.

Evaluating Derivatives at a Point

Once you have the derivative of a function, evaluating it at a specific point is simply a matter of substitution. This allows you to find the rate of change of the function at that exact point.

As seen in the exercise, the derivative \( \frac{dy}{dx} = \frac{1}{x^2} \) was evaluated at \( x = \sqrt{3} \). You substitute this value into the derivative expression:

• First, calculate \((\sqrt{3})^2\), which equals 3.
• Then, substitute to get \( \frac{1}{3} \).

This process gives the slope of the tangent line to the curve at \( x = \sqrt{3} \), making it a powerful tool for understanding how the function behaves locally. Evaluating derivatives enables practical applications such as finding instantaneous velocities or understanding economic trends.

Calculating the Derivative of a Function

Calculating the derivative of a function involves applying the rules of differentiation to find an expression that describes its rate of change. Derivatives are foundational to calculus, as they provide insights into the behavior and shape of graphs.

When calculating the derivative of a function, it is essential to:

• Identify each component of the function, such as constants, variables, and exponents.
• Apply relevant differentiation rules, such as the power rule, product rule, or quotient rule, to each component separately.
• Simplify the resulting expression to find the derivative in its most straightforward form.

In the example from the original exercise, the function \( y = 1 - \frac{1}{x} \) was differentiated by recognizing that \( \frac{1}{x} \) can be rewritten as \( x^{-1} \), allowing for straightforward application of the power rule. Understanding the process behind each step in derivative calculation allows students not only to solve problems but also to gain a deeper comprehension of the function's behavior.