Problem 34 - University Calculus: Early Transcendentals

Question

Find the slope of the tangent to the curve \(y=1 / \sqrt{x}\) at the point where \(x=4\).

Step-by-Step Solution

Answer

The slope is \(-\frac{1}{16}\).

Step 1: Find the derivative of the function

To find the slope of the tangent line to the curve at a particular point, we need the derivative of the given function \(y = \frac{1}{\sqrt{x}}\). Rewrite the function in terms of a power of \(x\) to use the power rule: \(y = x^{-1/2}\). The derivative is found using the power rule \(\frac{d}{dx}[x^n] = nx^{n-1}\). Thus, the derivative \(y'\) of \(y = x^{-1/2}\) is \(y' = -\frac{1}{2}x^{-3/2}\).

Step 2: Evaluate the derivative at the given point

Now that we have the derivative \(y' = -\frac{1}{2}x^{-3/2}\), we will evaluate it at \(x = 4\) to find the slope of the tangent line. Substitute \(x = 4\) into the derivative: \(y'(4) = -\frac{1}{2}(4)^{-3/2}\).

Step 3: Simplify the expression

Simplify the expression \(-\frac{1}{2}(4)^{-3/2}\). First, calculate \((4)^{-3/2}\): it is the same as \((\sqrt{4})^{-3} = (2)^{-3} = \frac{1}{8}\). Now substitute back: \(y'(4) = -\frac{1}{2} \times \frac{1}{8} = -\frac{1}{16}\). Thus, the slope of the tangent line at \(x = 4\) is \(-\frac{1}{16}\).

Key Concepts

Derivative of a Function

The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. Essentially, the derivative gives us the slope of the tangent line to the curve of the function at a specific point.

• A function's derivative is often denoted by the symbol \( f'(x) \) or \( dy/dx \).
• Finding the derivative involves applying various rules and techniques to determine how the function behaves at a very small interval, practically tending towards zero, around any point.

The process of deriving a function typically involves using rules like the power rule, product rule, or chain rule. In this case, we simplify the given function \( y = \frac{1}{\sqrt{x}} \) to \( y = x^{-1/2} \), making it easier to apply the power rule for differentiation. Understanding how these rules work in tandem is key to mastering derivatives in calculus.

Power Rule in Calculus

The power rule is one of the simplest and most commonly used rules to find derivatives. It states that if you have a function \( y = x^n \), its derivative is \( \frac{d}{dx}[x^n] = nx^{n-1} \). This rule makes it easy to differentiate polynomial expressions or any function that can be expressed in terms of powers of \( x \).

• The power rule applies directly to any term of the form \( x^n \). It's especially useful since it reduces the complexity of finding derivatives.
• In our exercise, we converted \( y = \frac{1}{\sqrt{x}} \) to \( y = x^{-1/2} \). Applying the power rule, we derive that the slope of the function's tangent line, or its derivative, is \( y' = -\frac{1}{2}x^{-3/2} \).

This method allows us to quickly ascertain how the function behaves as \( x \) changes, giving crucial insights into its graph and tangents.

Evaluating Derivatives

Once we have the derivative of a function, the next step often involves evaluating it at a specific point to find the slope of the tangent line there. This process provides the immediate rate of change of the function at that particular point.

• In our exercise, after finding the derivative \( y' = -\frac{1}{2}x^{-3/2} \), we need to evaluate it at \( x = 4 \).
• Substituting \( x = 4 \) into the derivative gives \( y'(4) = -\frac{1}{2}(4)^{-3/2} \).
• Simplifying this expression, we find that \( (4)^{-3/2} = \frac{1}{8} \), leading to a slope of \( -\frac{1}{16} \).

Evaluating derivatives in this manner allows us to determine precisely how steep or flat the curve is at any chosen point, helping in understanding the function's behavior and graph line.