Problem 22 - University Calculus: Early Transcendentals

Question

Find the slope of the curve at the point indicated. $$y=\frac{x-1}{x+1}, \quad x=0$$

Step-by-Step Solution

Answer

The slope of the curve at \( x = 0 \) is 2.

Step 1: Understand the Problem

We need to find the slope of the curve given by the function \( y = \frac{x-1}{x+1} \) at the point where \( x = 0 \). The slope of a curve at a point is given by the derivative of the function at that point.

Step 2: Find the Derivative

To find the slope of the curve, we first need to find the derivative of the function \( y = \frac{x-1}{x+1} \). We'll use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \).Here, \( u = x-1 \) and \( v = x+1 \). Therefore, \( u' = 1 \) and \( v' = 1 \). Applying the quotient rule:\[y' = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x + 1 - x + 1}{(x+1)^2} = \frac{2}{(x+1)^2}.\]

Step 3: Evaluate the Derivative at x = 0

With the derivative \( y' = \frac{2}{(x+1)^2} \), we now substitute \( x = 0 \) into the derivative to find the slope at that point:\[y'(0) = \frac{2}{(0+1)^2} = \frac{2}{1} = 2.\]So, the slope of the curve at \( x = 0 \) is 2.

Key Concepts

Slope of a Curve

The slope of a curve at any given point is a concept that helps us understand how steep the curve is at that specific location. It's similar to finding the slope of a straight line, but because curves can change steepness, it's a bit more complex.

To find the slope of a curve, we look at the derivative of the function. The derivative gives us a function that describes the slope of the original function at any point. For example, in the exercise, we're working with the function \( y = \frac{x-1}{x+1} \). At the point where \( x = 0 \), we need to calculate the derivative and then evaluate it at that point.

• Derivatives tell us about changing rates.
• They are key in understanding how the function behaves at different points.
• Finding the slope of a curve involves taking the derivative and plugging in the specific \( x \) value.

In short, the derivative is crucial as it allows us to pinpoint the instantaneous rate of change at a particular spot on the curve.

Quotient Rule

The quotient rule is an invaluable tool when differentiating functions that are presented as one function divided by another. If you have a fraction-like function, where both the numerator and the denominator are themselves functions, the quotient rule is your friend.

Using the formula \( y = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \), the derivative \( y' \) is given by:\[y' = \frac{u'v - uv'}{v^2}\]This formula helps break down the process into manageable parts. Let's see how it applies to our example problem.

For the equation \( y = \frac{x-1}{x+1} \), we identify:

• \( u = x-1 \) and \( v = x+1 \)
• \( u' = 1 \) since the derivative of \( x-1 \) is 1
• \( v' = 1 \) since the derivative of \( x+1 \) is 1

Putting these into the quotient rule formula, we calculate:\[y' = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}\]By carefully applying the rule, we successfully differentiated the function to find its derivative, which gives us a function representing the slope of the curve across all \( x \) values.

Differentiation Techniques

Differentiation is the cornerstone of calculus, providing ways to find the derivative of almost any function. Understanding different differentiation techniques is essential for tackling a wide range of problems. Here’s a basic rundown:

• The Power Rule: Easily solve derivatives of functions like \( x^n \). Just multiply by the exponent and reduce it by one.
• The Product Rule: Use this when dealing with two functions multiplied together. If \( y = u \cdot v \), then the derivative is \( y' = u'v + uv' \).
• The Quotient Rule: As discussed, this is used for functions divided by each other.
• The Chain Rule: This covers composite functions, allowing you to differentiate a function within another function.

In practice, often multiple rules are combined to find a derivative. For the exercise, the quotient rule was necessary as the function was defined by a fraction of two different expressions. By mastering these techniques, you’re equipped to solve not only our specific exercise but any complex problem involving derivatives.