Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to \(y=\left(x^{2}+1\right)^{-2}\) at \(\left(1, \frac{1}{4}\right)\)

Step-by-Step Solution

Answer

The equation is \( y = -\frac{1}{2}x + \frac{3}{4} \).

Step 1: Differentiate the Function

To find the equation of the tangent line, we first need the derivative of the function to find the slope of the tangent. The function is given as \( y = (x^2 + 1)^{-2} \). Using the chain rule, differentiate \( y \) with respect to \( x \). The derivative is \( y' = -2 (x^2 + 1)^{-3} \, \cdot\, 2x = -4x(x^2 + 1)^{-3} \).

Step 2: Evaluate the Derivative at the Given Point

We need the slope of the tangent line at the point \((1, \frac{1}{4})\). Evaluate the derivative we found in Step 1 at \( x = 1 \). Substitute \( x = 1 \) into \( y' = -4x(x^2 + 1)^{-3} \) to get \( y'(1) = -4(1)((1)^2 + 1)^{-3} = -4(1)(2)^{-3} = -\frac{1}{2} \).

Step 3: Use the Point-Slope Form to Find the Tangent Line Equation

With the slope \( m = -\frac{1}{2} \) and the point \((1, \frac{1}{4})\), use the point-slope form of a line equation \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point on the line. Plug in \( m = -\frac{1}{2} \), \( x_1 = 1 \), and \( y_1 = \frac{1}{4} \) to get \( y - \frac{1}{4} = -\frac{1}{2}(x - 1) \).

Step 4: Simplify the Equation to Slope-Intercept Form

Distribute \(-\frac{1}{2}\) on the right side to get \( y - \frac{1}{4} = -\frac{1}{2}x + \frac{1}{2} \). Add \( \frac{1}{4} \) to both sides to solve for \( y \), giving \( y = -\frac{1}{2}x + \frac{1}{2} + \frac{1}{4} \). Simplify to \( y = -\frac{1}{2}x + \frac{3}{4} \).

Key Concepts

Derivative

The derivative of a function represents the rate at which the function's value changes as its input changes. For a curve like \( y = (x^2 + 1)^{-2} \), the derivative tells us the slope of the tangent line at any point \( x \). Knowing the slope of the tangent line is crucial to writing its equation. To find this derivative, one must utilize differentiation rules such as the power rule, the product rule, or the chain rule, depending on the complexity of the function at hand. The derivative we found, \( y' = -4x(x^2 + 1)^{-3} \), is crucial because it provides the slope of the tangent line when evaluated at a specific point, such as \( x = 1 \).

Derivatives are an essential part of calculus, helping us understand how functions behave and change. They are foundational in fields that require analysis of anything that changes in relation to something else.

Chain Rule

The chain rule is a powerful tool in calculus used when differentiating composite functions. A composite function is essentially a function within another function, like \( y = (x^2 + 1)^{-2} \). To differentiate such a function, the chain rule instructs us to first differentiate the outer function while keeping the inner function unchanged, and then multiply by the derivative of the inner function.

Let's break this down with our given function. The outer function is \((u)^{-2}\), where \( u = x^2 + 1 \). Differentiate \((u)^{-2}\) with respect to \(u\) (which gives us \(-2u^{-3}\)), and then multiply by the derivative of \( u \) with respect to \( x \) (which is \( 2x \)). Combining these steps, we derive \( y' = -4x(x^2 + 1)^{-3} \). This method simplifies differentiation of complex functions, making it manageable.

Point-Slope Form

The point-slope form of a linear equation is an essential tool when you know the slope of a line and one point on the line. It's expressed as \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the known point and \( m \) is the slope. In our case, once we determined the slope at \( (1, \frac{1}{4}) \) using the derivative, we found it to be \( m = -\frac{1}{2} \).

Using the point \( (1, \frac{1}{4}) \), we plug these into the point-slope form to get \( y - \frac{1}{4} = -\frac{1}{2}(x - 1) \). This form is convenient because it directly provides a linear equation using the derivative's result and the given point. From here, converting to the slope-intercept form (if needed) becomes straightforward.

Slope-Intercept Form

The slope-intercept form, \( y = mx + b \), is a widely used linear equation form where \( m \) represents the slope and \( b \) is the y-intercept. Starting from the point-slope form, \( y - \frac{1}{4} = -\frac{1}{2}(x - 1) \), we can simplify to obtain the slope-intercept form.

First, distribute \(-\frac{1}{2}\) to \( (x - 1) \) gives \( y - \frac{1}{4} = -\frac{1}{2}x + \frac{1}{2} \). Then, solving for \( y \) by adding \( \frac{1}{4} \) to both sides, we get \( y = -\frac{1}{2}x + \frac{1}{2} + \frac{1}{4} \), which simplifies to \( y = -\frac{1}{2}x + \frac{3}{4} \). This form makes it easy to identify both the slope and the y-intercept of the tangent line, thus giving clear insights into the line's behavior on a graph.