Problem 16 - Thomas Calculus

Question

In Exercises \(13-16,\) differentiate the functions and find the slope of the tangent line at the given value of the independent variable. $$ y=(x+1)^{3}, \quad x=-2 $$

Step-by-Step Solution

Answer

The slope of the tangent line at \( x = -2 \) is 3.

Step 1: Differentiate the function

Given the function \( y = (x+1)^3 \), we need to find the derivative to determine the slope of the tangent line. The derivative \( y' \) is obtained using the chain rule. Let \( u = x+1 \), then \( y = u^3 \). The derivative \( y' = \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). We have: \( \frac{du}{dx} = 1 \), and \( \frac{dy}{du} = 3u^2 \). Thus, \( y' = 3(x+1)^2 \cdot 1 = 3(x+1)^2 \).

Step 2: Evaluate the derivative at the given point

The next step is to find the slope of the tangent line at \( x = -2 \). To do this, substitute \( x = -2 \) into the derivative we found. So, \( y' = 3(-2+1)^2 = 3(-1)^2 = 3 \cdot 1 = 3 \). Thus, the slope of the tangent line at \( x = -2 \) is 3.

Key Concepts

Understanding the Tangent Line

When you have a curve defined by a function, at any specific point on this curve, there is a straight line that just "touches" the curve. This line is called the tangent line. In everyday terms, it's like a line that just skims the surface without cutting through it.
For example, think about a hill. If you are at a specific point on the hill, you could imagine the direction you'd roll if you put a ball down. This direction aligns with the tangent line at that point. The slope of this tangent line tells us how steep the hill is at that exact place.

In calculus, the slope of the tangent line is mathematically calculated by finding the derivative of the function at the point of interest. In our problem, we found that the slope of the tangent line at the point where \( x = -2 \) for the function \( y = (x+1)^3 \) is 3. This information can tell us about the rate at which the function is changing at that point.

Applying the Chain Rule

The chain rule is a fundamental tool in calculus, especially useful when dealing with composite functions. A composite function is essentially a function of a function. In simpler terms, it's like having layers of operations and you need to peel them back to find the derivative.
Consider the function from our exercise: \( y = (x+1)^3 \). Here, we can see it as a composition of two functions – the "inner function" \( u = x+1 \), and the "outer function" \( y = u^3 \).
To find the derivative of \( y \) with respect to \( x \), the chain rule tells us to first differentiate the outer function with respect to the inner function (\( y = u^3 \)), then multiply it by the derivative of the inner function with respect to \( x \) (\( u = x+1 \)). This gives us:

• Derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \).
• Derivative of \( u = x+1 \) with respect to \( x \) is 1.

So, the derivative \( y' = 3(x+1)^2 \cdot 1 \), simplifying to \( y' = 3(x+1)^2 \).
This rule is called the "chain" rule because of how it links or "chains" together the differentiation of each layer of functions.

Derivative Evaluation and Finding Slope

After finding the derivative using the chain rule, the next step is to evaluate it at a given value to determine the slope of the tangent line at that specific point on the function.
In our example, once we differentiate \( y = (x+1)^3 \) to find \( y' = 3(x+1)^2 \), we substitute \( x = -2 \).
This gives us:

• Substitute \( x = -2 \) into \( y' \), which is \( 3(-2+1)^2 \).
• Simplify: This becomes \( 3(-1)^2 \) which equals \( 3 \times 1 = 3 \).

So, the slope of the tangent line at \( x = -2 \) is 3.
The process of derivative evaluation is essential because it offers a snapshot of the function's behavior at any desired point. This tells you how steep your function is, or how rapidly it is increasing or decreasing, which can be very insightful for understanding the graph's nature.