Problem 49 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine an equation of the tangent line to the function at the given point. $$ y=x \ln x $$ $$ (1,0) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(y=x \ln x\) at point (1,0) is \(y = x - 1\).

Step 1: Calculate the derivative of the function

We recognize that the function \(y = x \ln x\) is a product of two differentiable functions \(u(x) = x\) and \(v(x) = \ln x\). Using the product rule of differentiation: \((u.v)' = u'.v + u.v'\). Differentiate to get: \(y' = 1 . \ln x + x . 1/x = \ln x + 1\).

Step 2: Evaluate the derivative at the given point

The value of the derivative at point (1,0) gives the slope of the tangent at that point. Substitute \(x = 1\) into the derivative: \(y'(1) = \ln 1 + 1 = 0 + 1 = 1\).

Step 3: Apply the point-slope form

We use the point-slope form of a line to find the equation of the tangent line. The point-slope form is \(y - y_1 = m(x - x_1)\), where m is the slope of the line and \((x_1, y_1)\) is a point on the line. We have m = 1 (from Step 2) and point (1, 0). So, substitute these into the equation, we get: \(y - 0 = 1(x - 1)\). Simplify to get: \(y = x - 1\).

Key Concepts

Understanding the Derivative

Let's dive into how derivatives help us find the slope of a tangent line. When you take the derivative of a function, you're essentially finding the rate at which the function changes at any given point. This is key for determining the slope of the tangent line.

In this exercise, the function given was \(y = x \ln x\). To get the derivative, we first need to understand:

• \(u(x) = x\): The simple linear part.
• \(v(x) = \ln x\): The logarithmic part.

The derivative of \(u(x)\) is 1, and the derivative of \(v(x)\) is \(1/x\). Together, they help us apply the product rule effectively.

Mastering the Point-Slope Form

The point-slope form is a handy formula for finding the equation of a line when you know a point on the line and its slope. The formula is \(y - y_1 = m(x - x_1)\), where:

• \((x_1, y_1)\) is a known point on the line.
• \(m\) is the slope of the line.

This exercise used the point-slope form to find the tangent at the point (1,0) with a slope of 1. By plugging these into the formula, we simplify it to \(y = x - 1\). This represents the equation of the tangent line elegantly.

Applying the Product Rule

The product rule is a technique in calculus used for differentiation. When dealing with products of two functions, the derivative isn't as straightforward as just taking the derivative of each part. Instead, the product rule states that: \[(u \cdot v)' = u'v + uv'\] This means that you take the derivative of the first function \(u\), multiply it by the second function \(v\), then add the product of the first function \(u\) and the derivative of the second function \(v'\).

For \(y = x \ln x\), you get \(y' = 1 \cdot \ln x + x \cdot 1/x\), which simplifies to \(\ln x + 1\). This elegant solution quickly gives us the slope needed for the tangent line.