Problem 43 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 43 through 46 , find an equation for the tangent line to the given curve at the specified point. 43\. \(y=x \ln x^{2}\) where \(x=1\)

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 2x - 2\).

Step 1: Find the derivative of the function

Given the function is \(y = x \ln x^2\). To find the equation of the tangent line, first find the derivative \(\frac{dy}{dx}\).

Step 2: Apply the product and chain rules

Rewrite the function as \(y = x \ln (x^2)\). We apply the product rule where \(u = x\) and \(v = \ln (x^2)\). The product rule states \(uv' + vu'\).

Step 3: Simplify the logarithmic term

Simplify \(u' = 1\) and \(v' = \frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2} = \frac{2}{x}\). Thus, \(uv' + vu' = x \cdot \frac{2}{x} + \ln(x^2)\cdot 1 = 2 + 2\ln(x)\). So, \(\frac{dy}{dx} = 2 + 2\ln(x)\).

Step 4: Evaluate the derivative at the given point

To find the slope of the tangent line at \(x=1\), compute \(\frac{dy}{dx}\) at \(x=1\). The derivative evaluated at \(x=1\) is \(2 + 2 \ln(1) = 2 + 2 (0) = 2\). Hence, the slope \(m\) of the tangent line at \(x=1\) is 2.

Step 5: Find the y-coordinate of the point on the curve

Substitute \(x=1\) into the original function to find the y-coordinate. \(y = 1 \ln(1^2) = 1 \cdot 0 = 0\). Hence, the point of tangency is (1, 0).

Step 6: Find the equation of the tangent line

Use the point-slope form \(y - y_1 = m (x - x_1)\) with \(m = 2\), \(x_1 = 1\), and \(y_1 = 0\). Thus, the equation of the tangent line is \(y - 0 = 2(x - 1)\) which simplifies to \(y = 2x - 2\).

Key Concepts

derivative

To understand how to find the equation of a tangent line, first, we need to understand derivatives. A derivative represents the rate at which a function is changing at any given point. More formally, it is the slope of the function at that point. If we have a function given by \(y = f(x)\), the derivative, written as \(\frac{dy}{dx}\) or \(f'(x)\), tells us how much \(y\) changes with a small change in \(x\). For example, in the problem, we have the function \(y = x \ln(x^2)\). The goal is to find how this function changes as \( x \) changes, which requires us to compute \(\frac{dy}{dx}\). This derivative helps us later to find the slope of the tangent line at a given point.

product rule

The product rule is a useful technique for differentiating functions that are products of two or more functions. If we have two functions, \(u(x)\) and \(v(x)\), the product rule states that the derivative of their product is given by \( (uv)' = u'v + uv' \). In the exercise, our function can be broken down into \( u = x \) and \( v = \ln(x^2) \). By applying the product rule:
- Differentiate \( u(x) = x \) to get \( u' = 1 \).
- Differentiate \( v(x) = \ln(x^2) \) using the chain rule: \( v'(x) = \frac{2}{x} \).
This gives us: \( x \cdot \frac{2}{x} + \ln(x^2) \cdot 1 \), which simplifies to \(2 + 2\ln(x)\). Therefore, the derivative \( \frac{dy}{dx} = 2 + 2\ln(x) \).

chain rule

The chain rule is another crucial rule for finding derivatives, especially when dealing with compositions of functions. If we have a function \( y = f(g(x)) \), the chain rule states that the derivative of this function is the derivative of \( f \) evaluated at \( g(x) \) multiplied by the derivative of \( g(x) \), or formally \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). In our exercise, the function \( v = \ln(x^2) \) is composed of \( \ln(u) \) where \( u = x^2 \). To differentiate, we use the chain rule:
- Differentiate the outer function \( \ln(u) \), which gives \( \frac{1}{u} \), and then multiply by the derivative of the inner function \( x^2 \), which is \( 2x \). So, \( \frac{d}{dx}( \ln(x^2) ) = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \).

point-slope form

The point-slope form is a way to express the equation of a line when we know the slope and a point on the line. It is given by \( y - y_1 = m (x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. In the exercise, we already calculated the slope of the tangent line at the point \( x = 1 \) as 2. We also determined that the point of tangency is \( (1, 0) \). Using these values, we substitute \( m = 2 \), \( x_1 = 1 \), and \( y_1 = 0 \) into the point-slope form to get:
\( y - 0 = 2 (x - 1) \)
This simplifies to the equation of the tangent line: \( y = 2x - 2 \).