Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\), then find the equation of the tangent line at the indicated \(x\) -value. $$y=\frac{x^{x}}{x+1}, \quad x=1$$

Short answer

\(\frac{dy}{dx} = \frac{1}{4}\) and the tangent line is \(y = \frac{1}{4}x + \frac{1}{4}\).

Step-by-step solution

Apply Logarithmic Differentiation

First, take the natural logarithm of both sides of the equation to apply logarithmic differentiation. Let \( u = x^x \). This gives:\[ \ln y = \ln \left( \frac{x^x}{x+1} \right) \]Using the property of logs, this is equivalent to:\[ \ln y = \ln x^x - \ln(x+1) \]Notice that \( \ln x^x = x \ln x \). So, the equation becomes:\[ \ln y = x \ln x - \ln(x+1) \]

Differentiate Using Implicit Differentiation

Differentiate both sides with respect to \( x \). Start with the left side:\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \]Now differentiate the right side:\[ \frac{d}{dx}(x \ln x - \ln(x+1)) = (1 \cdot \ln x + x \cdot \frac{1}{x}) - \frac{1}{x+1} \]Simplifying the above:\[ \frac{d}{dx}(x \ln x - \ln(x+1)) = \ln x + 1 - \frac{1}{x+1} \]

Solve for \(\frac{dy}{dx}\)

Set the derivatives equal and solve for \( \frac{dy}{dx} \):\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1 - \frac{1}{x+1} \]Multiply through by \( y \):\[ \frac{dy}{dx} = y \left( \ln x + 1 - \frac{1}{x+1} \right) \]

Evaluate \(\frac{dy}{dx}\) at \(x=1\)

Substitute \( x = 1 \) into the equation to find \( \frac{dy}{dx} \):Firstly, calculate \( y \) at \( x = 1 \): \[ y = \frac{1^1}{1 + 1} = \frac{1}{2} \]Now evaluate the derivative:\[ \frac{dy}{dx} = \frac{1}{2} \left( \ln 1 + 1 - \frac{1}{1+1} \right) = \frac{1}{2} (0 + 1 - 0.5) = \frac{1}{2} (0.5) = \frac{1}{4} \]

Find the Equation of the Tangent Line

The general equation of the tangent line is \( y = mx + b \), where \( m \) is the slope at the given point. The slope \( m = \frac{1}{4} \) and at \( x = 1 \), \( y = \frac{1}{2} \).We have: \[ \frac{1}{2} = \frac{1}{4} \cdot 1 + b \]Solving for \( b \):\[ \frac{1}{2} = \frac{1}{4} + b \]\[ b = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4} \]Thus, the equation is:\[ y = \frac{1}{4}x + \frac{1}{4} \]

Key concepts

Implicit Differentiation

Implicit Differentiation is a technique used in calculus when dealing with equations that define a function implicitly rather than explicitly. Sometimes, functions are not clearly solved for one variable with respect to another, such as the equation we worked with earlier, where the function is defined implicitly by \[ \ln y = x \ln x - \ln(x+1) \]. Implicit differentiation is beneficial in studying these more challenging relationships.
When we take the derivative of both sides of an implicit equation with respect to one variable, we have to apply the chain rule carefully. For example, the derivative of \( \ln y \) with respect to \( x \) is \( \frac{1}{y} \cdot \frac{dy}{dx} \) due to the chain rule, where \( \frac{dy}{dx} \) is the differential we want to find.
By working through the chain of derivatives, we're able to isolate \( \frac{dy}{dx} \), which is the derivative we seek. This step is foundational in our larger process of finding the derivative using logarithmic differentiation.

Tangent Line

The concept of the tangent line is geometrically important when analyzing a curve at a specific point. A tangent line touches a curve at exactly one point. It is considered the 'best linear approximation' to the curve at that individual point.
To find the tangent line equation at a given point \((x_{0}, y_{0})\) on a curve \(y=f(x)\), we utilize the derivative \(\frac{dy}{dx}\), which provides the slope \(m\) of the tangent line at \(x_{0}\). For the example provided, after computing the derivative and evaluating it at \(x=1\), we determined \(m = \frac{1}{4}\).
With the point-slope form of a line, \(y - y_{0} = m(x - x_{0})\), we rearrange it to find the full equation, yielding \(y = \frac{1}{4}x + \frac{1}{4}\) for the tangent line. This equation perfectly describes the line that just barely touches the graph of the function at the point \( (1, \frac{1}{2}) \).

Derivative Calculation

Calculating derivatives forms the core of calculus and is essential in understanding how a function is changing at any point. In the exercise, the function \(y=\frac{x^{x}}{x+1}\) is complex and involves both quotient and implicit differentiation. To manage this, logarithmic differentiation is used.
First, the derivative of \(x \ln x\) is derived as \(\ln x + 1\), using the product rule. For the logarithmic function, we differentiate \(-\ln(x+1)\) as \(-\frac{1}{x+1}\). Together, these expressions give us the simplified derivative of the logarithmic function as follows: \[ \ln x + 1 - \frac{1}{x+1} \].
After this, finding \(\frac{dy}{dx}\) requires multiplying by \(y\) due to the implicit nature of our differentiation. Finally, substituting the given \(x\) value provides the numerical derivative \(\frac{1}{4}\) at the specific point, capturing the function's slope at that slice.