Question

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

Short answer

The derivative is \(1 - 2\ln(2)\), and the tangent line is \(y = (1 - 2\ln(2))x + 2\ln(2) + 1\)."

Step-by-step solution

Take the natural logarithm of both sides

Start with the function \( y = (1+x)^{1/x} \). To use logarithmic differentiation, take the natural logarithm of both sides: \( \ln(y) = \ln((1+x)^{1/x}) \). This simplifies using the logarithmic identity \( \ln(a^b) = b\ln(a) \), to \( \ln(y) = \frac{1}{x} \ln(1+x) \).

Differentiate implicitly with respect to x

Differentiate both sides with respect to \( x \). The derivative of \( \ln(y) \) is \( \frac{1}{y} \frac{dy}{dx} \). For the right side \( \frac{1}{x} \ln(1+x) \), use the product rule: \( \frac{d}{dx} \left( \frac{1}{x} \ln(1+x) \right) = \frac{1}{x} \cdot \frac{d}{dx}(\ln(1+x)) + \ln(1+x) \cdot \frac{d}{dx}\left(\frac{1}{x}\right) \).

Simplify the derivatives

For \( \frac{1}{x} \cdot \frac{d}{dx}(\ln(1+x)) \), use the chain rule for \( \ln(1+x) \) yielding \( \frac{1}{1+x} \cdot 1 \) so it becomes \( \frac{1}{x(1+x)} \). For \( \ln(1+x) \cdot \frac{d}{dx}(\frac{1}{x}) \), \( \frac{d}{dx}(\frac{1}{x}) \) is \( -\frac{1}{x^2} \). Therefore, the expression becomes \( \frac{1}{x(1+x)} - \frac{\ln(1+x)}{x^2} \).

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

Set the derivatives equal: \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x(1+x)} - \frac{\ln(1+x)}{x^2} \). Multiply through by \( y \) to get \( \frac{dy}{dx} = y \left( \frac{1}{x(1+x)} - \frac{\ln(1+x)}{x^2} \right) \). Substitute \( y = (1+x)^{1/x} \) into the expression for \( \frac{dy}{dx} \).

Evaluate the derivative at \( x=1 \)

Substitute \( x = 1 \) into \( y = (1+x)^{1/x} \) so it becomes \( y = 2 \). Substitute \( x = 1 \) into \( \frac{dy}{dx} \) to get \( \frac{dy}{dx} = 2 \left( \frac{1}{2} - \frac{\ln(2)}{1} \right) = 1 - 2\ln(2) \).

Find the equation of the tangent line

Use the point \((1, 2)\) and the slope \(1 - 2\ln(2)\) in the point-slope form of a line, \( y - y_1 = m(x - x_1) \). This becomes \( y - 2 = (1 - 2\ln(2))(x - 1) \). Simplify this equation to find the tangent line: \( y = (1 - 2\ln(2))x + 2\ln(2) + 1 \).

Key concepts

Derivative Calculation

Understanding how to compute derivatives is crucial in calculus, especially when dealing with complex functions. In this exercise, we use a technique called logarithmic differentiation to find the derivative of a given function. The function is \( y = (1+x)^{1/x} \), which at first glance seems difficult to differentiate directly.

Logarithmic differentiation involves taking the natural logarithm of both sides of an equation, transforming a product or quotient into a sum or difference that is easier to differentiate. For this function, we apply the logarithmic identity \( \ln(a^b) = b \ln(a) \), rewriting it as \( \ln(y) = \frac{1}{x} \ln(1+x) \).

Once transformed, implicit differentiation allows us to handle the derivative of a variable multiplied by or divided by itself, which is essential for more complicated structures like this one. By doing so, we simplify the process of finding \( \frac{dy}{dx} \), making the calculation manageable.

Implicit Differentiation

In calculus, implicit differentiation is an essential tool for finding derivatives when the equation of a function is not explicitly solved for one variable. In the case of our function, once we take the natural logarithm, we end up differentiating with respect to \( x \) when \( y \) is still part of the expression.

To perform implicit differentiation, we take derivatives on both sides of the transformed equation \( \ln(y) = \frac{1}{x} \ln(1+x) \). The left side, \( \ln(y) \), differentiates to \( \frac{1}{y} \frac{dy}{dx} \), thanks to the chain rule. On the right side, we apply the product rule to differentiate \( \frac{1}{x} \ln(1+x) \), breaking it into simpler parts.

• The first part, \( \frac{1}{x} \), differentiates to \( -\frac{1}{x^2} \) using basic derivative rules.
• The second part, \( \ln(1+x) \), differentiates to \( \frac{1}{1+x} \).

By combining these, we effectively break down the original function into components simple enough for straightforward calculus techniques, showcasing the power of implicit differentiation in complex problems.

Equation of Tangent Line

Finding the equation of a tangent line involves two key pieces: the slope of the function at a particular point and the point itself. After finding the derivative using logarithmic differentiation and implicit differentiation, we evaluate the derivative at the specific \( x \)-value given—in this case, \( x = 1 \).

The derivative evaluation at \( x = 1 \) gives us the slope \( \frac{dy}{dx} = 1 - 2\ln(2) \). Knowing both the slope and the point \((1, 2)\), we use the point-slope form of a line to formulate the equation. This form is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the given point and \( m \) is the slope.

Substituting the values into this formula, we find the equation becomes: \[y = (1 - 2\ln(2))x + 2\ln(2) + 1\]Effectively, this equation describes the line that just touches the curve at \( x = 1 \) without cutting through it, giving us the tangent at that specific point, which is often crucial in understanding the instantaneous rate of change.