Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\), then find the equation of the tangent line at the indicated \(x\) -value. $$y=\frac{x+1}{x+2}, \quad x=1$$

Short answer

The derivative is \(\frac{1}{9}\) at \(x=1\), and the tangent line is \(y = \frac{1}{9}x + \frac{5}{9}\)."

Step-by-step solution

Apply Logarithmic Differentiation

First, take the natural logarithm of both sides to simplify the expression. So we have:\[ \ln y = \ln \left( \frac{x+1}{x+2} \right) \]Using the properties of logarithms, we can rewrite it as:\[ \ln y = \ln (x+1) - \ln (x+2) \]

Differentiate Both Sides

Differentiate both sides with respect to \(x\). Recall that the derivative of \(\ln y\) with respect to \(x\) is \( \frac{1}{y} \frac{dy}{dx} \):\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x+1} - \frac{1}{x+2} \]Multiplying both sides by \(y\) gives:\[ \frac{dy}{dx} = y \left( \frac{1}{x+1} - \frac{1}{x+2} \right) \]

Substitute Original Function

Substitute back the original function \(y = \frac{x+1}{x+2}\):\[ \frac{dy}{dx} = \frac{x+1}{x+2} \left( \frac{1}{x+1} - \frac{1}{x+2} \right) \]

Simplify the Derivative

Now, simplify the expression:\[ \frac{dy}{dx} = \frac{x+1}{x+2} \left( \frac{1}{x+1} - \frac{1}{x+2} \right) = \frac{x+1}{x+2} \left( \frac{x+2 - (x+1)}{(x+1)(x+2)} \right) \]\[ = \frac{x+1}{x+2} \cdot \frac{1}{(x+1)(x+2)} = \frac{1}{(x+2)^2} \]

Evaluate the Derivative at x=1

Find \( \frac{dy}{dx} \) at \(x = 1\):\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1}{(1+2)^2} = \frac{1}{9} \]

Find the Equation of the Tangent Line

The slope of the tangent line at \(x=1\) is \(\frac{1}{9}\). The \(y\) value of the function at \(x=1\) is:\[ y = \frac{(1+1)}{(1+2)} = \frac{2}{3} \]Using the point-slope form: \(y - y_1 = m(x - x_1)\), where \((x_1,y_1) = (1, \frac{2}{3})\) and \(m = \frac{1}{9}\):\[ y - \frac{2}{3} = \frac{1}{9}(x - 1) \]

Simplify the Equation of the Tangent Line

Simplify the equation to find:\[ y - \frac{2}{3} = \frac{1}{9}x - \frac{1}{9} \]Adding \(\frac{2}{3}\) to both sides:\[ y = \frac{1}{9}x - \frac{1}{9} + \frac{6}{9} \]\[ y = \frac{1}{9}x + \frac{5}{9} \]

Key concepts

Tangent Line Equation

A tangent line is a straight line that just touches, or is "tangent" to, a curve at a certain point. In calculus, finding the equation of the tangent line at a given point provides a snapshot of how the curve behaves at that specific moment. The slope of this tangent line represents how steep the curve is at the point where it touches. This slope is derived from the derivative of the function evaluated at the given point.
To find the equation of the tangent line for a specific function, you'll use the point-slope form. Once you have the slope from the derivative and the exact point on the curve, plug these values into the point-slope formula to get the tangent line's equation.

Derivatives

Derivatives are a central concept in calculus. They measure how a function changes as its input changes. In essence, they provide the slope of the function at any point. For a given curve described by the function \(y=f(x)\), the derivative, \(\frac{dy}{dx}\), gives the rate of change of \(y\) with respect to \(x\).
When you differentiate a function, you're essentially finding the function's slope, or steepness, at every point along its curve. This helps us understand the function's behavior in terms of increasing, decreasing, or remaining constant. Differentiation can be done using various rules, including the product, quotient, and chain rules, to simplify calculations.

Properties of Logarithms

The properties of logarithms are crucial tools for simplifying and managing expressions involving logarithms. They enable us to transform products, quotients, and powers into simpler sums or differences. Here are a few key properties:

• Product property: \( \ln(ab) = \ln a + \ln b \)
• Quotient property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• Power property: \( \ln(a^b) = b \ln a \)

Using these properties, complex expressions can be broken down into simpler terms, making derivatives more accessible, as seen when applying logarithmic differentiation. They also illustrate how logarithms simplify mathematical relationships, particularly those involving growth and decay.

Point-Slope Form

The point-slope form is a helpful way to write the equation of a line when you know a single point on the line and its slope. The form is given by: \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the line, and \(m\) is the slope.
This form is particularly useful when deriving the equation for the tangent line. If you know the slope of the tangent (from your derivative) and the coordinates of the point of tangency, you can easily plug these values into the point-slope form to get the line’s equation. This method provides an efficient way to describe lines without immediately rearranging them into the slope-intercept form, \(y = mx + b\).