Problem 46 - Essential Calculus Early Transcendentals

Question

\(45-46=\) Find an equation of the tangent line to the curve at the given point. $$y=e^{x} / x, \quad(1, e)$$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = e \).

Step 1: Find the Derivative of the Function

To find the equation of the tangent line, we first need the derivative of the function. The function is given by \( y = \frac{e^x}{x} \). We will use the quotient rule to differentiate it. The quotient rule states \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \), where \( u = e^x \) and \( v = x \).ewline Derivative of \( u \) is \( u' = e^x \) and derivative of \( v \) is \( v' = 1 \).ewline Applying the quotient rule, the derivative \( \frac{dy}{dx} = \frac{e^x \cdot x - e^x}{x^2} = \frac{e^x(x - 1)}{x^2} \).

Step 2: Evaluate the Derivative at the Given Point

Next, we evaluate the derivative at the point \((1, e)\). Substitute \( x = 1 \) into the derivative:ewline \( \frac{dy}{dx} \bigg|_{x=1} = \frac{e^1(1 - 1)}{1^2} = \frac{e(0)}{1} = 0 \).ewline Hence, the slope of the tangent line at \( x = 1 \) is 0.

Step 3: Use Point-Slope Form to Find the Equation of the Tangent Line

The equation of the tangent line can be found using the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \).ewline Here, \( m \) is the slope, which we found to be 0, and \( (x_1, y_1) \) is the point \((1, e)\).ewline Substitute these values into the equation: \( y - e = 0(x - 1) \).ewline Simplifying gives \( y = e \).ewline Therefore, the equation of the tangent line is \( y = e \).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It provides the slope of the tangent line to the curve of a function at a particular point.
To find the derivative of a function, we determine the limit of the difference quotient as the change in the input approaches zero. In mathematical terms, if we have a function \( f(x) \), the derivative \( f'(x) \) can be expressed as: \\( f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \).
For the function \( y = \frac{e^x}{x} \), we use calculus rules for finding derivatives, like the quotient rule. Derivatives help us understand a function's behavior, such as where it increases or decreases, and identify local maxima and minima.
In this context, derivatives allow us to find the slope of a tangent line, which is crucial for applications like physics, where velocity and acceleration are derivatives of position with respect to time.

Quotient Rule

The quotient rule is a critical formula used to differentiate functions that are expressed as the ratio of two simpler functions. If we have a function \( y = \frac{u}{v} \), the quotient rule states that the derivative is:

• \( \left( \frac{d}{dx} \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).

Here, \( u \) and \( v \) are functions of \( x \), while \( u' \) and \( v' \) are their respective derivatives.
In our specific exercise, with \( u = e^x \) and \( v = x \), applying the quotient rule gives us the expression \( \frac{e^x(x-1)}{x^2} \) for the derivative.
Using the quotient rule is essential when dealing with rational functions, as it helps simplify the differentiation process and provides accurate results.

Point-Slope Form

To find the equation of a tangent line, we often use the point-slope form, a straightforward method. Given a point \((x_1, y_1)\) and a slope \(m\), the point-slope form is:

• \( y - y_1 = m(x - x_1) \)

In this method, \( m \) represents the slope of the tangent line derived from the derivative value at the given point.
For the given exercise, the slope was found to be 0 at point \((1, e)\). Substituting these into the point-slope form formula results in the tangent line equation \( y - e = 0(x - 1) \), which simplifies to \( y = e \).
Point-slope form is highly useful when you're given a specific point and slope, helping you quickly find the linear equation of the tangent line or any other line efficiently.