Problem 50 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine an equation of the tangent line to the function at the given point. $$ y=\frac{\ln x}{x} $$ $$ \left(e, \frac{1}{e}\right) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(y=\frac{\ln x}{x}\) at the point \((e, \frac{1}{e})\) is \(y=\frac{1}{e}\).

Step 1: Differentiate the function

To find the slope of the tangent line, differentiate the given function using the quotient rule, which states that the derivative of \(\frac{u}{v} = \frac{vu' - uv'}{v^2}\), where \(u = \ln(x)\) and \(v = x\). The derivative of \(u\) is \(1/x\), and the derivative of \(v\) is 1. Substituting these values into the quotient rule gives \(y' = \frac{x(1/x) - (\ln(x))(1)}{x^2} = \frac{1 - \ln(x)}{x}\).

Step 2: Find the slope of the tangent line

The slope of the tangent line at the given point is obtained by evaluating the derivative at \(x = e\). Substituting \(x = e\) into \(y'\) gives \(m = \frac{1 - \ln(e)}{e} = \frac{1 - 1}{e} = 0\). So, the slope of the tangent line is zero.

Step 3: Use the point-slope form of a line to find the equation of the tangent line

The point-slope form of the equation of a line is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope of the line. Substituting \(m = 0, x_1 = e,\) and \(y_1 = \frac{1}{e}\) into the point-slope form gives \(y - \frac{1}{e} = 0(x - e)\), which simplifies to \(y = \frac{1}{e}\).

Key Concepts

Quotient Rule Differentiation

Understanding how to differentiate functions that are divided by one another is crucial, especially when the functions are not simple constants. The quotient rule is a technique used in calculus to find the derivative of the division of two functions. When we have a function defined as \(y = \frac{u(x)}{v(x)}\), the derivative of that function, denoted as \(y'\), is found using the following formula: \[ \frac{du}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\].

For example, if we take the function \(y = \frac{\ln(x)}{x}\), we let \(u(x) = \ln(x)\) and \(v(x) = x\). Differentiating these individually, we get \(u'(x) = \frac{1}{x}\) for the natural log function, and \(v'(x) = 1\) since the derivative of \(x\) with respect to \(x\) is 1. Applying the quotient rule gives us \(y' = \frac{x(1/x) - (\ln(x))(1)}{x^2}\), simplifying to \(y' = \frac{1 - \ln(x)}{x^2}\).

The quotient rule is vital for functions where variables are divided by each other, providing a systematic method to differentiate them accurately.

Point-Slope Form

Once the derivative of a function is known, we can use this information to find the equation of a tangent line at a specific point. The point-slope form is the go-to equation for doing just that and is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \(\left(x_1, y_1\right)\) represents a point the line passes through.

For our exercise, after finding that the slope \(m\) is zero at the point \(\left(e, \frac{1}{e}\right)\), we plug the values into the point-slope formula, which simplifies significantly since the slope is zero. This gives us a horizontal line, \(y = \frac{1}{e}\), which is the equation of our tangent line. The point-slope form is highly beneficial when needing to write the equation of a line quickly, given a point and a slope.

Natural Log Function

The natural logarithm function, denoted as \(\ln(x)\), is the inverse of the exponential function \(e^x\). In calculus, understanding the \(\ln(x)\) function is important as it appears in various integrations and differentiations. When differentiating \(\ln(x)\), the result is \(\frac{1}{x}\), a fact that is crucial to solving problems involving the quotient rule and determining the slope of tangent lines as we did in our exercise.

The natural log has a constant base of \(e\), the irrational number approximately equal to 2.71828, which is the rate of growth shared by all continually growing processes. When \(x = e\) in \(\ln(x)\), the value is 1 because \(e\) is the number such that the area under the curve \(y = \frac{1}{x}\) from 1 to \(e\) is 1. This property simplifies our derivative calculated in the exercise when we substitute \(x = e\) and plays a key role in mathematical analysis and number theory.