Question

A line with slope \(m\) passes through the origin and is tangent to \(y=\ln (x / 3) .\) What is the value of \(m ?\)

Short answer

The value of \(m\) is \(1/3\).

Step-by-step solution

Find the Derivative of the Function

The derivative of \(y = \ln(x/3)\) can be found using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function. So, derivative of \(y\), \(y'\), is \((1/(x/3)) * (1/3) = 1/x\).

Find the Point where the Line is Tangent to the Function

The line passes through the origin and is tangent to \(y = \ln(x/3)\), so it must cross the x-axis at the point where \(y=0\), which is when \(x=3\).

Evaluate the Derivative at the Tangent Point

Evaluate the derivative at the point \(x=3\) gives \(1/3\).

Set the Derivative Equal to the Slope

The slope of the tangent line is the same as the derivative of the function at the tangent point. Therefore, \(m = y'(3) = 1/3\).

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus, particularly important when finding the derivatives of composite functions. A composite function is formed when one function is nested inside another, for example, if we have an outer function like the natural logarithm, \( \ln(x) \), and an inner function, such as \( x/3 \). To differentiate such a composite function, the chain rule guides us to take the derivative of the outer function and multiply it by the derivative of the inner function.

Using the chain rule, we can efficiently tackle calculus problems involving compounded expressions. It states that if we have a function \( h(x) = f(g(x)) \), then its derivative \( h'(x) = f'(g(x)) \cdot g'(x) \). This rule becomes especially useful when dealing with more complex functions where recognizing the layers – or 'the functions within functions' – allows us to differentiate step by step, ensuring accuracy in our calculations.

For example, in the textbook problem, \(y = \ln(x / 3)\), the derivative is found by recognizing \(x / 3\) as the inner function and \(\ln(x)\) as the outer function. Thus, we apply the chain rule as demonstrated in the steps provided.

Derivative of Logarithmic Functions

Logarithmic functions often appear in calculus problems, and understanding their derivatives is essential. The natural logarithm function, represented by \( \ln(x) \), has a relatively simple derivative of \( 1/x \). However, when the logarithm function includes a transformation, such as \( \ln(x/3) \), we must adjust our approach to find the derivative.

The derivative of a logarithmic function that is more complex than simply \( \ln(x) \) often requires the use of the chain rule. In our aforementioned exercise, the derivative of \( \ln(x/3) \) is not just \(1/x\) due to the presence of the division by 3. To correctly find the derivative, we apply the chain rule, which results in \(1/(x/3) \cdot (1/3) = 1/x\), simplifying the process and ensuring we do not overlook any components of the function.

Finding Tangent Slope

When solving calculus problems involving tangents to curves, finding the slope of the tangent line at a particular point is a common task. The slope of the tangent line to a curve at a certain point is simply the value of the derivative of the function at that point. This is because the derivative represents the rate of change of the function, which is exactly what the slope of a line represents - the steepness or incline.

In the exercise, we were tasked to find the slope of a line tangent to \( y = \ln(x / 3) \) that passes through the origin. Since the line passes through the origin, it intersects the graph of the function when \( y = 0 \), which leads us to find the corresponding \( x \) value. Upon finding the \( x \) value where the line touches the curve, we evaluate the derivative at this point, which gives us the slope of the tangent line at that point. In this case, evaluating the derivative \( y' = 1/x \) at \( x = 3 \) provides us with the slope \( m \), confirming that \( m = 1/3 \). This technique is widely applicable for various functions beyond logarithmic ones, making it an indispensable tool in a mathematician's arsenal.