Problem 37 - Technical Mathematics with Calculus

Question

Tangent to a Curve Find the slope of the tangent at the point indicated. $$y=\log (4 x-3) \text { at } x=2$$

Step-by-Step Solution

Answer

The slope of the tangent to the curve \(y = \log(4x - 3)\) at \(x = 2\) is \(\frac{4}{5}\).

Step 1: Differentiate the function

To find the slope of the tangent line to the curve at a particular point, we need to first find the derivative of the given function with respect to x. The function is given by \(y = \log(4x - 3)\). The derivative of \(y\) with respect to \(x\) using the chain rule is \(\frac{dy}{dx} = \frac{1}{4x - 3} \cdot \frac{d}{dx}(4x - 3) = \frac{4}{4x - 3}\).

Step 2: Evaluate the derivative at the given point

Next, we need to evaluate the derivative at the point where \(x = 2\). Substituting \(x = 2\) into the derivative, we get \(\frac{dy}{dx}\bigg|_{x=2} = \frac{4}{4(2) - 3} = \frac{4}{8 - 3} = \frac{4}{5}\).

Step 3: Interpret the result

The value \(\frac{4}{5}\) represents the slope of the tangent line to the curve at the point where \(x = 2\). Therefore, the slope of the tangent is \(\frac{4}{5}\).

Key Concepts

Derivative of Logarithmic Functions

Understanding the derivation of logarithmic functions is essential in calculus, particularly when dealing with the slopes of tangent lines. The logarithmic function, typically written as \(y = \log_b(x)\), where \(b\) is the base, has a derivative that follows a certain rule. The general formula for the derivative of a logarithmic function with base \(b\) with respect to \(x\) is \(\frac{1}{x \ln(b)}\).

In the scenario where we deal with the natural logarithm, that is \(\log_e(x)\) or \(\ln(x)\), the derivative simplifies to \(\frac{1}{x}\) because \(\ln(e) = 1\). However, for our example with function \(y = \log(4x - 3)\), which implies a natural logarithm (since the base \(e\) is typically omitted), the derivative involves applying the chain rule which we'll lay out in the next section. The idea is to differentiate the outer function, which is \(\log(x)\), and multiply it by the derivative of the inner function, \(4x - 3\), resulting in a final answer of \(\frac{4}{4x - 3}\) for the derivative of \(y\). This concept is fundamental when we want to find slopes of tangents to curves at a given point.

Chain Rule in Calculus

The chain rule in calculus is a pivotal concept for computing the derivative of composite functions. When a function \(y\) is composed of an outer function \(f\) and an inner function \(g(x)\), i.e., \(y = f(g(x))\), the chain rule states that the derivative of \(y\) with respect to \(x\) is the product of the derivative of \(f\) with respect to \(g\) and the derivative of \(g\) with respect to \(x\), notated as \(\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

Applying this to our example, the derivative of \(y = \log(4x - 3)\) involves differentiating the outer function \(\log(u)\), where \(u = 4x - 3\), giving us \(\frac{1}{u}\) as a result. Then, we multiply this by the derivative of the inner function \(u\), which is \(\frac{du}{dx} = 4\). The multiplication of \(\frac{1}{u}\) and \(4\) yields the derivative \(\frac{4}{4x - 3}\), precisely showing how the chain rule is used to find the derivative of composite functions.

Evaluating Derivatives at a Point

Evaluating derivatives at a specific point is a fundamental skill in calculus, necessary for finding slopes of tangents, rates of change, and other values dependent on a specific instance. Once you've found the derivative of a function, as we have done using the chain rule in the previous example, the next step is to 'plug in' the value of \(x\) for the point of interest.

In our exercise, we want to evaluate the derivative at \(x = 2\). By substituting \(x = 2\) into the derivative \(\frac{dy}{dx} = \frac{4}{4x - 3}\), we calculate the derivative at that point, which simplifies to \(\frac{4}{5}\). This value is the slope of the curve at that precise point, meaning that for our logarithmic function, a line tangent to the curve at \(x = 2\) would rise 4 units for every 5 units it runs horizontally. Evaluating derivatives at specific points allows us to construct tangents and understand the behavior of functions at those points.