Problem 47 - Essential Calculus

Question

Find \(F^{\prime}(x)\). $$ F(x)=\int_{1}^{3 x} \frac{1}{t} d t $$

Step-by-Step Solution

Answer

The derivative of the given function is \( F^{\prime}(x) = 1/x \).

Step 1: Apply the Fundamental Theorem of Calculus

Applying the Fundamental Theorem of Calculus, we rewrite the integral function as follows: \( F(x) = ln|3x| - ln|1| \), which simplifies to \( F(x) = ln|3x| \).

Step 2: Derive the Resulting Function

Now, we need to find the derivative of \( F(x) \), or \( F^{\prime}(x) \). Using the chain rule, which says if we have a composition of functions (say, \( f(g(x)) \)), the derivative of that would be \( f^{\prime}(g(x)) \cdot g^{\prime}(x) \), we obtain \( F^{\prime}(x) = (1/(3x)) \cdot 3 = 1/x \).

Key Concepts

Understanding Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function. It can be viewed as the reverse operation of differentiation. In the given exercise, you are asked to evaluate the integral \( F(x) = \int_{1}^{3x} \frac{1}{t} \, dt \).
The task is to find the derivative of this integral, essentially reversing the process of integration. Here, integration is used to solve for an indefinite form which involves variable limits of integration. This is an example of what's known as the "Fundamental Theorem of Calculus". This theorem connects the processes of differentiation and integration.

The Fundamental Theorem states two primary things:

• It establishes a connection between differentiation and integration, asserting that integration can be reversed by differentiation.
• If \( F(x) \) is the integral from \( a \) to \( b \) of \( f(t) \, dt \), then the derivative \( F^{\prime}(x) = f(x) \) when the variable of integration is part of the upper limit.

Understanding this relationship allows for simplifying many calculus problems by transforming integrals into derivatives.

The Role of the Chain Rule

The chain rule is a critical technique when working with composite functions. It plays an essential role in differentiating functions that are composed of other functions, which is common in calculus problems involving integration.

When you have a function that includes another function inside it, such as \( F(x) = \ln|3x| \), using the chain rule becomes necessary. According to the chain rule, the derivative of a composite function \( f(g(x)) \) is given by \( f^{\prime}(g(x)) \cdot g^{\prime}(x) \).
In the original solution, once integration has simplified \( F(x) \) to a form suitable for differentiation, the next step involves applying the chain rule to find \( F^{\prime}(x) \).

• First, recognize the outer function, which here is \( \ln|u| \), with \( u = 3x \).
• The derivative of \( \ln|3x| \) is \( \frac{1}{3x} \), reflecting the natural logarithm rule.
• Multiply this by the derivative of the inner function \( u = 3x \), which is simply 3.

This results in \( F^{\prime}(x) = \frac{1}{x} \), effectively employing the chain rule to address the composition of functions.

Exploring Derivatives

Derivatives are a cornerstone of calculus, dealing with how functions change. A derivative provides instant rates of change and can describe how a function behaves at any point.
In this exercise, we initially compute a definite integral and then look for its derivative. Essentially, you could consider this reversing the integration to understand how \( F(x) \) changes with respect to \( x \).

Here are a few points to remember about derivatives:

• Derivatives tell us the slope of the function at a given point. It is expressed as \( f^{\prime}(x) \) or \( \frac{df}{dx} \).
• By differentiating \( F(x) = \ln|3x| \), we find that \( F^{\prime}(x) = \frac{1}{x} \), showing the rate of change of \( F(x) \) with respect to \( x \).
• Finding derivatives of logarithmic functions like \( \ln(x) \), involves using their basic rule: \( \frac{d}{dx}[\ln|x|] = \frac{1}{x} \).

This exercise underscores how knowledge of derivatives allows not just the calculation of rates but also aids in solving and simplifying various calculus problems.