Problem 44 - Calculus Single Variable

Question

Calculate the first and second derivatives of \(F(x)=\int_{a}^{u(x)} f(t) d t\) for the given functions \(u\) and \(f\) \(u(x)=\log _{2}(x) \quad f(t)=1 / t\)

Step-by-Step Solution

Answer

The first derivative is \(F'(x) = \frac{1}{x \log_2(x) \ln(2)}\) and the second derivative is \(F''(x) = -\frac{1 + \log_2(x)}{x^2 (\log_2(x))^2 \ln(2)}\).

Step 1: Understand the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus relates differentiation and integration. If \(F(x) = \int_{a}^{u(x)} f(t) \, dt\), the derivative of \(F(x)\) with respect to \(x\) is \(F'(x) = f(u(x)) \cdot u'(x)\).

Step 2: Find the First Derivative

Substitute \(f(t) = \frac{1}{t}\) and \(u(x) = \log_2(x)\). The derivative \(u'(x)\) is \(\frac{d}{dx}[\log_2(x)] = \frac{1}{x\ln(2)}\). Therefore, the first derivative is \[F'(x) = \frac{1}{\log_2(x)} \cdot \frac{1}{x\ln(2)} = \frac{1}{x\log_2(x)\ln(2)}\].

Step 3: Differentiate Again for the Second Derivative

Apply the quotient rule to differentiate \(F'(x) = \frac{1}{x\log_2(x)\ln(2)}\). Let \(g(x) = x\log_2(x)\ln(2)\), then \(g'(x)\) can be found using the product rule. After calculating, \(g'(x) = (1 + \log_2(x)) \frac{\ln(2)}{x}\). Then, the second derivative is \[F''(x) = -\frac{g'(x)}{(g(x))^2} = -\frac{\ln(2)(1 + \log_2(x))}{(x\ln(2)\log_2(x))^2} = -\frac{1 + \log_2(x)}{x^2(\log_2(x))^2\ln(2)}\].

Step 4: Simplify if Possible

Examine if the expression for \(F''(x)\) can be simplified further. Here \(F''(x)\) is already in its simplest form given the logarithmic and algebraic components involved.

Key Concepts

Derivatives

Derivatives help us understand how a function changes as its input changes. In this exercise, calculating derivatives of the function \( F(x) = \int_{a}^{u(x)} f(t) \, dt \) involves using the Fundamental Theorem of Calculus. Such theorem provides a way to relate an integral function with its derivative. In simple terms, it says the derivative of our integral function \( F(x) \) is \( f(u(x)) \cdot u'(x) \). In this case,

• We used \( f(t) = \frac{1}{t} \).
• The function \( u(x) \) is \( \log_2(x) \).

So, after substitution and solving, the first derivative of \( F(x) \) turned out to be \( \frac{1}{x\log_2(x)\ln(2)} \). Handy, isn't it?

Logarithmic Functions

Logarithmic functions are essential in this problem. They help us understand exponential growth and decay. When we say `logarithms`, we refer to functions like \( \log_2(x) \), which means finding the power to which a base (here, 2) must be raised to produce a particular number \( x \).

Here, we use \( \log_2(x) \) to find the derivative. The derivative of a log function \( \log_b(x) \) follows a special rule: it is \( \frac{d}{dx}[\log_b(x)] = \frac{1}{x\ln(b)} \). This transformation is crucial in finding \( u'(x) \) in our problem. Understanding how these log rules work is vital because they often surface in exams or real-world problems.

Integration

Integration is about finding areas under curves or, more abstractly, 'accumulating' quantities. In this problem, integration acts in reverse of differentiation. Use it to build functions from their derivatives. The Fundamental Theorem of Calculus is the key player when connecting integration with derivatives.

The given function \( F(x) = \int_{a}^{u(x)} f(t) \, dt \) means we are calculating an integral with changing upper limits \( u(x) = \log_2(x) \). Once differentiated, it converts to a derivative problem as explored earlier. Notice how identifying integration boundaries, and how they change concerning \( x \), can seamlessly connect with differentiation. Thus, integration, coupled with differentiation, allows us to solve complex variable problems efficiently.