Problem 48 - 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 \(f\) \(u(x)=\log _{2}(x) \quad f(t)=\log _{3}(t)\)

Step-by-Step Solution

Answer

\( F'(x) = \frac{\log_{3}(\log_{2}(x))}{x \ln(2)} \), \( F''(x) = \frac{1}{x^2 \ln^2(2) \cdot \log_{2}(x) \ln(3)} - \frac{\log_{3}(\log_{2}(x))}{x^2 \ln(2)} \)

Step 1: Understand the Problem

We need to find the first and second derivatives of the function \( F(x) = \int_{a}^{u(x)} f(t) \, dt \) where \( u(x) = \log_{2}(x) \) and \( f(t) = \log_{3}(t) \). This is an application of the Fundamental Theorem of Calculus and the chain rule.

Step 2: Apply the Fundamental Theorem of Calculus for First Derivative

The derivative of \( F(x) = \int_{a}^{u(x)} f(t) \, dt \) with respect to \( x \) is given by \( F'(x) = f(u(x)) \cdot u'(x) \). Here it implies we need to evaluate \( f(\log_{2}(x)) \cdot (\log_{2}(x))' \).

Step 3: Calculate \( f(u(x)) \)

Substitute \( u(x) = \log_{2}(x) \) into \( f(t) = \log_{3}(t) \) to get \( f(u(x)) = \log_{3}(\log_{2}(x)) \).

Step 4: Differentiate \( u(x) = \log_{2}(x) \)

The derivative of \( \log_{2}(x) \) with respect to \( x \) is calculated using the change of base formula: \( \frac{1}{x \ln(2)} \).

Step 5: Calculate First Derivative

Combine the results: \( F'(x) = \log_{3}(\log_{2}(x)) \cdot \frac{1}{x \ln(2)} \). This is the first derivative of \( F(x) \).

Step 6: Prepare for Second Derivative

To find the second derivative \( F''(x) \), apply the product rule to \( F'(x) = \log_{3}(\log_{2}(x)) \cdot \frac{1}{x \ln(2)} \). Let \( g(x) = \log_{3}(\log_{2}(x)) \) and \( h(x) = \frac{1}{x \ln(2)} \). The product rule states \( (gh)' = g'h + gh' \).

Step 7: Differentiate \( g(x) \)

To find \( g'(x) = (\log_{3}(\log_{2}(x)))' \), use chain rule: \( \frac{d}{dx} \log_{3}(v) = \frac{1}{v \ln(3)} \cdot v' \), where \( v = \log_{2}(x) \) and thus \( v' = \frac{1}{x \ln(2)} \). Therefore, \( g'(x) = \frac{1}{\log_{2}(x) \ln(3)} \cdot \frac{1}{x \ln(2)} \).

Step 8: Differentiate \( h(x) \)

\( h(x) = \frac{1}{x \ln(2)} \) differentiates to \( h'(x) = -\frac{1}{x^2 \ln(2)} \) using the power rule \( \left(\frac{1}{x}\right)' = -\frac{1}{x^2} \).

Step 9: Calculate Second Derivative

Apply the product rule: \( F''(x) = g'(x)h(x) + g(x)h'(x) \). This becomes: \[ F''(x) = \frac{1}{\log_{2}(x) \ln(3)} \cdot \frac{1}{x \ln(2)} \cdot \frac{1}{x \ln(2)} + \log_{3}(\log_{2}(x)) \cdot \left( -\frac{1}{x^2 \ln(2)} \right) \].

Step 10: Simplify Second Derivative

Simplify \( F''(x) \) to express it in a more compact form: \[ F''(x) = \frac{1}{x^2 \ln^2(2) \cdot \log_{2}(x) \ln(3)} - \frac{\log_{3}(\log_{2}(x))}{x^2 \ln(2)} \]. This represents the second derivative of \( F(x) \).

Key Concepts

Chain Rule

The chain rule is a crucial tool in calculus to differentiate composite functions. When a function is composed of multiple inner functions, the chain rule helps in finding their derivatives. In this exercise, the chain rule is essential for differentiating functions like \( \log_3(\log_2(x)) \). Here is how it works:

• First, identify the inner and outer functions. In this case, \( \log_2(x) \) is the inner function and \( \log_3 \) is the outer function.
• The chain rule states: if \( y = u(v(x)) \), then the derivative \( \frac{dy}{dx} = \frac{du}{dv}\cdot\frac{dv}{dx} \).
• Apply it by differentiating the outer function and multiplying it by the derivative of the inner function.

So, for \( \log_3(\log_2(x)) \), you first differentiate \( \log_3 \) with respect to its argument, and then multiply by the derivative of \( \log_2(x) \).
Understanding the chain rule simplifies the complex process of differentiating nested functions and is invaluable for tackling problems involving compositions like these.

Product Rule

The product rule is designed for differentiating products of two functions. In basic terms, if you have two functions \( g(x) \) and \( h(x) \) multiplied together, the product rule helps find their derivative. The rule is defined as follows:

• For functions \( g(x) \) and \( h(x) \), the derivative is \( (gh)' = g'h + gh' \).
• This means you differentiate the first function, multiply it by the second unchanged, and add it to the derivative of the second function multiplied by the first unchanged.
• In this exercise, it was applied to the first derivative \( F'(x) = \log_3(\log_2(x)) \cdot \frac{1}{x\ln(2)} \).

By using the product rule, you can differentiate each component separately and sum up these individual parts. Understanding this rule can greatly simplify the calculation of derivatives when functions are multiplied together as seen in this problem.

Derivative Calculation

Derivative calculation is a foundational concept in calculus that determines the rate of change of a function with respect to a variable. In this exercise, calculating both the first and second derivatives required the application of both the chain and product rules together with fundamental derivative techniques.

• The first derivative of \( F(x) \) was found using the fundamental theorem of calculus, turning the integral into a composite function problem that requires the chain rule.
• For the second derivative, the problem involved applying the product rule to the already differentiated parts.
• The calculations showed how derivatives compound upon one another — requiring earlier steps' derivatives to be incorporated into the next.

Mastering derivative calculations involves recognizing which rules to use and when. This exercise demonstrates these skills in practice by seamlessly integrating the chain and product rules to differentiate complex compositions.

Logarithmic Functions

Logarithmic functions, represented as \( \log_b(x) \), are functions inversely related to exponential functions and play a crucial role in various branches of mathematics. In this exercise, understanding how to differentiate logarithmic expressions is key.

• The derivative of \( \log_2(x) \) involves using the change of base formula: \( \frac{1}{x \ln(2)} \).
• For operations involving \( \log_3(x) \), similar principles apply: transforming it where necessary into natural logs for simplicity.
• In derivative calculations involving logarithms, remember to manage the base with natural logarithms for clarity.

Logarithms convert multiplicative processes into additive ones, simplifying the derivation process of more complex expressions.
Recognizing patterns and applying derivatives to these functions is necessary to solve integrals and differentiations involving varied bases, as illustrated in this problem.