Problem 46 - 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 \(f\) \(u(x)=2^{x} \quad f(t)=2^{t}\)

Step-by-Step Solution

Answer

The first derivative is \( F'(x) = 2^{(2^x)} \cdot 2^x \cdot \ln(2) \); the second is \( F''(x) = 2^{(2^x)} \cdot 2^{2x} \cdot \ln^3(2) + 2^{(2^x + x)} \cdot \ln^2(2) \)."

Step 1: Understanding the Fundamental Theorem of Calculus and Chain Rule

The function given is of the form \( F(x) = \int_{a}^{u(x)} f(t) \, dt \), where the fundamental theorem of calculus says that \( \frac{d}{dx} \int_{a}^{u(x)} f(t) dt = f(u(x))u'(x) \). We will be applying the chain rule along with this theorem to calculate the derivative of the function with respect to \( x \).

Step 2: Calculate the First Derivative

To find the first derivative of \( F(x) = \int_{a}^{2^x} 2^t \, dt \), we apply the fundamental theorem of calculus and the chain rule:\[F'(x) = f(u(x)) \cdot u'(x) = 2^{(2^x)} \cdot \frac{d}{dx}(2^x).\]We also know \( \frac{d}{dx}(2^x) = 2^x \ln(2) \). Therefore,\[ F'(x) = 2^{(2^x)} \cdot 2^x \cdot \ln(2).\]

Step 3: Setting Up the Second Derivative

The second derivative \( F''(x) \) is the derivative of \( F'(x) = 2^{(2^x)} \cdot 2^x \cdot \ln(2) \). We need to apply the product rule and the chain rule to differentiate \( F'(x) \). Let\[F'(x) = g(x) \cdot h(x),\]where \( g(x) = 2^{(2^x)} \) and \( h(x) = 2^x \ln(2) \).

Step 4: Differentiate Using Product and Chain Rules

First, find \( g'(x) \) using the chain rule:\[g'(x) = 2^{(2^x)} \cdot \ln(2) \cdot 2^x \cdot \ln(2).\]Next, differentiate \( h(x) \):\[h'(x) = 2^x \ln(2) \cdot \ln(2).\]Apply the product rule:\[F''(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) = (2^{(2^x)} \ln(2) \cdot 2^x \ln(2)\cdot 2^x \ln(2)) + (2^{(2^x)} \cdot 2^x \ln(2) \cdot \ln(2)).\]This simplifies to:\[F''(x) = 2^{(2^x)} \cdot (2^x \ln^2(2)) \cdot 2^x + 2^{(2^x)} \cdot 2^x \ln^2(2).\]

Key Concepts

Chain Rule

The Chain Rule is an essential tool in calculus, especially when dealing with compositions of functions. It helps us find the derivative of a composite function, that is, a function made up of two or more functions linked together. When you have a function \( y = f(g(x)) \), the derivative \( y' \), is found by differentiating the outer function \( f \) at \( g(x) \) and multiplying it by the derivative of \( g(x) \).
This can be expressed as:

• \( y' = f'(g(x)) \cdot g'(x) \)

In the context of the original exercise, we applied the Chain Rule when finding the first derivative of the integral function. Specifically, after applying the Fundamental Theorem of Calculus, we used the Chain Rule on \( 2^x \) to find its derivative, which is vital for computing \( F'(x) \).
The formula for the Chain Rule ensures each part is differentiated systematically, maintaining the integrity of leveraged operations in calculus.

Product Rule

The Product Rule is used when you need to differentiate two functions that are multiplied together. If you have functions \( u(x) \) and \( v(x) \), the derivative of their product \( uv \) is found using:

• \( (uv)' = u'v + uv' \)

This formula allows you to break down complex differentiations into simpler tasks, calculating each part separately before combining them. In the exercise, it was essential for finding the second derivative of the function \( F(x) \). When differentiating \( F'(x) = g(x) \cdot h(x) \), we needed the Product Rule to express \( F''(x) \) by finding \( g'(x) \) and \( h'(x) \) separately before plugging them into the formula. This rule ensures a systematic differentiation of product components.

First Derivative

The first derivative of a function provides us with the rate at which the function’s value is changing at any given point, commonly referred to as the function's slope. In calculus, this derivative tells us how fast the function is growing or declining. In the earlier exercise, to find the first derivative \( F'(x) \), we used the Fundamental Theorem of Calculus together with the Chain Rule.
We started with:

• \( F(x) = \int_{a}^{u(x)} f(t) \, dt \)
• \( u(x) = 2^x \) and \( f(t) = 2^t \)

When applying the theorem, we derived:

• \( F'(x) = f(u(x)) \cdot u'(x) = 2^{(2^x)} \cdot 2^x \ln(2) \)

Understanding the first derivative is crucial for knowing the behaviour of functions and understanding motion, trends, or any dynamic changes they exhibit.

Second Derivative

The second derivative of a function offers insights into its concavity and the acceleration of its rate of change. Essentially, it’s about how the rate itself is changing. For the exercise, calculating the second derivative \( F''(x) \) required more than just applying simple rules; it involved combining both the Chain Rule and the Product Rule.
When you have \( g(x) = 2^{(2^x)} \) and \( h(x) = 2^x \ln(2) \), the derivative of \( F'(x) \) required:

• Differentiation of each function separately
• Using results from the product of these derivatives

The result was a more complex expression:

• \( F''(x) = 2^{(2^x)} \cdot (2^x \ln^2(2)) \cdot 2^x + 2^{(2^x)} \cdot 2^x \ln^2(2) \)

Second derivatives are invaluable for understanding deeper aspects of the function’s graph, such as inflection points and the nature of its curvature, key to interpreting physical or conceptual dynamics reflected by the function.