Problem 62 - Calculus: Early Transcendentals

Question

Find the derivative of the function. \( F(x) = \displaystyle \int^{2x}_{\sqrt{x}} \arctan t \, dt \)

Step-by-Step Solution

Answer

The derivative is \( F'(x) = 2\arctan(2x) - \frac{1}{2\sqrt{x}}\arctan(\sqrt{x}) \).

Step 1: Recognize the Fundamental Theorem of Calculus

To find the derivative of a function defined as an integral, we use the Fundamental Theorem of Calculus Part 1, which states that if \( G(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), then the derivative \( G'(x) \) is \( f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \).

Step 2: Identify the Function and Limits

In the given function, \( F(x) = \int_{\sqrt{x}}^{2x} \arctan t \, dt \), identify \( f(t) = \arctan t \), the upper limit \( b(x) = 2x \), and the lower limit \( a(x) = \sqrt{x} \).

Step 3: Apply the Derivative Formula

Using the formula from Step 1, calculate the derivative: \[ F'(x) = \arctan(2x) \cdot \frac{d}{dx}(2x) - \arctan(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \]

Step 4: Differentiate the Upper and Lower Limits

Calculate the derivatives of the limits: \( \frac{d}{dx}(2x) = 2 \) and \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \).

Step 5: Substitute the Derivatives Back

Substitute back into the equation from Step 3:\[ F'(x) = \arctan(2x) \cdot 2 - \arctan(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

Step 6: Simplify the Expression

Finally, simplify the expression: \[ F'(x) = 2 \arctan(2x) - \frac{1}{2\sqrt{x}} \arctan(\sqrt{x}) \].

Key Concepts

Derivative of Integral

The concept of finding the derivative of an integral is beautifully addressed by the Fundamental Theorem of Calculus. This theorem links the concept of differentiation and integration, two core operations in calculus. When we have an integral like \( F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), where both the limits of integration might be functions of \( x \), we can find its derivative by using the following key steps:

• First, identify the function you are integrating, denoted \( f(t) \).
• Next, acknowledge the functions \( a(x) \) and \( b(x) \) that represent the lower and upper limits of the integral respectively.
• Finally, according to the Fundamental Theorem of Calculus: \( F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \).
  This formula helps us compute the derivative by multiplying the function \( f(t) \) evaluated at the limits and those limits’ derivatives.

This approach allows us to effectively take the role of both differentiation and integration across changing limits.

Upper and Lower Limits Differentiation

When dealing with integrals where the limits are not constant but rather functions of a variable like \( x \), we need to apply differentiation rules to these variable limits. In the problem given, the upper limit is \( b(x) = 2x \), and the lower limit is \( a(x) = \sqrt{x} \).

• The first task is to compute \( b'(x) \), the derivative of the upper limit, which leads to \( \frac{d}{dx}(2x) = 2 \).
• Similarly, compute \( a'(x) \), the derivative of the lower limit, to find \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \).

By computing these derivatives, we incorporate the rate at which the limits change as the variable \( x \) alters. This recognition is crucial as it considers the real-world aspect where boundaries are not static. Utilizing these values as part of the derivative formula will reinforce the understanding that integrals with variable limits require a broader view complimented by differentiation.

Arctan Function in Calculus

The arctan function, or the inverse tangent function, frequently appears in calculus, particularly in integration and differentiation contexts. In our exercise, the nested function inside the integral is \( \arctan t \), which is essential to consider while applying the derivative formula.

• Recognize that \( f(t) = \arctan(t) \) is used directly in the fundamental theorem for evaluating the integral's contribution at the changing limits.
• For example, at the upper limit \( b(x) = 2x \), we directly evaluate the function as \( \arctan(2x) \).
• Similarly, for the lower limit \( a(x) = \sqrt{x} \), we use \( \arctan(\sqrt{x}) \).

This grasp of the arctan function’s behavior is pivotal as it determines how the integral's outcome changes, reflected in the derivative. Pay special attention to these functions within your calculations, as they directly impact the integral's value and subsequent analyses.