Problem 13 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle h(x) = \int^{e^x}_1 \ln t \,dt \)

Step-by-Step Solution

Answer

The derivative is \( e^x \cdot x \).

Step 1: Recall Part 1 of the Fundamental Theorem of Calculus

Part 1 of the Fundamental Theorem of Calculus states that if \( F(x) \) is defined as \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). This principle is useful for finding derivatives of integrals with variable limits of integration.

Step 2: Identify the Function and the Limits

The function given is \( h(x) = \int_{1}^{e^x} \ln t \, dt \). The integral is with respect to \( t \), with a lower limit of 1 and an upper limit of \( e^x \).

Step 3: Apply the Chain Rule

Since the upper limit of the integral is \( e^x \) and not simply \( x \), we'll need to apply the chain rule. We recognize \( h(x) \) as a composite function.

Step 4: Differentiate Using the Chain Rule

First, consider \( u = e^x \), then \( h(x) = \int_{1}^{u} \ln t \, dt \). By the Fundamental Theorem of Calculus, \( \frac{d}{du} \int_{1}^{u} \ln t \, dt = \ln u \). To find \( \frac{dh}{dx} \), we multiply by \( \frac{du}{dx} \).

Step 5: Calculate the Derivative

The derivative of \( u = e^x \) with respect to \( x \) is \( e^x \). Hence, \( \frac{dh}{dx} = \ln(e^x) \cdot e^x \).

Step 6: Simplify the Expression

Since \( \ln(e^x) = x \), the expression simplifies to \( x \cdot e^x \). Therefore, the derivative is \( e^x \cdot x \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to compute derivatives of composite functions. A composite function is essentially a function inside another function, like having layers of an onion. The chain rule enables us to differentiate these layers effectively.

In mathematical terms, if you have a composite function denoted as \( f(g(x)) \), the chain rule helps find its derivative as \( f'(g(x)) \cdot g'(x) \).

• First, you differentiate the outer function \( f \) with respect to its input \( g(x) \).
• Then, you multiply this by the derivative of the inner function \( g \) with respect to \( x \).

In our exercise, we used this rule because the integral's upper limit, \( e^x \), depended on \( x \). This made \( h(x) = \int_{1}^{e^x} \ln t \, dt \) a composite function. We set \( u = e^x \) making it necessary to apply the chain rule, leading us to multiply by \( e^x \), the derivative of \( u \).

Derivative of Integrals

Understanding how to differentiate integrals is crucial, especially when using the Fundamental Theorem of Calculus. The theorem connects differentiation and integration, showing how they are inverse processes.

Part 1 of the Fundamental Theorem of Calculus states that if you have a function \( F(x) = \int_{a}^{x} f(t) \, dt \), then its derivative is \( F'(x) = f(x) \). This tells us that the derivative of the integral from \( a \) to \( x \) of a function \( f \) is the function itself evaluated at \( x \).

• In cases where the limits of integration vary with \( x \), adjustments, such as the chain rule, might be necessary.
• This principle simplifies such integral differentiation tasks significantly.

In the given problem, the complexity arose because the upper limit of the integral was not just \( x \), but \( e^x \). By recognizing this, and with help from the chain rule, we differentiate properly and find the result.

Composite Functions

Composite functions arise when the output of one function becomes the input for another. They are everywhere in calculus and are often written in the form \( f(g(x)) \).

When dealing with composite functions, it's essential to understand both individual functions involved.

• You start by identifying the inner function \( g(x) \).
• Then, take note of the outer function \( f \), which encompasses \( g(x) \).

Inside the exercise, the problem turned the integral into a composite function due to the upper limit being \( e^x \). This required us to apply the chain rule for derivation. Recognizing the nature of composite functions can significantly simplify calculus challenges, as it allows leveraging rules appropriately to solve complex derivatives like integrating functions with variable limits.