Problem 157 - Calculus Volume 1

Question

In the following exercises, use the Fundamental Theorem of Calculus, Part \(1,\) to find each derivative. $$ \frac{d}{d x} \int_{1}^{x^{2}} \frac{\sqrt{t}}{1+t} d t $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2x^2}{1+x^2} \).

Step 1: Identify the Function and Limits

The problem requires us to find the derivative of an integral \( \frac{d}{dx} \int_{1}^{x^2} \frac{\sqrt{t}}{1+t} \, dt \). The upper limit is \( x^2 \), which is a function of \( x \), and the lower limit is a constant, \( 1 \).

Step 2: Apply the Fundamental Theorem of Calculus, Part 1

The theorem states that if \( F(x) = \int_{a}^{g(x)} f(t) \, dt \), then the derivative \( \frac{d}{dx}F(x) = f(g(x)) \cdot g'(x) \). In this case, \( f(t) = \frac{\sqrt{t}}{1+t} \), \( g(x) = x^2 \), and \( g'(x) = 2x \).

Step 3: Substitute and Differentiate

Substitute \( g(x) = x^2 \) into \( f(t) \) to get \( f(g(x)) = \frac{\sqrt{x^2}}{1+x^2} = \frac{x}{1+x^2} \) since \( \sqrt{x^2} = x \) for \( x \geq 0 \). Then, multiply by \( g'(x) = 2x \).

Step 4: Calculate the Derivative

The derivative is \( f(g(x)) \cdot g'(x) = \frac{x}{1+x^2} \cdot 2x = \frac{2x^2}{1+x^2} \).

Key Concepts

Derivative of Integral

The Fundamental Theorem of Calculus is a powerful tool that connects differentiation and integration. Part 1 of this theorem helps us find the derivative of an integral with a variable as one of its limits. In our exercise, the task is to compute the derivative of the integral \(\int_{1}^{x^2} \frac{\sqrt{t}}{1+t} \, dt\). Here are some key steps:

• The integral function is considered over the interval from 1 to \(x^2\), with \(x^2\) as the upper limit.
• The goal is to differentiate this integral with respect to \(x\).

According to the theorem, when we have an integral of the form \(F(x) = \int_{a}^{g(x)} f(t) \, dt\), the derivative is given by \(\frac{d}{dx}F(x) = f(g(x)) \cdot g'(x)\). Notice how the functions \(f(t)\) and \(g(x)\) depict the integral’s inner function and upper limit respectively. We apply this directly to find the derivative.

Upper Limit as Function

In many calculus problems, the upper limit of the integral is not simply \(x\) or a constant, but rather a function of \(x\). This is exactly the case in our problem where the upper limit is \(x^2\). Here’s why this is important:

• A function as an upper limit implies the use of the chain rule during differentiation.
• To manage the variable upper limit, we need to consider how it affects our differentiation steps.

Quickly substitute \(g(x) = x^2\) into the integrand and find that \(\frac{\sqrt{x^2}}{1+x^2} = \frac{x}{1+x^2}\). This expression is important as it gets multiplied by the derivative of the upper limit function \(x^2\), which is \(2x\). This multiplication arises from the chain rule.

Chain Rule for Differentiation

The chain rule is a fundamental principle used in calculus to differentiate composite functions. It becomes particularly handy when dealing with functions that have other functions nested within. In relation to our task:

• The derivative is \(\frac{d}{dx}F(x) = f(g(x)) \cdot g'(x)\).
• It involves differentiating the outer function and multiplying by the derivative of the inner function, \(g(x)\).

In our problem, after substituting \(g(x) = x^2\) into \(f(t) = \frac{\sqrt{t}}{1+t}\), it is crucial to also differentiate \(g(x)\), yielding \(g'(x) = 2x\). Therefore, our chain rule application results in the derivative \(\frac{2x^2}{1+x^2}\). This showcases how integral tasks in calculus often require the interplay of different rules to arrive at a solution.