Problem 67 - Calculus Early Transcendentals

Question

Simplify the following expressions. $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} d t$$

Step-by-Step Solution

Answer

Question: Find the simplified form of the expression $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} dt$$. Answer: The simplified form of the expression is $$2\sqrt{1+x^2}$$.

Step 1: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if F(x) is an antiderivative of the function f(x) on the interval [a, b], then $$\int_a^b f(x) dx = F(b) - F(a)$$ Here, we have $$F(x) = \int \sqrt{1+t^2} dt$$ We are given the expression $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} dt = \frac{d}{d x} (F(x) - F(-x))$$

Step 2: Apply the chain rule for differentiation

We will apply the chain rule for differentiation, which says that $$\frac{d}{d x} (F(x) - F(-x)) = F'(x) - (-1) \cdot F'(-x)$$ In other words, $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} dt = F'(x) + F'(-x)$$ Now, recall that F'(x) is equal to the integrand function, which in our case, is $$F'(x) = \sqrt{1+x^2}$$ So, we can substitute this back into the expression: $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} dt = \sqrt{1+x^2} + \sqrt{1+(-x)^2}$$

Step 3: Simplify the expression

Since both terms under the square root have the same value inside, we can write $$\sqrt{1+x^2} + \sqrt{1+(-x)^2} = \sqrt{1+x^2} + \sqrt{1+x^2} = 2\sqrt{1+x^2}$$ So the simplified expression is: $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} dt = 2\sqrt{1+x^2}$$

Key Concepts

Antiderivative

An antiderivative of a function is essentially the reverse of taking a derivative. If you have a function \( f(x) \), its antiderivative \( F(x) \) is a new function whose derivative becomes \( f(x) \). In simpler terms, when you differentiate \( F(x) \), you return to \( f(x) \).

For instance, if \( F(x) = x^2 \), the antiderivative of \( f(x) = 2x \), then differentiating \( F(x) \) gives us \( f(x) \). This concept helps lay the foundation for integrating functions over specific intervals, which can be seen in the Fundamental Theorem of Calculus.

• Used for determining areas under curves
• Critical in solving differential equations

Chain Rule

The chain rule is a method for finding the derivative of a composite function. A composite function is formed when one function is applied inside another. Think of it like peeling an onion, where each layer represents a different function.

Mathematically, if you have a composite function \( g(f(x)) \), the chain rule states:
\[ \frac{d}{dx} g(f(x)) = g'(f(x)) \cdot f'(x) \]

This rule is indispensable when differentiating expressions where one part depends on another. In our exercise, we applied the chain rule to differentiate \( F(x) - F(-x) \) effectively by recognizing the dependency between the functions. Understanding this concept helps with examining how variables are influenced by changes in other variables, especially in complex calculations.

• Helps in dealing with nested functions
• Crucial for optimizing equations in calculus

Derivative

Derivatives express how a function changes as its input changes. They give the rate of change or the slope of the function at any point.

For example, if \( f(x) = x^2 \), its derivative \( f'(x) = 2x \) tells us the slope of the curve at any point \( x \). In simpler terms, it reveals how steep the curve is at every point.

Derivatives are fundamental in calculus because they provide critical insights:

• How fast something is changing at a given moment
• Identifying maximum and minimum values of functions
• Solving optimization problems

They are the building blocks for many concepts and are extensively used across different fields such as physics, engineering, and economics. By understanding how derivatives work, students can analyze a wide range of problems involving rates of change and slopes of functions.