Problem 2 - Calculus of a Single Variable

Question

Choosing an Antiderivative In Exercises \(1-4,\) select the correct antiderivative. \(\frac{d y}{d x}=\frac{x}{x^{2}+1}\) $$ \begin{array}{ll}{\text { (a) } \ln \sqrt{x^{2}+1}+c} & {\text { (b) } \frac{2 x}{\left(x^{2}+1\right)^{2}}+c} \\ {\text { (c) arctan } x+c} & {\text { (d) } \ln \left(x^{2}+1\right)+c}\end{array} $$

Step-by-Step Solution

Answer

The correct antiderivative is option (a) \( \ln \sqrt{x^{2}+1}+c \)

Step 1: Identify the derivative of each option

Find the derivative of each option a, b, c and d because the antiderivative of a function \( f \) is a function whose derivative is \( f \). In other words, if \( F' = f \) then \( F \) is an antiderivative of \( f \).

Step 2: Derivative of option (a)

When differentiating the function option (a) \( \ln \sqrt{x^{2}+1}+c \), we get \( \frac{x}{x^{2}+1} \), which matches the given function.

Step 3: Derivative of option (b)

Differentiating the function option (b) \( \frac{2 x}{\left(x^{2}+1\right)^{2}}+c \), we do not get the given function \( \frac{x}{x^{2}+1} \), so this option is incorrect.

Step 4: Derivative of option (c)

For function option (c) arctan \( x+c \), when we differentiate, we do not get the given function \( \frac{x}{x^{2}+1} \), so this option is also incorrect.

Step 5: Derivative of option (d)

Finally, differentiating the function option (d) \( \ln \left(x^{2}+1\right)+c \), we do not get the given function \( \frac{x}{x^{2}+1} \), meaning this option too, is incorrect.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with the study of change and motion. Through its two main operations, differentiation and integration, calculus provides methods for analyzing and predicting the behavior of functions. Differentiation is the process of finding the rate at which a function is changing at any given point, while integration is essentially the reverse process, summing up small pieces to find the whole.

Calculus is widely used in science, economics, and engineering to solve problems involving change and motion. For example, in physics, calculus is used to calculate the trajectory of a moving object subject to forces. In the exercise provided, we use calculus to determine the antiderivative, a concept that arises from integration. Understanding the basics of calculus is crucial to tackle more advanced problems in mathematics and the sciences.

Integration

Integration is a core concept in calculus that refers to the process of finding the antiderivative or the integral of a function. It's the counterpart to differentiation, and you can think of it as the operation that tells us the 'accumulated quantity'—like finding the total distance traveled by an object when given its speed over time.

There are two main types of integration: definite and indefinite. Definite integration gives us the actual total quantity between two limits, while indefinite integration, as seen in the exercise, does not specify limits and results in a general form of the antiderivative plus a constant of integration, denoted as 'C'. Integration is applied in various fields, from calculating areas under curves to determining the distribution functions in probability.

Indeterminate Integration

Indeterminate integration, also known as indefinite integration, involves finding a function when its derivative is given, without any specified limits. The result is not a single function but a family of functions that differ by a constant. That's why you'll always see an '+C' at the end of an indefinite integral; this 'C' represents an unknown constant because a derivative doesn’t reveal any information about the original function’s y-intercept.

The exercise presented is a classic example of indeterminate integration. You are given a derivative and are asked to find the set of all possible antiderivatives. This is fundamental in calculus, as it allows us to reconstruct functions given their rates of change. The process requires knowledge of common derivatives and often, some algebraic manipulation.

Inverse of Differentiation

The 'inverse of differentiation' refers to the process of finding the original function from its derivative, which is exactly what the operation of integration does. This inverse process is not straightforward, as many different functions can have the same derivative. Hence, the antiderivative must be found by considering the family of functions that all differ by a constant, as differentiation strips away constants.

When solving for an antiderivative, as in the exercise, one must work backwards, applying knowledge of common derivatives to deduce the original function. This requires a solid understanding of basic differentiation rules. The correct antiderivative can often be found by considering what function, when differentiated, would yield the given derivative. In the provided problem, the function aggregates changes to produce a quantity — the antiderivative — that represents what the function might have been before differentiation took place.