Problem 34 - Thomas Calculus

Question

Find all possible functions with the given derivative. \begin{equation}\quad \text { a. }y^{\prime}=\frac{1}{2 \sqrt{x}} \quad \text { b. } y^{\prime}=\frac{1}{\sqrt{x}} \quad \text { c. } y^{\prime}=4 x-\frac{1}{\sqrt{x}}\end{equation}

Step-by-Step Solution

Answer

The functions are: (a) \( y = \sqrt{x} + C \), (b) \( y = 2\sqrt{x} + C \), and (c) \( y = 2x^2 - 2\sqrt{x} + C \), with \( C \) being an integration constant.

Step 1: Integrate the First Derivative (a)

Given the derivative \( y' = \frac{1}{2\sqrt{x}} \), we need to find the original function \( y \) by integrating with respect to \( x \). The integral is:\[y = \int \frac{1}{2\sqrt{x}} \, dx = \int \frac{1}{2} x^{-1/2} \, dx.\]The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \). Therefore, the result is:\[y = \frac{1}{2} \cdot 2x^{1/2} + C = \sqrt{x} + C.\]

Step 2: Verify the Result (a)

Differentiate \( y = \sqrt{x} + C \) to confirm the solution. Since the derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \) and the constant \( C \) vanishes, the result matches the original derivative. This verifies that \( y = \sqrt{x} + C \) is correct.

Step 3: Integrate the First Derivative (b)

For \( y' = \frac{1}{\sqrt{x}} \), integrate with respect to \( x \) to find \( y \):\[y = \int \frac{1}{\sqrt{x}} \, dx = \int x^{-1/2} \, dx.\]The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \). Therefore, the integral is:\[y = 2x^{1/2} + C.\]

Step 4: Verify the Result (b)

Differentiate \( y = 2\sqrt{x} + C \) to confirm it matches the derivative \( y' = \frac{1}{\sqrt{x}} \). The derivative of \( 2\sqrt{x} \) is \( \frac{1}{\sqrt{x}} \), and the derivative of \( C \) is zero. This confirms the correctness of \( y = 2\sqrt{x} + C \).

Step 5: Integrate the First Derivative (c)

For \( y' = 4x - \frac{1}{\sqrt{x}} \), integrate each term separately. Integrate the first term:\[\int 4x \, dx = 2x^2\] and the second term:\[\int -\frac{1}{\sqrt{x}} \, dx = -2x^{1/2}.\]Combine these with the constant of integration \( C \), resulting in:\[y = 2x^2 - 2\sqrt{x} + C.\]

Step 6: Verify the Result (c)

Differentiate \( y = 2x^2 - 2\sqrt{x} + C \). The derivative of \( 2x^2 \) is \( 4x \) and the derivative of \( -2\sqrt{x} \) is \( -\frac{1}{\sqrt{x}} \). Since the derivative of \( C \) is zero, this confirms that our solution is correct.

Key Concepts

Integration Techniques

To tackle calculus problems involving differentiation and integration, knowing various integration techniques is crucial. Integration, essentially the reverse operation of differentiation, is used to find the original function from its derivative, known as the antiderivative. A common method to integrate a function is to use substitution, which simplifies complex integrals.
Another technique is partial fraction decomposition, which is useful for rational functions. You break the fraction into simpler parts to integrate easily. The integration by parts method is similar to the product rule in differentiation and helps when you need to integrate products of functions.

• Substitution: Simplifies integration by changing variables.
• Partial Fractions: Useful for integrating rational expressions.
• By Parts: Applies to products of functions.

Even for simple functions like those in our exercise, being familiar with these strategies allows you to effectively solve a wide range of problems.

Antiderivatives

An antiderivative of a function is another function whose derivative equals the original function. When integrating a function, we are essentially looking for its antiderivative.
Every function has infinitely many antiderivatives, differing by a constant, because the derivative of a constant is zero. This leads to the general solution that includes a constant of integration, often denoted by \( C \). For example, if you integrate \( y' = \frac{1}{\sqrt{x}} \), you find the antiderivative \( y = 2\sqrt{x} + C \). Here are some key points about antiderivatives:

• Every antiderivative is a solution of a differential equation.
• The constant \( C \) represents all possible vertical shifts.
• Knowing basic derivatives helps in determining antiderivatives.

Recognizing common derivatives can quicken the process of finding their antiderivatives, making problem-solving smoother.

Calculus Problem Solving

Solving calculus problems like finding indefinite integrals requires a clear strategy. Begin by analyzing what is given, such as finding a function whose derivative you already know. Much like our exercise, validate your solution by differentiating the result to check if it matches the original derivative given.
Steps to approach any calculus problem include:

• Understand the Problem: Identify what is being asked.
• Plan a Strategy: Choose the best integration technique.
• Solve: Execute your plan carefully and find the antiderivative.
• Verify: Differentiate your solution to ensure accuracy.

Using this approach ensures a systematic and efficient process to tackle a variety of problems, reinforcing your understanding and application of calculus concepts.