Problem 5 - University Calculus: Early Transcendentals

Question

Find the first and second derivatives. $$y=\frac{4 x^{3}}{3}-x+2 e^{x}$$

Step-by-Step Solution

Answer

The first derivative is \( y' = 4x^2 - 1 + 2e^x \), and the second derivative is \( y'' = 8x + 2e^x \).

Step 1: Differentiate the First Term

The first term of the function is \( \frac{4x^3}{3} \). Use the power rule for differentiation, which states \( \frac{d}{dx} x^n = nx^{n-1} \). Therefore, the derivative of \( \frac{4x^3}{3} \) is \( 4x^2 \).

Step 2: Differentiate the Second Term

The second term is \( -x \). The derivative of \( x \) with respect to \( x \) is \( 1 \). Thus, the derivative of \( -x \) is \( -1 \).

Step 3: Differentiate the Third Term

The third term is \( 2e^x \). The derivative of \( e^x \) is \( e^x \). Therefore, the derivative of \( 2e^x \) is \( 2e^x \) itself.

Step 4: Combine First Derivative Results

Combine the derivatives from Steps 1, 2, and 3 to find the first derivative of \( y \). So, \( y' = 4x^2 - 1 + 2e^x \).

Step 5: Differentiate the First Derivative

To find the second derivative, differentiate each term of the first derivative \( y' = 4x^2 - 1 + 2e^x \). Start with the first term: the derivative of \( 4x^2 \) is \( 8x \) (using the power rule).

Step 6: Differentiate Constant and Exponential Terms

The derivative of the constant \(-1\) is \(0\). The derivative of the exponential term \(2e^x\) is \(2e^x\).

Step 7: Combine Second Derivative Results

Combine the results from Step 5 and Step 6 to find the second derivative of \( y \). Thus, \( y'' = 8x + 2e^x \).

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus, particularly when it comes to differentiation. Simply put, differentiation measures how a function changes as its input changes, and the power rule is a tool that helps us find this change for powers of x. For any function in the form of \( x^n \), the power rule states that the derivative is \( nx^{n-1} \). This rule makes it quick and easy to differentiate functions that are polynomials.

• The power rule applies to any term where x is raised to a constant power.
• In our example, \( \frac{4x^3}{3} \) becomes \( 4x^2 \).
• The rule simplifies complex polynomial derivatives into quicker solutions.

Remember, the power rule only works if we can express the function in the \( x^n \) form. If you have constants or coefficients multiplied by these terms, like \( \frac{4}{3} \) in the example, you differentiate by separately applying the power rule just to the \( x^3 \) part.

Exponential Functions

Exponential functions, such as \( e^x \), are crucial in calculus due to their unique properties. These functions appear frequently across different problems, especially those modeling growth and decay processes.One unique feature of exponential functions is their derivatives closely resemble the original function. For \( e^x \), the derivative is \( e^x \) itself. This makes them very straightforward to deal with when differentiating.

• They are especially important in situations involving rates of change and natural growth/decay processes.
• Exponential functions are characterized by their constant relative rates of change.

In our exercise, the term \( 2e^x \) remains unchanged after differentiation, reinforcing the simple yet powerful nature of exponential growth functions. Differentiating these functions is a neat opportunity to appreciate their mathematical elegance.

Calculus Problem-Solving

Solving calculus problems often involves a systematic approach that breaks down into manageable steps. Each step serves as a building block for reaching the ultimate solution, just as we saw in the differentiation exercise provided. When tackling a calculus problem:

• Identify and separate each term in the equation.
• Apply calculus rules, such as the power rule or rules for derivatives of constants and exponentials, to each term individually.
• Combine the results for an overall solution.

These steps help break down complex problems methodically. It becomes crucial especially when transitioning from simple functions to composite ones or when dealing with second derivatives. In our problem, after finding the first derivative by applying the appropriate rules step-by-step, we further differentiated to find the second derivative. This structured approach is essential in calculus as it helps in solving both simple and complex functions more transparently and efficiently.