Problem 6 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

First derivative: \( y' = x^2 + x - e^{-x} \); Second derivative: \( y'' = 2x + 1 + e^{-x} \).

Step 1: Differentiate the Polynomial Terms

Given function: \( y = \frac{x^3}{3} + \frac{x^2}{2} + e^{-x} \). Differentiate the first term \( \frac{x^3}{3} \). Apply the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \):\( \frac{d}{dx}(\frac{x^3}{3}) = \frac{1}{3} \cdot 3x^{2} = x^{2} \).Differentiate the second term \( \frac{x^2}{2} \):\( \frac{d}{dx}(\frac{x^2}{2}) = \frac{1}{2} \cdot 2x^{1} = x \).

Step 2: Differentiate the Exponential Term

Differentiate the last term \( e^{-x} \): Apply the chain rule. The derivative of \( e^{-x} \) is \(-e^{-x}\) because the derivative of \(-x\) with respect to \(x\) is \(-1\).Thus, the derivative is \(-e^{-x} \).

Step 3: Sum the First Derivatives

Combine the derivatives found in steps 1 and 2:\( y' = x^{2} + x - e^{-x} \).

Step 4: Differentiate the First Derivative (First Term)

Differentiate the first term of the derivative \( y' = x^{2} + x - e^{-x} \):\( \frac{d}{dx}(x^2) = 2x \).

Step 5: Differentiate the First Derivative (Second Term)

Differentiate the second term \( x \):\( \frac{d}{dx}(x) = 1 \).

Step 6: Differentiate the First Derivative (Third Term)

Differentiate the third term \( -e^{-x} \):\( \frac{d}{dx}(-e^{-x}) = -(-1)e^{-x} = e^{-x} \).

Step 7: Sum the Second Derivatives

Combine the second derivatives found in steps 4, 5, and 6:\( y'' = 2x + 1 + e^{-x} \).

Key Concepts

The Power Rule

The power rule is one of the simplest but most important rules in calculus for finding derivatives. It comes in handy when you need to differentiate polynomial expressions. The general formula for the power rule is:

• If you have a term of the form \( x^n \), where \( n \) is a constant, the derivative is \( nx^{n-1} \).

To see it in action, let's look at the term \( \frac{x^3}{3} \). Using the power rule, we get:

• \( \frac{d}{dx}\left(\frac{x^3}{3}\right) = \frac{1}{3} \cdot 3x^{2} = x^{2} \).

The power rule was also applied to the term \( \frac{x^2}{2} \):

• \( \frac{d}{dx}\left(\frac{x^2}{2}\right) = \frac{1}{2} \cdot 2x^{1} = x \).

When you apply the power rule, always remember to multiply by the coefficient of the original term and reduce the exponent by one. This simple rule is incredibly powerful for swiftly cracking through polynomial derivatives.

The Chain Rule

The chain rule is a critical tool for derivatives involving composite functions. When a function is nested within another function, the chain rule helps us out. It tells us to differentiate the outer function and then multiply by the derivative of the inner function. Here's the formula:

• If you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).

Let's use the chain rule on the term \( e^{-x} \):

• Treat the outer function as \( e^u \) and the inner function as \( u = -x \).
• The derivative of the outer function \( e^u \) is \( e^u \), and we multiply it by the derivative of \( u \), which is \(-1\).

Therefore, the derivative of \( e^{-x} \) is

• \( e^{-x} \times (-1) = -e^{-x} \).

This step reconciles both the complexities of exponential functions and those of composition, simplifying intricate calculations.

Exponential Functions

Exponential functions are a special class of functions where the variable appears in the exponent, such as \( e^x \). Here's what's particularly appealing about their derivatives:

• The derivative of \( e^x \) is simply \( e^x \) again, because the rate of change of exponential growth is proportional to the function itself.
• For functions like \( e^{-x} \), the derivative involves the chain rule but follows a similar process.

For instance, our example term \( e^{-x} \) requires multiplying by the derivative of the exponent:

• First, differentiate \( e^{-x} \) using the rule for exponential functions.
• Apply the chain rule since the exponent is a function of \( x \): the derivative of \(-x\) is \(-1\).
• This gives us \( -e^{-x} \).

Therefore, mastering derivatives of exponential functions allows us to handle growth models and decay processes efficiently and accurately, making them incredibly useful across many fields.