Problem 95 - Calculus and its applications

Question

Differentiate. $$ y=\frac{e^{x}}{x^{2}+1} $$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2} \).

Step 1: Recognize the Rule to Apply

The function \( y=\frac{e^{x}}{x^{2}+1} \) is a quotient of two functions. To differentiate this function, we'll apply the Quotient Rule.

Step 2: State the Quotient Rule

The Quotient Rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u}{v} \) is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]In this case, let \( u = e^x \) and \( v = x^2 + 1 \).

Step 3: Differentiate the Numerator

Differentiate the numerator function \( u(x) = e^x \). The derivative of \( e^x \) is \( e^x \), so \( u' = e^x \).

Step 4: Differentiate the Denominator

Differentiate the denominator function \( v(x) = x^2 + 1 \). The derivative of \( x^2 \) is \( 2x \) and the derivative of constant 1 is 0, so \( v' = 2x \).

Step 5: Plug Values into the Quotient Rule

Using the Quotient Rule: \[ y' = \frac{(e^x)(x^2 + 1) - (e^x)(2x)}{(x^2 + 1)^2} \]

Step 6: Simplify the Expression

Simplify the expression obtained:\[ (e^x)(x^2 + 1) - (e^x)(2x) = e^x(x^2 + 1 - 2x) = e^x(x^2 - 2x + 1) \]So, \[ y' = \frac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2} \]

Step 7: Write the Final Derivative

The derivative of the function \( y = \frac{e^x}{x^2 + 1} \) is:\[ y' = \frac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2} \]

Key Concepts

Quotient Rule

The Quotient Rule is a method for differentiating a function represented as the division of two other functions. When you are dealing with a function like \( y = \frac{e^x}{x^2 + 1} \), where one function is divided by another, you employ the Quotient Rule. This rule is essential because it allows us to find the derivative of the whole expression efficiently.The Quotient Rule formula is given by:

• If \( u(x) \) is the numerator and \( v(x) \) is the denominator, then the derivative of the division is:

\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]This formula tells us to:

• Multiply the derivative of the numerator by the denominator.
• Subtract the product of the numerator and the derivative of the denominator.
• All of this is then divided by the square of the denominator.

All these steps help us focus on finding the exact slope of the function at any given point.

Exponential Functions

Exponential functions, such as \( e^x \), are unique mathematical expressions where the variable appears in the exponent. They are highly significant in calculus, especially due to their natural growth and decay processes in real-world scenarios. For example, \( e^x \) is the base of natural logarithms and has a distinct property: its derivative is itself (\( e^x \)).This property simplifies differentiation: when you see \( e^x \) in a function like the one in the original exercise, you know:

• The derivative of \( e^x \) is simply \( e^x \).
• No complex changes are required to determine \( u' \) when you apply the Quotient Rule.

Through this straightforward rule, exponential functions become easier to handle within differentiation and integration processes.

Derivatives of Rational Functions

Rational functions are expressions where a polynomial is divided by another polynomial. Deriving these functions can sometimes seem complex, but rules like the Quotient Rule streamline the process. In our exercise, the function is divided into two parts: \( e^x \) as the numerator and \( x^2 + 1 \) as the denominator.To differentiate rational functions effectively:

• Identify each part of the function, labeling them as numerator \( u(x) \) and denominator \( v(x) \).
• Calculate the derivative of each part individually. Here, \( u' = e^x \) and \( v' = 2x \).
• Apply the Quotient Rule to combine these derivatives into the neat formula provided:

\[ y' = \frac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2} \]By breaking down these steps, calculating the derivative of a rational function becomes manageable and methodical. This process emphasizes the importance of organizing each component separately before combining them into the final derivative form.