Problem 98 - Calculus and its applications

Question

Differentiate. $$ f(x)=\frac{x e^{-x}}{1+x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{e^{-x}(1-x+x^2-x^3)}{(1+x^2)^2} \).

Step 1: Identify the Derivative Rule Needed

The function \( f(x) = \frac{x e^{-x}}{1+x^2} \) is a quotient of two functions. To differentiate it, we need to use the quotient rule, which is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x e^{-x} \) and \( v = 1+x^2 \).

Step 2: Differentiate the Numerator

Identify \( u = x e^{-x} \) and use the product rule to differentiate it, which is \( (uv)' = u'v + uv' \). For \( u \):- \( u_1' = x' = 1 \)- \( u_2 = e^{-x} \)- \( (x e^{-x})' = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x} - xe^{-x} \). Therefore, \( u' = e^{-x} - xe^{-x} \).

Step 3: Differentiate the Denominator

The denominator \( v = 1 + x^2 \) is a simple polynomial. Differentiate it:- \( (1 + x^2)' = 0 + 2x = 2x \).Therefore, \( v' = 2x \).

Step 4: Apply the Quotient Rule

Now apply the quotient rule using the derivatives calculated:\[ \left( \frac{x e^{-x}}{1+x^2} \right)' = \frac{(e^{-x} - xe^{-x})(1+x^2) - (x e^{-x})(2x)}{(1+x^2)^2} \]

Step 5: Simplify the Expression

Expand the terms in the numerator:- \((e^{-x} - xe^{-x})(1 + x^2) = e^{-x} + e^{-x}x^2 - xe^{-x} - x^3e^{-x} \)- \( (x e^{-x})(2x) = 2x^2 e^{-x} \)Combine them to get:\[ e^{-x} + e^{-x}x^2 - xe^{-x} - x^3e^{-x} - 2x^2 e^{-x} \]Simplify it:\[ e^{-x} - xe^{-x} - x^3e^{-x} + x^2 e^{-x} \]Thus, the simplified derivative is:\[ \frac{e^{-x} - xe^{-x} + x^2 e^{-x} - x^3 e^{-x}}{(1+x^2)^2} = \frac{e^{-x}(1-x+x^2-x^3)}{(1+x^2)^2}\]

Step 6: Final Answer

The derivative of \( f(x) = \frac{x e^{-x}}{1+x^2} \) is: \[ f'(x) = \frac{e^{-x}(1-x+x^2-x^3)}{(1+x^2)^2} \]

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental method in calculus for finding the derivative of a function defined as a quotient of two simpler functions. In our exercise, the function is given as \( f(x) = \frac{x e^{-x}}{1 + x^2} \). This function can be thought of as \( \frac{u}{v} \), where \( u \) is the numerator \( x e^{-x} \) and \( v \) is the denominator \( 1 + x^2 \).
To differentiate a function of this form, the quotient rule provides a systematic approach:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
Here's how it works:

• Start by finding the derivative of the numerator \( u' \).
• Next, find the derivative of the denominator \( v' \).
• Substitute these derivatives into the formula above to compute the overall derivative.
• Remember that the denominator in the quotient is squared, helping maintain the function's original structure.

This rule greatly simplifies handling complex fractions like the one in our problem, ensuring you can differentiate them accurately.

Product Rule

The Product Rule is used when you need to differentiate a product of two functions. This is exactly what we encounter when dealing with the numerator \( u = x e^{-x} \) in the exercise.
The Product Rule states:
\[ (uv)' = u'v + uv' \]
In this scenario, think of:

• \( u_1 = x \) and its derivative \( u_1' = 1 \)
• \( u_2 = e^{-x} \) and its derivative employs the exponential function derivative, \( u_2' = -e^{-x} \)

Using the Product Rule, you combine these derivatives:
\[ (x e^{-x})' = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x} - xe^{-x} \]
The rule is a lifesaver for handling derivatives of products, especially when they incorporate indicators that exemplify exponential behavior like \( e^{-x} \). This makes the math behind such problems more straightforward.

Exponential Function

The exponential function \( e^x \) is unique and highly significant in calculus due to its constant rate of growth. Differentiating exponential functions is remarkably straightforward—they remain basically unchanged under differentiation.
For an exponential function like \( e^{-x} \), use the chain rule to differentiate it:

• The derivative of \( e^{-x} \) is \( -e^{-x} \). This is because \( e^x \)'s derivative is the function itself, and the negative sign arises from the chain rule handling the inner function \( -x \).

The exponential function plays a crucial role in our exercise since it appears in both the numerator and when executing the Product Rule.
Understanding its behavior helps significantly when simplifying the expressions in calculus and provides insight into the rapid yet smooth way these functions change. Remember, the distinct nature of the exponential function ensures it pops up frequently in both theoretical and practical applications of calculus.