Problem 52 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=\frac{e^{-x} \cos ^{2} x}{x^{2}+x+1} $$

Step-by-Step Solution

Answer

Use the quotient rule to differentiate the function.

Step 1: Identify the components of the function

The function is given as \( y = \frac{e^{-x} \cos^{2} x}{x^2 + x + 1} \). This function is a division of two parts: the numerator \( e^{-x} \cos^{2} x \) and the denominator \( x^2 + x + 1 \).

Step 2: Apply the quotient rule for differentiation

The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative is \( y' = \frac{u'v - uv'}{v^2} \). We identify \( u = e^{-x} \cos^{2} x \) and \( v = x^2 + x + 1 \).

Step 3: Differentiate the numerator (u)

To differentiate \( u = e^{-x} \cos^{2} x \), apply the product rule because it is a product of \( e^{-x} \) and \( \cos^{2} x \). Use:- The derivative of \( e^{-x} \) is \( -e^{-x} \).- The derivative of \( \cos^{2} x \) is \( 2 \cos x (-\sin x) = -2 \cos x \sin x \).Thus, \( u' = (-e^{-x}) \cos^{2} x + e^{-x} (-2 \cos x \sin x) \).

Step 4: Differentiate the denominator (v)

Differentiate \( v = x^2 + x + 1 \) to get \( v' = 2x + 1 \).

Step 5: Substitute into the quotient rule

Now substitute \( u' \), \( u \), \( v' \), and \( v \) into the quotient rule:\[ y' = \frac{((-e^{-x}) \cos^{2} x + e^{-x} (-2 \cos x \sin x))(x^2 + x + 1) - (e^{-x} \cos^{2} x)(2x + 1)}{(x^2 + x + 1)^2} \].

Step 6: Simplify the expression

Simplify the derivative expression by expanding and combining like terms in the numerator. This might involve careful algebraic manipulation, and organizing terms to make the expression more manageable. However, simplify as much as needed depending on the specific task requirements, but leave it in readable form.

Key Concepts

Quotient Rule

The Quotient Rule is a method of differentiating functions that are defined as a quotient of two differentiable functions. If you have a function expressed as a ratio \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions, the quotient rule states that the derivative, \( y' \), is given by: \[ y' = \frac{u'v - uv'}{v^2}\] To apply the quotient rule, it's essential to:

• Identify the numerator (\( u \)) and the denominator (\( v \)).
• Differentiate both \( u \) and \( v \) separately to get \( u' \) and \( v' \).
• Substitute these into the formula.

In our exercise, \( u = e^{-x} \cos^{2} x \) and \( v = x^2 + x + 1 \). By using the quotient rule, we manage to find the derivative of the given complex function. It is a powerful technique especially when dealing with rational functions, making the process systematic and structured.

Product Rule

The Product Rule is a key tool when differentiating a product of two functions. When faced with a function \( u \cdot v \), where both \( u \) and \( v \) are differentiable, the derivative \( (uv)' \) is given by: \[ (uv)' = u'v + uv'\] Applying the product rule requires:

• Identifying each function in the product.
• Taking the derivative of each function separately.
• Substituting these into the product rule formula.

In our problem, within the numerator of the quotient \( e^{-x} \cos^{2} x \), we applied the product rule because it is expressed as the product of \( e^{-x} \) and \( \cos^{2} x \). Differentiating \( e^{-x} \) gives \( -e^{-x} \), while \( \cos^{2} x \) requires the chain rule leading to \( -2 \cos x \sin x \). Combining these results helps simplify and compute the necessary parts for the quotient rule.

Logarithmic Differentiation

Logarithmic Differentiation is a technique that can simplify the differentiation process, especially when dealing with complex products or powers of functions. It involves taking the natural logarithm of both sides of the function before differentiating. This method capitalizes on the property of logarithms that turns multiplicative functions into additive ones: - When you have a function expressed like \( y = f(x) \cdot g(x) \), by taking the log, it becomes \( \ln y = \ln f(x) + \ln g(x) \). Logarithmic differentiation steps include:

• Taking \( \ln \) of both sides of the equation.
• Using log properties to simplify.
• Differentiating implicitly with respect to \( x \).
• Solving for \( y' \).

In the exercise, though a straightforward application wasn't used, knowing the function's complexity could suggest using logarithmic differentiation. This isn't always necessary with standard rules like the quotient or product rule but is critical for functions expressed as products or powers with complicated forms. It broadens the toolkit, offering versatile approaches for various types of functions.