Problem 6 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\frac{\sin x}{x^{2}+\sin x} $$

Step-by-Step Solution

Answer

\(f^{\prime}(x) = \frac{x^2 \cos x - 2x \sin x}{(x^2 + \sin x)^2}\)

Step 1: Identify the Rule

The function to differentiate is a quotient of two functions, so we will use the quotient rule: if \( u(x) \) and \( v(x) \) are functions, then the derivative of \( \frac{u}{v} \) is given by \( \left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime} v - uv^{\prime}}{v^2} \).

Step 2: Define Functions and Derivatives

Let \( u(x) = \sin x \) and \( v(x) = x^2 + \sin x \). Then, the derivatives are \( u^{\prime}(x) = \cos x \) and \( v^{\prime}(x) = 2x + \cos x \).

Step 3: Apply the Quotient Rule

Substitute into the quotient rule formula:\[ f^{\prime}(x) = \frac{(\cos x)(x^2 + \sin x) - (\sin x)(2x + \cos x)}{(x^2 + \sin x)^2} \].

Step 4: Simplify the Derivative

Simplify the expression by distributing:\[ f^{\prime}(x) = \frac{\cos x \cdot x^2 + \cos x \cdot \sin x - 2x \cdot \sin x - \sin x \cdot \cos x}{(x^2 + \sin x)^2} \]Cancel out the \( \cos x \cdot \sin x \) terms:\[ f^{\prime}(x) = \frac{\cos x \cdot x^2 - 2x \cdot \sin x}{(x^2 + \sin x)^2} \]

Key Concepts

Quotient Rule

The quotient rule is a formula used in calculus for finding the derivative of a function that is the quotient, or division, of two other functions. When we have a function defined as \( f(x) = \frac{u(x)}{v(x)} \), we can use the quotient rule to find its derivative. The formula is:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]

This formula emphasizes the need to differentiate both the numerator \( u(x) \) and the denominator \( v(x) \). Here's a simple way to remember it:

• Differentiating the top function \( u(x) \), you get \( u'(x) \).
• Multipy it by the bottom function \( v(x) \).
• Subtract the result of the bottom's derivative \( v'(x) \) times the top function.
• Finally, divide everything by the square of \( v(x) \).

This method allows us to handle functions that don't divide cleanly, providing a valuable tool in differentiation.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are basic elements of calculus because they exhibit unique and continuous behavior. The sine function, \( \sin x \), oscillates between -1 and 1 and is periodic with a period of \( 2\pi \). Its derivative, \( \cos x \), reflects the rate of change of \( \sin x \), also periodic with the same period and oscillation range.

Understanding these properties is essential when applying rules like the quotient rule. In our problem, \( u(x) = \sin x \), so the derivative is simply \( u'(x) = \cos x \). Similarly, for mixed functions like \( x^2 + \sin x \), you combine derivatives:

• The derivative of \( x^2 \) is \( 2x \).
• When combined with \( \sin x \), the derivative \( v'(x) \), is \( 2x + \cos x \).

Recognizing how these trigonometric derivatives interact with power functions is key to effective calculus problem-solving.

Derivative Simplification

Once we apply the quotient rule to find the derivative, simplifying this derivative is often necessary. Simplification involves breaking down the expression to its most straightforward form to make it easier to understand and use. Initially, our derivative is given by a large fraction with several terms in the numerator:\[ f'(x) = \frac{\cos x \cdot x^2 + \cos x \cdot \sin x - 2x \cdot \sin x - \sin x \cdot \cos x}{(x^2 + \sin x)^2} \]

By distributing and combining like terms, we simplify the numerator. Specifically, terms \( \cos x \cdot \sin x - \sin x \cdot \cos x \) cancel out because they are equal and opposite. This cancellation leaves us with the simplified form:\[ f'(x) = \frac{\cos x \cdot x^2 - 2x \cdot \sin x}{(x^2 + \sin x)^2} \]
In calculus, simplifying derivatives is crucial for understanding their behavior and making further computations easier. Clear and simplified expressions also help in identifying any potential critical points or points of interest in a function's graph.