Problem 6 - Calculus - AP Edition

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

To differentiate the given function, identify which rule to use. The function is a fraction, so we use the quotient rule. The quotient rule is: \( (\frac{u}{v})^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2} \), where \( u = \sin x \) and \( v = x^2 + \sin x \).

Step 2: Differentiate the Numerator

Differentiate the numerator \( u = \sin x \). The derivative is \( u^{\prime} = \cos x \).

Step 3: Differentiate the Denominator

Differentiate the denominator \( v = x^2 + \sin x \). The derivative is \( v^{\prime} = 2x + \cos x \).

Step 4: Apply the Quotient Rule

Use the quotient rule: \( f^{\prime}(x) = \frac{(\cos x)(x^2 + \sin x) - (\sin x)(2x + \cos x)}{(x^2 + \sin x)^2} \).

Step 5: Simplify the Expression

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

Step 6: Final Step: State Your Result

The derivative of the function \( f(x) = \frac{\sin x}{x^2 + \sin x} \) is: \[ f^{\prime}(x) = \frac{x^2 \cos x - 2x \sin x}{(x^2 + \sin x)^2} \].

Key Concepts

Understanding the Quotient Rule

The quotient rule allows us to differentiate functions that are given as a fraction, where we have one function divided by another. It's a vital tool in calculus, especially when dealing with complex fractions. Here's how it works: when you have a function of the form \( \frac{u}{v} \), its derivative \( (\frac{u}{v})^{\prime} \) is given by \( \frac{u^{\prime}v - uv^{\prime}}{v^2} \). This formula tells us to:

• Differentiate the numerator function \( u \), resulting in \( u^{\prime} \).
• Differentiate the denominator function \( v \), resulting in \( v^{\prime} \).
• Multiply the derivative of the numerator by the original denominator.
• Subtract the product of the original numerator and the derivative of the denominator.
• Divide the entire expression by the square of the denominator.

Using the quotient rule can seem complicated at first, but with practice, it becomes a straightforward, methodical process for working with fraction-based functions in calculus.

Differentiation Techniques Simplified

Differentiation is a core concept in calculus that involves finding the derivative of a function, which is essentially the rate at which the function changes. There are several techniques to help find derivatives quickly and accurately:

• Power Rule: For a function \( x^n \), the derivative is \( nx^{n-1} \).
• Product Rule: When two functions multiply, \( (uv)^{\prime} = u^{\prime}v + uv^{\prime} \).
• Chain Rule: Used for composite functions, \((f(g(x)))^{\prime} = f^{\prime}(g(x))g^{\prime}(x) \).
• Quotient Rule: Specifically for functions that are ratios, vital for functions like \( \frac{u}{v} \).

Each technique serves a purpose, and careful selections ensure accurate results. Recognizing which rule to apply is key; for example, for a fraction such as \( \frac{\sin x}{x^2 + \sin x} \), the quotient rule is applicable. With practice, these differentiation techniques become intuitive and a powerful toolset for any calculus student.

Trigonometric Functions in Calculus

Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) are common in calculus and have specific differentiation rules. Understanding these rules helps in solving more complex problems.

• Sine Function: The derivative of \( \sin x \) is \( \cos x \).
• Cosine Function: The derivative of \( \cos x \) is \( -\sin x \).
• Tangent Function: The derivative of \( \tan x \) is \( \sec^2 x \).

When working with functions like \( \frac{\sin x}{x^2 + \sin x} \), it's essential to know how to differentiate these trigonometric functions as components of the larger calculus problem. Each trigonometric function has a distinct derivative that can significantly affect the outcome of the calculations. Mastery of these basic trigonometric derivatives is crucial for successfully tackling more in-depth calculus challenges.