Problem 12 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\csc x \cot x $$

Step-by-Step Solution

Answer

The derivative is \(-\csc x (\csc^2 x + \cot^2 x)\).

Step 1: Identify the Functions

Given the function, we have \( f(x) = \csc x \cot x \). We identify two trigonometric functions involved: \( \csc x \) and \( \cot x \).

Step 2: Recall Derivatives of Trigonometric Functions

We need the derivatives of \( \csc x \) and \( \cot x \). Recall:\[ \frac{d}{dx} (\csc x) = -\csc x \cot x \]\[ \frac{d}{dx} (\cot x) = -\csc^2 x \]

Step 3: Apply the Product Rule

The product rule states that \( \,\frac{d}{dx}(u \cdot v) = u\frac{dv}{dx} + v\frac{du}{dx}\). Let \( u = \csc x \) and \( v = \cot x \). Then, \[ \frac{d}{dx}(\csc x \cdot \cot x) = \csc x \frac{d}{dx}(\cot x) + \cot x \frac{d}{dx}(\csc x) \].

Step 4: Differentiate Each Part Using Known Derivatives

Differentiate each function:\[ \csc x \frac{d}{dx}(\cot x) = \csc x (-\csc^2 x) = -\csc^3 x \]\[ \cot x \frac{d}{dx}(\csc x) = \cot x (-\csc x \cot x) = -\csc x \cot^2 x \].

Step 5: Combine the Results

Combine the results from Step 4 into the product rule formula:\[ \frac{d}{dx}(\csc x \cot x) = -\csc^3 x - \csc x \cot^2 x \].

Step 6: Simplify the Expression

Factor out \( -\csc x \) from both terms:\[ f^{\prime}(x) = -\csc x (\csc^2 x + \cot^2 x) \].

Key Concepts

Trigonometric Functions

Trigonometric functions are vital in calculus, representing relationships between angles and lengths in right triangles. In the function given, we are working with two specific trigonometric functions: cosecant (\( \csc x \)) and cotangent (\( \cot x \)).

• The cosecant function is the reciprocal of the sine function: \( \csc x = \frac{1}{\sin x} \).
• The cotangent function is the reciprocal of the tangent function: \( \cot x = \frac{\cos x}{\sin x} \).

It is important to memorize these relationships and the unit circle, as these functions are used frequently in calculus and other areas of mathematics. With these functions, we can determine sides of triangles and solve for unknowns in various types of equations.

Derivatives

Derivatives represent the rate at which a function is changing at any given point. They are a cornerstone of calculus and are used to find slopes of tangent lines, rates of change, and optimize problems involving trigonometric functions.When computing derivatives of trigonometric functions, it’s useful to remember these rules:

• The derivative of \( \csc x \) is \(-\csc x \cot x\) which captures how the cosecant function changes in relation to changes in \( x \).
• The derivative of \( \cot x \) is \(-\csc^2 x \). This tells us how the cotangent function decreases as \( x \) increases.

Knowing these derivatives allows us to differentiate more complex expressions involving these functions, like in our exercise.

Product Rule

The product rule is a useful tool in differentiation when dealing with the product of two functions. It states that if you have two functions, \( u \) and \( v \), their derivative is given by \( u \frac{dv}{dx} + v \frac{du}{dx} \).In our given function, \( f(x) = \csc x \cot x \), \( u = \csc x \) and \( v = \cot x \). The product rule applies because the original function is the product of these two separate trigonometric functions.Through the product rule:

• Differentiate \( \csc x \) to get \(-\csc x \cot x \).
• Differentiate \( \cot x \) to get \(-\csc^2 x \).
• Substitute back into the product rule formula: \( \csc x(-\csc^2 x) + \cot x(-\csc x \cot x) \).

This calculation requires handling each function separately before combining them, which is the essence of using the product rule effectively.