Problem 13 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\frac{\cot x}{1+\csc x} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Given Function

We are given the function \( f(x) = \frac{\cot x}{1 + \csc x} \). To differentiate it, we will need to apply the quotient rule because it is a fraction.

Step 2: Recall the Quotient Rule

The quotient rule for differentiation is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = \cot x \) and \( v = 1 + \csc x \). We will find their derivatives \( u' \) and \( v' \).

Step 3: Differentiate \( u = \cot x \)

Find \( u' \):\( \cot x = \frac{\cos x}{\sin x} \), and its derivative is \( \frac{d}{dx}\cot x = -\csc^2 x \). Thus, \( u' = -\csc^2 x \).

Step 4: Differentiate \( v = 1 + \csc x \)

Find \( v' \):\( \csc x = \frac{1}{\sin x} \), and its derivative is \( \frac{d}{dx}\csc x = -\csc x \cot x \). So, \( v' = -\csc x \cot x \).

Step 5: Apply the Quotient Rule

Substitute \( u, v, u', v' \) into the quotient rule formula:\[ \frac{d}{dx} \left( \frac{\cot x}{1 + \csc x} \right) = \frac{-\csc^2 x (1 + \csc x) - (\cot x)(-\csc x \cot x)}{(1 + \csc x)^2} \]

Step 6: Simplify the Expression

Start by expanding the terms in the numerator:\[ -\csc^2 x (1 + \csc x) + \cot x \cdot \csc x \cdot \cot x \]Simplify:\[ = -\csc^2 x - \csc^3 x + \csc x \cot^2 x \]The derivative is:\[ f'(x) = \frac{-\csc^2 x - \csc^3 x + \csc x \cot^2 x}{(1 + \csc x)^2} \]

Step 7: Final Answer

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

Key Concepts

Quotient Rule

When you need to differentiate a function that is a fraction, like \( \frac{\cot x}{1+\csc x} \), the quotient rule comes in handy. This method simplifies the process of finding derivatives of ratios of two functions.
The quotient rule formula is:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)

Here, \( u \) is the numerator and \( v \) is the denominator. To find the derivative using this rule, follow these steps:

• Differentiate the top function \( u \) to get \( u' \)
• Differentiate the bottom function \( v \) to get \( v' \)
• Substitute \( u \), \( v \), \( u' \), and \( v' \) into the formula above
• Simplify the result

Exercise an extra degree of care in simplifying, as this step can get slightly tedious. The quotient rule is particularly useful when both the numerator and denominator are themselves functions, not just numbers.

Trigonometric Functions

Trigonometric functions like \( \sin x, \cos x, \tan x, \cot x, \csc x, \) and \( \sec x \) are fundamental in calculus. These functions often show up in differentiation and integration.
In our exercise, the main trigonometric functions we deal with are \( \cot x \) (cotangent) and \( \csc x \) (cosecant). Understanding these functions helps to navigate differentiation problems more smoothly:

• \( \cot x = \frac{\cos x}{\sin x} \), and its derivative is \( -\csc^2 x \)
• \( \csc x = \frac{1}{\sin x} \), and its derivative is \( -\csc x \cot x \)

Knowing these derivatives by heart enables you to apply the quotient rule effectively when these functions are involved. Trigonometric derivatives often need careful attention to detail due to the multiple functions involved in their expressions.

Derivative Calculation

Calculating derivatives is a core function in calculus, particularly to find the rate of change of a function. Let's follow this through with our problem, using the quotient rule.
Here's how to calculate the derivative of \( f(x) = \frac{\cot x}{1 + \csc x} \):

• Identify \( u = \cot x \) and \( v = 1 + \csc x \)
• Find the derivative of the numerator, \( u \), which satisfies \( u' = -\csc^2 x \)
• Calculate the derivative of the denominator, \( v \), which leads to \( v' = -\csc x \cot x \)
• Substitute these into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{\cot x}{1 + \csc x} \right) = \frac{-\csc^2 x(1 + \csc x) + \cot x \cdot \csc x \cdot \cot x}{(1 + \csc x)^2} \]
• Simplify further by expanding and simplifying the terms in the numerator

Ensuring each step is executed correctly can clarify complex problems, and though the calculation can be lengthy, it's systematic with practice. By doing this, you can better comprehend how different calculus rules apply to finding derivatives of tricky expressions.