Problem 18 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\frac{\left(x^{2}+1\right) \cot x}{3-\cos x \csc x} $$

Step-by-Step Solution

Answer

Use the quotient rule; solution involves differentiating and applying the quotient formula.

Step 1: Identify the differentiation rule

This is a quotient function, meaning it has a numerator and a denominator. To find the derivative, we must use the quotient rule: \[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\] where \(u\) is the numerator and \(v\) is the denominator.

Step 2: Define the numerator and denominator

From the function \(f(x)=\frac{(x^2+1)\cot x}{3-\cos x \csc x}\), the numerator \(u\) is \((x^2+1)\cot x\), and the denominator \(v\) is \(3-\cos x \csc x\).

Step 3: Differentiate the numerator

The numerator \(u=(x^2+1)\cot x\). To find \(u'\), use the product rule:\[(uv)' = u'v + uv'\] where \(u_1 = x^2+1\) and \(v_1 = \cot x\).- Derivative of \(u_1\): \(2x\)- Derivative of \(v_1\): \(-\csc^2 x\)Applying the product rule: \[u' = (2x)\cot x + (x^2+1)(-\csc^2 x)\]

Step 4: Differentiate the denominator

The denominator is \(v=3-\cos x \csc x\). Use the product rule for \(-\cos x \csc x\) which involves \(u_2 = \cos x\) and \(v_2 = \csc x\).- Derivative of \(u_2\): \(-\sin x\)- Derivative of \(v_2\): \(-\csc x \cot x\)The derivative of the product is:\[v' = 0 - (\sin x)\csc x - (\cos x)(-\csc x \cot x)\] Simplifying yields: \[v' = -1 + \csc x \tan x\]

Step 5: Apply the Quotient Rule

Combine the results using the quotient rule: \[f'(x) = \frac{u'v - uv'}{v^2}\]Substitute \(u'\), \(v\) and \(v'\):- \(u' = 2x \cot x - (x^2+1)\csc^2 x\)- \(v = 3 - \cos x \csc x\)- \(v' = -1 + \csc x \tan x\)\[f'(x) = \frac{[(2x \cot x - (x^2+1) \csc^2 x)(3 - \cos x \csc x)] - [(x^2+1) \cot x (-1 + \csc x \tan x)]}{(3 - \cos x \csc x)^2} \]

Step 6: Simplification is optional

For the purpose of differentiation, the derivative is complete by the quotient rule application. Simplifying the expression is optional and might not yield a significantly simpler form.

Key Concepts

Quotient Rule

When you are tasked with finding the derivative of a function that consists of a numerator and a denominator, you can use the Quotient Rule. This is especially helpful when dealing with rational functions where one part is divided by another. The formula for the Quotient Rule is \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] which tells us how to differentiate a function of the form \ \( \frac{u}{v} \ \).

• **Differentiate** both the numerator \ \( u \ \) and the denominator \ \( v \ \) separately to get \ \( u' \ \) and \ \( v' \ \).
• **Substitute** these into the Quotient Rule formula.
• **Simplify** if necessary.

The use of the Quotient Rule is crucial when a direct or simple differentiation approach isn't viable, especially when there are complex expressions involved in the numerator and the denominator.

Product Rule

Differentiating a function where two parts are multiplied together involves the Product Rule. This rule is essential for breaking down the derivative of a product of two functions. The formula for the Product Rule is: \[ (uv)' = u'v + uv' \] where \ \( u \ \) and \ \( v \ \) are functions of \ \( x \ \).Using the Product Rule:

• **Identify** the two parts of the product function, \ \( u \ \) and \ \( v \ \).
• **Find** the derivatives of both, \ \( u' \ \) and \ \( v' \ \).
• **Apply** the formula \ \( (uv)' = u'v + uv' \ \).
• **Combine** and **simplify** the result if needed.

For example, let's say \ \( u = x^2+1 \ \) and \ \( v = \cot x \ \). Their derivatives are \ \( u' = 2x \ \) and \ \( v' = -\csc^2 x \ \). So using the Product Rule results in the derivative \[ (2x)\cot x + (x^2+1)(-\csc^2 x). \] This step is a critical part of breaking down complicated derivative problems.

Derivative of Trigonometric Functions

Trigonometric functions like \ \( \sin x, \cos x, \tan x \ \), and their derivatives are fundamental in calculus. Knowing the derivatives of these functions is crucial.Here are the basic derivatives:

• \ \( (\sin x)' = \cos x \ \)
• \ \( (\cos x)' = -\sin x \ \)
• \ \( (\tan x)' = \sec^2 x \ \)
• \ \( (\cot x)' = -\csc^2 x \ \)
• \ \( (\sec x)' = \sec x \tan x \ \)
• \ \( (\csc x)' = -\csc x \cot x \ \)

To differentiate a trigonometric function, apply these rules where needed. For example, when differentiating \ \( \cot x \ \) in a product, use \ \( (\cot x)' = -\csc^2 x \ \) during calculations. Understanding these derivatives aids in handling complex calculus problems that involve trigonometric expressions.