Problem 18 - Calculus - AP Edition

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: Differentiate numerator and denominator, then apply formula.

Step 1: Identify Parts of the Quotient

The function given is a quotient: \( f(x) = \frac{u(x)}{v(x)} \), where \( u(x) = (x^2 + 1) \cot x \) and \( v(x) = 3 - \cos x \csc x \). We will use the quotient rule to differentiate. The quotient rule states: \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

Step 2: Differentiate the Numerator \(u(x)\)

First, find the derivative of \( u(x) = (x^2 + 1) \cot x \). Use the product rule: \((uv)' = u'v + uv'\). Here, \( u = x^2 + 1 \) and \( v = \cot x \).

Step 3: Differentiate \((x^2 + 1)\)

The derivative of \( x^2 + 1 \) is \( 2x \).

Step 4: Differentiate \( \cot x \)

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

Step 5: Apply the Product Rule to \(u(x)\)

Using the results from the previous steps, the derivative \( u'(x) = (x^2 + 1)(-\csc^2 x) + 2x \cot x \).

Step 6: Differentiate the Denominator \(v(x)\)

Differentiate \( v(x) = 3 - \cos x \csc x \). Use the product rule for \( u = \cos x \) and \( v = \csc x \).

Step 7: Differentiate \(\cos x\) and \(\csc x\)

\( \cos x \)'s derivative is \( -\sin x \), and \( \csc x \) derivative is \( -\csc x \cot x \).

Step 8: Apply the Product Rule to \(v(x)\)

The derivative \( v'(x) = -[\sin x \csc x + \cos x (-\csc x \cot x)] = -\sin x \csc x + \cos x \csc x \cot x \).

Step 9: Substitute into Quotient Rule Formula

Substitute the results: \( f'(x) = \frac{[(x^2 + 1)(-\csc^2 x) + 2x \cot x](3 - \cos x \csc x) - (x^2 + 1) \cot x [-\sin x \csc x + \cos x \csc x \cot x]}{(3 - \cos x \csc x)^2} \).

Step 10: Simplify the Expression (Optional)

Simplify the expression for \( f'(x) \) if needed, by expanding and combining like terms. This step can involve significant algebraic manipulation.

Key Concepts

Quotient Rule

The quotient rule is an essential technique in calculus for differentiating functions that are expressed as a ratio of two other functions. It's particularly useful when the function you're working with is given as a fraction, where both the numerator and the denominator are themselves functions of the variable.

The quotient rule is formally given by the formula:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

Here, \( u(x) \) is the numerator function, \( v(x) \) is the denominator function, \( u'(x) \) is the derivative of the numerator, and \( v'(x) \) is the derivative of the denominator.

To apply the quotient rule:

• First, differentiate the numerator to find \( u'(x) \).
• Next, differentiate the denominator to get \( v'(x) \).
• Then, apply the formula by plugging in these derivatives and the original functions.

This rule helps in finding the derivative of complex functions written as a quotient in a structured way, preventing mistakes common with more manual approaches.

Product Rule

The product rule is another fundamental tool in calculus, used to differentiate functions that are the product of two other functions. It is crucial when dealing with polynomials, trigonometric functions, or any scenario where functions are multiplied.

The product rule states:

• \( (uv)' = u'v + uv' \)

Here, \( u \) and \( v \) are functions of \( x \). Specifically:

• \( u' \) is the derivative of \( u \).
• \( v' \) is the derivative of \( v \).
• The derivative of the product \( uv \) is found by adding \( u'v \) and \( uv' \).

When applying the product rule, you start by differentiating one of the functions while keeping the other constant and then switch roles.

For example, if you have a function \( u(x) = (x^2 + 1) \cot x \):

• Differentiating \( x^2 + 1 \) gives \( 2x \).
• Differentiating \( \cot x \) gives \( -\csc^2 x \).
• Combining these gives \( u'(x) = (x^2 + 1)(-\csc^2 x) + 2x \cot x \).

The product rule simplifies the process of finding derivatives in cases of multiplied functions, eliminating confusion and error.

Trigonometric Functions

Trigonometric functions, such as \( \sin x, \cos x, \tan x, \cot x, \sec x, \csc x \), are often involved in calculus problems, particularly those dealing with mechanical and periodic motion analyses.

Differentiation of trigonometric functions requires understanding their specific derivatives:

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)
• \( \frac{d}{dx} \tan x = \sec^2 x \)
• \( \frac{d}{dx} \cot x = -\csc^2 x \)
• \( \frac{d}{dx} \sec x = \sec x \tan x \)
• \( \frac{d}{dx} \csc x = -\csc x \cot x \)

Understanding these derivatives is crucial when applying the product and quotient rules, as seen in functions combining these trigonometric expressions.

For example, when differentiating a function like \( \cot x \), knowing the derivative is \( -\csc^2 x \) speeds up the process. Similarly, applying these rules to \( \cos x \csc x \) involves using both product rules and trigonometric derivatives for accuracy.

Remember, trigonometric functions and their properties form a substantial part of calculus, making their derivatives indispensable in calculus problem-solving.