Problem 64 - Calculus for Biology and Medicine

Question

Use the quotient rule to show that $$ \frac{d}{d x} \csc x=-\csc x \cot x $$

Step-by-Step Solution

Answer

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

Step 1: Recall the Quotient Rule Formula

The quotient rule states that if you have a function that is the quotient of two other functions, say \( f(x) = \frac{g(x)}{h(x)} \), then its derivative is given by:\[\frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}\]

Step 2: Express Cosecant in Terms of Sine

Recall that \( \csc x = \frac{1}{\sin x} \). Let \( g(x) = 1 \) and \( h(x) = \sin x \). Thus, \( \csc x = \frac{g(x)}{h(x)} \).

Step 3: Differentiate Numerator and Denominator

Differentiate \( g(x) = 1 \):\[ g'(x) = 0 \]Now differentiate \( h(x) = \sin x \):\[ h'(x) = \cos x \]

Step 4: Apply the Quotient Rule

Using the quotient rule formula:\[\frac{d}{dx}\left(\frac{1}{\sin x}\right) = \frac{0 \cdot \sin x - 1 \cdot \cos x}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x}\]

Step 5: Simplify the Expression

Recognize that \( \sin x \cdot \csc x = 1 \) and \( \sin x \cdot \cot x = \cos x \). Therefore, the expression can be rewritten as:\[\frac{-\cos x}{\sin^2 x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\cot x \cdot \csc x\]

Step 6: Conclusion

Thus, we have shown that:\[\frac{d}{d x} \csc x = -\csc x \cot x\] This completes the proof using the quotient rule.

Key Concepts

Quotient Rule

The quotient rule is a powerful differentiation technique used for finding the derivative of a function that is the ratio of two other functions. If you have a function expressed as \[ f(x) = \frac{g(x)}{h(x)} \]then the quotient rule states that the derivative can be found using the formula:\[ \frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]This formula helps to break down the differentiation process by considering each part of the quotient separately. Notice that:

• \( g'(x) \) is the derivative of the numerator.
• \( h(x) \) is the original denominator.
• \( g(x) \) is the original numerator.
• \( h'(x) \) is the derivative of the denominator.

Using this method ensures accuracy in differentiating more complex functions. It's important to apply the rule carefully, paying attention to both the change in the numerator and the denominator. Make sure to simplify the results for clarity.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and cosecant, are fundamental in calculus, particularly when it comes to differentiation. The cosecant function, denoted as \( \csc x \), is the reciprocal of the sine function. It is defined as:\[ \csc x = \frac{1}{\sin x} \]Understanding how to differentiate trigonometric functions involves knowing their basic identities and interrelationships:

• Sine: \( \sin x \)
• Cosine: \( \cos x \)
• Cotangent: \( \cot x = \frac{\cos x}{\sin x} \)

The differentiation of these functions often requires recognizing and using these identities. As seen in the problem, knowing that \( \csc x = \frac{1}{\sin x} \) allows us to apply the quotient rule. For students, a key insight is to remember these reciprocal relationships, which can be extremely helpful in simplifying the differentiation process.

Differentiation Techniques

Differentiation is one of the core processes in calculus, allowing us to find the rate at which a function is changing. Various techniques are available for differentiating different types of functions, with the quotient rule being just one example. Here are some other common methods:

• Product Rule: Used for differentiating products of two functions.
• Chain Rule: Useful for functions composed of several layers, such as \( f(g(x)) \).
• Basic Differentiation: Applies to simple polynomial terms, like \( x^n \), where the derivative is \( nx^{n-1} \).

These techniques are like tools in a toolbox, and knowing when to use each one is key to solving complex calculus problems efficiently. In the given exercise, the quotient rule was the most suitable tool due to the division involved in the trigonometric function \( \csc x \). By mastering these differentiation techniques, solving calculus problems becomes more straightforward, making it easier to tackle a range of mathematical challenges.