Problem 69 - Precalculus with Limits

Question

The rate of change of the function \( f(x) = \sin x + \csc x \) with respect to change in the variable \( x \) is given by the expression \( \cos x - \csc x \cot x \). Show that the expression for the rate of change can also be \( -\cos x \cot^2 x \)

Step-by-Step Solution

Answer

The derivative of the function \( f(x) = \sin x + \csc x \) i.e., \( \cos x - \csc x \cot x \) can indeed be equivalently expressed as \( -\cos x \cot^2 x \)

Step 1: Differentiate the given function

Taking derivative of the given function \( f(x) = \sin{x} + \csc{x} \) using knowledge of differentiation of trigonometric functions.\n\n\( f'(x) = \cos{x} - \csc{x} \cot{x} \)

Step 2: Express \( \csc{x} \) and \( \cot{x} \) in terms of sine and cosine

Simplify the function by rewriting \( \csc{x} \) and \( \cot{x} \) in terms of sine and cosine. By doing that, we get:\n\n\( f'(x) = \cos{x} - \frac{1}{\sin{x}}\frac{\cos{x}}{\sin{x}} \)

Step 3: Simplify the expression

Carry out the division to simplify the expression more:\n\n\( f'(x) = \cos{x} - \frac{\cos{x}}{\sin^2{x}} \)

Step 4: Express \( \sin^2{x} \) using trigonometric relation

The expression for the derivative can be rewritten as:\n\n\( f'(x) = \cos{x} - \cos{x}\cot^2{x} \)

Step 5: Verification

Finally, reverting \( \cot^2{x} \) back to its original form results in the second form of the derivative:\n\n\( f'(x) = \cos{x} - \cos{x} \cot^2{x} \), which shows that the initial form of the derivative \( \cos x - \csc x \cot x \) and \( -\cos x \cot^2 x \) are equivalent.

Key Concepts

Trigonometric Functions in Differentiation

Trigonometric functions include sine (\(\sin\) ), cosine (\(\cos\) ), tangent (\(\tan\) ), cosecant (\(\csc\) ), secant (\(\sec\) ), and cotangent (\(\cot\) ). Each of these functions plays a crucial role in mathematics, especially in calculus.

When differentiating trigonometric functions, we apply rules that showcase these functions' periodic and oscillatory nature. For instance:

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

In the given exercise, we differentiate \(\sin x\) and \(\csc x\) . This results in specific rules being applied, such as chain and quotient rules. Understanding these derivative rules helps simplify complex expressions involving trigonometric functions. Differentiation transforms functions, allowing us to analyze how they change, which is particularly useful for finding the rate of change of a quantity represented by the function.

Understanding Rate of Change with Trigonometric Functions

The rate of change of a function reflects how a dependent variable, such as \(y\) , changes in relation to an independent variable, \(x\) . In calculus, this is commonly known as the derivative. It provides us insight into the behavior of the function, such as when it increases or decreases.

In trigonometric functions, the rate of change can oscillate between positive and negative values due to their periodic nature. For example, if \(f(x) = \sin x\) , as the input \(x\) varies, \(\cos x\) reflects how steeply the graph of \(\sin x\) slopes.

• When \(\cos x > 0\), the graph of\(\sin x\) is rising.
• When \(\cos x < 0\), the graph of\(\sin x\) is falling.

The exercise showcases how calculating the derivative reveals different expressions for the rate of change. Here, the rate of change is shown in two forms:\(\cos x - \csc x \cot x\) and \(-\cos x \cot^2 x\). These forms appear different, yet they result from applying trigonometric identities and differentiation techniques. This kind of knowledge is essential for solving complex calculus problems and understanding the behavior of functions under various transformations.

Derivatives in Trigonometric Functions

Derivatives are a cornerstone concept in calculus, representing an instant rate of change and the slope of a function at a particular point. When applied to trigonometric functions, derivatives help us understand their behavior over intervals and at specific points.

By differentiating trigonometric functions, we obtain mathematical expressions that describe how quickly these functions change as their inputs change. In the example provided, the function \(f(x) = \sin x + \csc x\) was differentiated:

• The derivative \(f'(x)\) is \(\cos x - \csc x \cot x\) which presents our initial understanding of the rate of change.
• Rewriting in the form \(-\cos x \cot^2 x\) enhances our comprehension of the relationship between these trigonometric components.

Differentiation requires the use of rules such as the product rule, quotient rule, and chain rule, especially when dealing with combined trigonometric functions. Recognizing these derivative transformations enables us to express complex relationships in simpler, more interpretable forms. Mastering these helps in predicting and analyzing function behaviors over their domain, aiding in both theoretical and practical applications.