Problem 207 - Calculus Volume 1

Question

Use the quotient rule to derive the given equations. $$ \frac{d}{d x}(\csc x)=-\csc x \cot x $$

Step-by-Step Solution

Answer

The derivative of \( \csc x \) is \(-\csc x \cot x\) using the quotient rule.

Step 1: Recall the Quotient Rule Formula

The quotient rule is used for differentiating functions that are the ratio of two differentiable functions. If we have a function \( f(x) = \frac{u(x)}{v(x)} \), its derivative \( f'(x) \) using the quotient rule is given by:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

Step 2: Rewrite the Function as a Quotient

Recall that the cosecant function \( \csc x \) is the reciprocal of the sine function, \( \csc x = \frac{1}{\sin x} \). Here, \( u(x) = 1 \) and \( v(x) = \sin x \), so \( \csc x = \frac{1}{\sin x} \).

Step 3: Identify Derivatives of Components

Find \( u' \) and \( v' \). Since \( u = 1 \), its derivative \( u' = 0 \). For \( v = \sin x \), its derivative \( v' = \cos x \).

Step 4: Apply the Quotient Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula:\[\frac{d}{dx}\left(\frac{1}{\sin x}\right) = \frac{\sin x \cdot 0 - 1 \cdot \cos x}{(\sin x)^2}\]Simplifying, we get:\[\frac{-\cos x}{\sin^2 x} = \frac{-1}{\sin x} \times \frac{\cos x}{\sin x} = -\csc x \cot x\]

Step 5: Verify the Result

From the quotient rule application and simplification, we derived \(-\csc x \cot x\) which matches the given formula. Therefore, the differentiation of \( \csc x \) is correct: \[ \frac{d}{dx}(\csc x) = -\csc x \cot x \].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. When we differentiate a function, we are essentially determining its derivative.

A derivative tells us the slope of the tangent line to the curve of the function at a point. This slope represents the function's rate of change.

• For a simple linear function, the derivative is constant because the rate of change is the same everywhere.
• For more complex functions, like polynomials and trigonometric functions, the derivative gives a function that tells us how the rate of change varies at every point.

In the context of our original exercise, we used the quotient rule to differentiate the cosecant function. This is because the cosecant function can be represented as a ratio, or quotient, of two simpler functions.

The quotient rule is a specific technique in differentiation used when dealing with two functions divided by each other. With a grasp of differentiation, understanding how functions behave and change becomes much more accessible.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and cosecant, play a crucial role in various areas of mathematics and applied sciences. They relate angles to the ratios of sides in right triangles and appear in wave patterns and oscillations.

Trigonometric functions have unique properties and are periodic, meaning they repeat their values in regular intervals.

• The sine function, denoted as \( \sin x \), is a basic trigonometric function often involved with its reciprocal, the cosecant.
• The reciprocal of sine is the cosecant, written as \( \csc x = \frac{1}{\sin x} \). This reflects how trigonometric functions can be expressed in terms of one another.

In differentiation, knowing the derivatives of trigonometric functions is essential. For example, the derivative of \( \sin x \) is \( \cos x \). This knowledge is vital for applying rules like the quotient rule, as seen in our exercise.

Understanding trigonometric functions and their properties helps in analyzing waves, circular motions, and other periodic phenomena.

Calculus

Calculus is the branch of mathematics that deals with continuous change, and it is divided mainly into two areas: differential calculus and integral calculus. Here, our focus is on differential calculus, which involves the concept of differentiation, as used in the original exercise.

Differential calculus allows us to create mathematical models that simulate real-world situations where change occurs.

• Through the process of differentiation, we can determine how functions change, enabling us to solve problems involving motion, growth, and decay.
• The other main branch, integral calculus, is about accumulation and finds applications where totals are needed (e.g., area under a curve).

Calculus provides tools to solve complex problems that algebra and geometry alone cannot solve. In calculus, techniques such as the quotient rule help differentiate complex functions like the cosecant function in our problem.

Mastery of calculus equips learners with a robust framework to approach a wide variety of problems, from physics to engineering and beyond.