Problem 22 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ x \cos x \csc x $$

Step-by-Step Solution

Answer

The derivative is \( -x - (x-1)\cot x \csc x \).

Step 1: Identify the Product Rule

The function given is a product of three functions: \( f(x) = x \), \( g(x) = \cos x \), and \( h(x) = \csc x \). To find the derivative of the product \( x \cos x \csc x \), apply the product rule for three functions, which states: if \( u(x) = f(x) \cdot g(x) \cdot h(x) \), then \( u'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \).

Step 2: Find Individual Derivatives

Calculate the derivatives of each individual component:1. \( f(x) = x \) so \( f'(x) = 1 \).2. \( g(x) = \cos x \) so \( g'(x) = -\sin x \).3. \( h(x) = \csc x \) so \( h'(x) = -\csc x \cot x \).

Step 3: Apply Product Rule

Substitute into the product rule:\[ \begin{align*} u'(x) &= f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \&= (1) \cos x \csc x + x (-\sin x) \csc x + x \cos x (-\csc x \cot x) \end{align*} \]

Step 4: Simplify the Expression

Simplify each term in the derivative:1. \( \cos x \csc x = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \csc x \).2. \( -x \sin x \csc x = -x \cdot \frac{\sin x}{\sin x} = -x \).3. \( -x \cos x \csc x \cot x = -x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -x \cot x \csc x \).Combining these, the expression becomes:\[ \cot x \csc x - x - x \cot x \csc x \].

Step 5: Combine Like Terms

The terms \( \cot x \csc x \) and \( -x \cot x \csc x \) can be combined. Therefore, the derivative simplifies to:\[ -x - (x-1)\cot x \csc x \].

Key Concepts

Product Rule for Differentiation

The product rule is a useful tool in calculus to find the derivative of a product of two or more functions. It's especially helpful when you encounter functions that are multiplied together, like in this problem where we have three functions: \( x \), \( \cos x \), and \( \csc x \). The main idea is that the derivative of a product is not simply the product of their derivatives.

• The rule for two functions is: \( (fg)' = f'g + fg' \).
• For three functions, like in our problem, it extends to: \( (fgh)' = f'gh + fg'h + fgh' \).

This allows us to break down a complex derivative into simpler parts, and ensures all combinations of the functions are considered for their impact on the derivative.
To apply this rule successfully, determine the derivative for each function individually first. Then plug them into the product rule formula.

Trigonometric Functions in Calculus

Trigonometric functions like sine, cosine, tangent, cosecant, secant, and cotangent often appear in calculus problems. In this exercise, we focus on \( \cos x \) and \( \csc x \), both critical trigonometric functions used in the product we are differentiating.Each of these functions has specific derivatives:

• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \csc x \) is a bit more complex: \(-\csc x \cot x \).

Understanding these derivatives is key to applying them correctly according to the product rule. When differentiating any trigonometric function, remember to mind their signs and transformations, as they can drastically change the resultant derivative.
These functions often transform the solution into forms involving cotangent and cosecant, as seen in our original solution, emphasizing the importance of these derivatives.

Understanding Derivatives

Derivatives reflect the rate of change of a function. When applying the product rule, as in our problem, derivatives of each function are needed before substitutions are made.In our exercise:

• For \( f(x) = x \), the derivative is straightforward: \( f'(x) = 1 \), representing a constant rate.
• Trigonometric functions require more careful differentiation, utilizing their derivatives as listed in the previous section. This involves applying negative signs and complex multiplier terms, like in \( \csc x \).

A derivative helps us understand how the function behaves as \( x \) changes, capturing its dynamics graphically and mathematically.
Thus, derivatives are foundational in calculus, enabling us to study the slope of curves, motion, growth rates, and other key aspects of varying quantities.