Problem 26 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ y=\frac{1}{x} \sin ^{-5} x-\frac{x}{3} \cos ^{3} x $$

Step-by-Step Solution

Answer

The derivative of the given function is \(-\frac{\sin^{-5}{x}}{x^2} - \frac{5 \cos{x}}{x \sin^6{x}} - \frac{1}{3} \cos^{3}{x} + x \cos^{2}{x} \sin{x}\).

Step 1: Differentiate the first term

Let's find the derivative of the first term, \( \frac{1}{x} \sin^{-5}{x} \). This term can be rewritten as \( x^{-1} \sin^{-5}{x} \). We will apply the product rule here. If we let \( u = x^{-1} \) and \( v = \sin^{-5}{x} \), then the derivative is \( u'v + uv' \).1. **Differentiate \( u = x^{-1} \):** \( u' = -x^{-2} \).2. **Differentiate \( v = \sin^{-5}{x} \):** - Use the chain rule where the outer function is \( z^{-5} \) and the inner function \( z = \sin{x} \). - \( v = (\sin{x})^{-5} \), thus \( v' = -5(\sin{x})^{-6} \cos{x} \).**Combine using product rule:** \[ \frac{d}{dx}(x^{-1} \sin^{-5}(x)) = (-x^{-2}) \sin^{-5}{x} + x^{-1}(-5 \sin^{-6}{x}) \cos{x} \].

Step 2: Differentiate the second term

Now, we differentiate the second term \( -\frac{x}{3} \cos^{3}{x} \). This is a product of \( -\frac{x}{3} \) and \( \cos^{3}{x} \). Again, we will use the product rule here with \( u = -\frac{x}{3} \) and \( v = \cos^{3}{x} \).1. **Differentiate \( u = -\frac{x}{3} \):** \( u' = -\frac{1}{3} \).2. **Differentiate \( v = \cos^{3}{x} \):** - Again, use the chain rule. - Outer function is \( z^3 \) and inner function \( z = \cos{x} \). - \( v' = 3 \cos^{2}{x} \cdot (-\sin{x}) = -3 \cos^{2}{x} \sin{x} \).**Combine using product rule:** \[ \frac{d}{dx}(-\frac{x}{3} \cos^{3}(x)) = -\frac{1}{3} \cos^{3}{x} + \left(-\frac{x}{3}\right)(-3 \cos^{2}{x} \sin{x}) \].

Step 3: Combine derivatives

Now, we simply add together the derivatives from Steps 1 and 2 to find the derivative of the entire expression:\[ \frac{d}{dx} \left( \frac{1}{x} \sin^{-5}{x} - \frac{x}{3} \cos^{3}{x} \right) = (-x^{-2}) \sin^{-5}{x} + x^{-1}(-5 \sin^{-6}{x}) \cos{x} - \frac{1}{3} \cos^{3}{x} + x \cos^{2}{x} \sin{x} \].

Step 4: Simplify the expression

Finally, let's simplify the expression from Step 3:1. First term: \( -x^{-2} \sin^{-5}{x} \).2. Second term: \( -5x^{-1} \sin^{-6}{x} \cos{x} \).3. Third term: \( -\frac{1}{3} \cos^{3}{x} \).4. Fourth term: \( x \cos^{2}{x} \sin{x} \).Thus, the simplified derivative is:\[ -\frac{\sin^{-5}{x}}{x^2} - \frac{5 \cos{x}}{x \sin^6{x}} - \frac{1}{3} \cos^{3}{x} + x \cos^{2}{x} \sin{x} \].

Key Concepts

Product Rule

Calculating derivatives of products in calculus can seem complex, but the product rule makes this task easier. When differentiating a function that is the product of two functions, we apply the product rule, which states: if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).

This formula is incredibly useful because it helps us take derivatives of expressions where two functions are multiplied together, like the terms we see in the given problem.

Let's explore the application:

• For the first term, \( u = x^{-1} \) and \( v = \sin^{-5}{x} \), the product rule gives us: \(-x^{-2}\cdot \sin^{-5}{x} + x^{-1}\cdot(-5 \sin^{-6}{x} \cos{x})\).
• For the second term, with \( u = -\frac{x}{3} \) and \( v = \cos^{3}{x} \), we apply the product rule again: \(-\frac{1}{3} \cos^{3}{x} + \left(-\frac{x}{3}\right)(-3 \cos^{2}{x} \sin{x})\).

Each step employs the product rule to break down the expressions and calculate their derivatives accurately. Mastering this rule is crucial for tackling more complex derivative problems.

Chain Rule

The chain rule is a powerful tool for differentiating composite functions, that is, functions within functions. It provides a method for separating the derivative of these functions by breaking them down into their outer and inner components. The basic idea is: if you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).

We see the chain rule in action in this problem when differentiating functions like \( \sin^{-5}{x} \) and \( \cos^{3}{x} \). For example:

• With \( v = \sin^{-5}{x} \), consider the outer function \( z^{-5} \) and inner function \( z = \sin{x} \). Then, \( v' = -5(\sin{x})^{-6} \cdot \cos{x} \).
• Similarly, for \( v = \cos^{3}{x} \), where \( v' = 3 \cos^{2}{x} \cdot (-\sin{x}) \), the outer function is \( z^3 \) and the inner function is \( \cos{x} \).

Utilizing the chain rule allows us to systematically approach these products within trigonometric powers by identifying and handling each layer of function individually.

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in calculus, especially when considering their derivatives in problems like the one presented here. Understanding their basic properties and how they change with differentiation is crucial for solving calculus problems.

Let's discuss the derivatives of the basic trigonometric functions:

• The derivative of \( \sin{x} \) is \( \cos{x} \).
• The derivative of \( \cos{x} \) is \(-\sin{x} \).

We apply these fundamental rules when differentiating more complex functions that involve powers of sine and cosine.

For example, when determining the derivative of \( \sin^{-5}{x} \), recognizing that \( \sin{x} \) is the inner part helps apply the chain rule efficiently. Similarly, \( \cos^{3}{x} \) requires using the basic derivative formula for \( \cos{x} \) and applying the chain rule.

Having a good grasp of these derivative rules for trigonometric functions will significantly enhance your ability to tackle derivatives involving products of these functions.