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Chapter 3: Problem 26

Find the derivatives of the functions in Exercises \(19-40\) $$y=\frac{1}{x} \sin ^{-5} x-\frac{x}{3} \cos ^{3} x$$

Short Answer

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

Step by step solution

01

Identify the Functions

In the given expression, we have two main components: 1. \( y_1 = \frac{1}{x} \sin^{-5}(x) \).2. \( y_2 = -\frac{x}{3} \cos^3(x) \).Let's find the derivative of each part separately and then combine the results.
02

Derivative of First Function

For \( y_1 = \frac{1}{x} \sin^{-5}(x) \), we need to use the product rule and the chain rule. The product rule states: \((fg)' = f'g + fg'\), where \( f = \frac{1}{x} \) and \( g = \sin^{-5}(x) \). 1. Derivative of \( f \): \( f' = -\frac{1}{x^2} \).2. Derivative of \( g \) using the chain rule: \( g = (\sin(x))^{-5} \). Let \( u = \sin(x) \), then \( g = u^{-5} \). The derivative, \( g' = -5u^{-6}u' \), where \( u' = \cos(x) \). So, \( g' = -5 \sin^{-6}(x) \cos(x) \).Substitute into the product rule:\( y_1' = \left(-\frac{1}{x^2}\right) \sin^{-5}(x) + \frac{1}{x} \left(-5 \sin^{-6}(x) \cos(x)\right) \).
03

Derivative of Second Function

For \( y_2 = -\frac{x}{3} \cos^3(x) \), use the product and chain rule.The product rule applies with \( f = -\frac{x}{3} \) and \( g = \cos^3(x) \).1. Derivative of \( f \): \( f' = -\frac{1}{3} \).2. Derivative of \( g \) using the chain rule: \( g = (\cos(x))^3 \). Let \( u = \cos(x) \), then \( g = u^3 \). The derivative, \( g' = 3u^2u' \), where \( u' = -\sin(x) \). So, \( g' = 3 \cos^2(x) (-\sin(x)) \) or \( g' = -3 \cos^2(x) \sin(x) \).Substitute into the product rule:\( y_2' = \left(-\frac{1}{3}\right) \cos^3(x) + \left(-\frac{x}{3}\right)(-3 \cos^2(x) \sin(x)) \).
04

Simplify Derivatives

Simplify both derivatives from Steps 2 and 3. 1. From Step 2: \[ y_1' = -\frac{1}{x^2} \sin^{-5}(x) - \frac{5 \cos(x)}{x} \sin^{-6}(x) \].2. From Step 3: \[ y_2' = -\frac{1}{3} \cos^3(x) + x \cos^2(x) \sin(x) \]. Simplifying the second term yields: \[ y_2' = -\frac{1}{3} \cos^3(x) + \frac{x \cos^2(x) \sin(x)}{3} \].
05

Combine Derivatives

Combine the derivatives found in Steps 2 and 3 to get the derivative of the original function:\[ y' = y_1' + y_2' \].Putting them together:\[ y' = -\frac{1}{x^2} \sin^{-5}(x) - \frac{5 \cos(x)}{x} \sin^{-6}(x) - \frac{1}{3} \cos^3(x) + \frac{x \cos^2(x) \sin(x)}{3} \].This is the complete derivative of the function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
When you have a function that is the product of two differentiable functions, you can use the product rule to find the derivative. The product rule formula is: \((fg)' = f'g + fg'\). This allows you to differentiate each function separately and then combine them.
  • Start by identifying the two functions involved in the product.
  • Differentiating one function while keeping the other as-is.
  • Then switch: differentiate the other function while keeping the first one unchanged.

In our exercise, we first divided the expression into two parts and found that for the first function, \(y_1\), we had \(f = \frac{1}{x}\) and \(g = \sin^{-5}(x)\). We need to apply the product rule to differentiate the product of these two functions.
  • The derivative of \(f = \frac{1}{x}\) is \(f' = -\frac{1}{x^2}\).
  • For \(g\), which is \(\sin^{-5}(x)\), we apply the chain rule next.
Chain Rule
The chain rule is used when you have a function inside another function, which is usually expressed as \(f(g(x))\). It helps you find the derivative by differentiating the outer function and multiplying it by the derivative of the inner function. This step is essential when dealing with composite functions like our \(g = u^{-5} = (\sin(x))^{-5}\).
  • Start by letting \(u = \sin(x)\), so \(g\) becomes \(u^{-5}\).
  • Differentiate the outer function: \((u^{-5})' = -5u^{-6}\).
  • Find the derivative of the inner function, \(u' = \cos(x)\).
  • Combine them to get \(g' = -5\sin^{-6}(x)\cos(x)\).

Repeating similar steps for the function \(y_2 = -\frac{x}{3} \cos^3(x)\), we let \(u = \cos(x)\). Again, we find \(g' = -3 \cos^2(x) \sin(x)\) by following the chain rule.
Simplification of Expressions
After finding derivatives using the product and chain rules, simplification is the next big task. The goal is to make the expression as straightforward and manageable as possible. This process often involves combining like terms or factoring.
A good place to start is by simplifying each derivative section separately before combining them. For example, with \(y_1'\), you find the expression is \(-\frac{1}{x^2} \sin^{-5}(x) - \frac{5 \cos(x)}{x} \sin^{-6}(x)\).
  • Check each term for common factors.
  • Simplify fractions when possible.

In our original problem, both formed derivatives were combined into \[y' = -\frac{1}{x^2} \sin^{-5}(x) - \frac{5 \cos(x)}{x} \sin^{-6}(x) - \frac{1}{3} \cos^3(x) + \frac{x \cos^2(x) \sin(x)}{3}\]. Ensuring the expression is fully simplified makes the final derivative more usable.

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Most popular questions from this chapter

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