Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Apply the Chain Rule more than once to find the indicated derivative. \(D_{u}\left[\cos ^{4}\left(\frac{u+1}{u-1}\right)\right]\)

Step-by-Step Solution

Answer

\(D_{u}\left[\cos^{4}\left(\frac{u+1}{u-1}\right)\right] = \frac{8\cos^{3}\left(\frac{u+1}{u-1}\right)\sin\left(\frac{u+1}{u-1}\right)}{(u-1)^{2}}\)

Step 1: Identify the Outer Function

The outermost function in the expression \( \cos^4\left(\frac{u+1}{u-1}\right) \) is \( y = x^4 \) where \( x = \cos\left(\frac{u+1}{u-1}\right) \). This means we need to differentiate \( y = x^4 \) with respect to \( x \).

Step 2: Differentiate the Outer Function

Apply the power rule to differentiate \( y = x^4 \). The derivative is \( 4x^3 \). This is the derivative of \( \cos^4 \left(\frac{u+1}{u-1}\right) \) with respect to \( \cos \left(\frac{u+1}{u-1}\right) \).

Step 3: Identify the Middle Function

Now, consider the middle function \( x = \cos\left(\frac{u+1}{u-1}\right) \). The derivative of the cosine function \( \cos(v) \) where \( v = \frac{u+1}{u-1} \) is \( -\sin(v) \).

Step 4: Differentiate the Middle Function

Differentiate \( x = \cos\left(v\right) \) to get \( \frac{dx}{dv} = -\sin\left(\frac{u+1}{u-1}\right) \).

Step 5: Identify the Inner Function

The innermost function is \( v = \frac{u+1}{u-1} \). Differentiate \( v \) with respect to \( u \) using the quotient rule.

Step 6: Differentiate the Inner Function using the Quotient Rule

The quotient rule states that if \( h(u) = \frac{f(u)}{g(u)} \), then \( h'(u) = \frac{f'(u)g(u) - f(u)g'(u)}{g(u)^2} \). Here, \( f(u) = u+1 \) and \( g(u) = u-1 \). Calculate the derivatives: \( f'(u) = 1 \) and \( g'(u) = 1 \). Thus, the derivative \( v'(u) = \frac{(1)(u-1) - (u+1)(1)}{(u-1)^2} = \frac{u-1-u-1}{(u-1)^2} = \frac{-2}{(u-1)^2} \).

Step 7: Apply the Chain Rule

Combine the derivatives using the chain rule: \[ D_{u} \left[ \cos^4 \left( \frac{u+1}{u-1} \right) \right] = \left(4 \cos^3 \left(\frac{u+1}{u-1}\right)\right) \left(-\sin\left(\frac{u+1}{u-1}\right)\right) \left(\frac{-2}{(u-1)^2}\right) \].

Step 8: Simplify the Expression

Simplify the expression obtained in the previous step. The derivative simplifies to:\[ D_{u} \left[ \cos^4 \left( \frac{u+1}{u-1} \right) \right] = \frac{8 \cos^3 \left( \frac{u+1}{u-1} \right) \sin\left(\frac{u+1}{u-1}\right)}{(u-1)^2} \].

Key Concepts

Understanding Derivatives

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In simpler terms, it's like capturing the 'rate of change' or the 'slope' of a function at any specific point. When solving complex derivatives, you often have to apply multiple rules, such as the chain rule, to break down each component effectively.

For instance, when you're given a composite function, like in our example problem, identifying each layer of the function is crucial. This allows you to systematically find the derivative by working from the outermost function to the innermost one. Understanding derivatives is not just about applying formulas but discerning how different mathematical elements are connected and change together.

Using the Power Rule

The power rule is one of the simplest and most frequently used tools for finding derivatives. It's applicable for functions in the form of \( x^n \), where its derivative becomes \( nx^{n-1} \). This rule is essential for differentiating functions like \( y = x^4 \). In our example, this principle applies when differentiating the outer function \( ext{cos}^4(x) \) with respect to \( ext{cos}(x) \).

When using the power rule, remember that each time you differentiate, you're finding how the 'power' parts of the function affect its rate of change. It's quick, effective, and forms the backbone of tackling polynomial components in derivatives. Its simplicity allows us to focus more energy on solving complex inner functions.

Applying the Quotient Rule

The quotient rule is used for finding the derivative of a quotient of two functions. If you have \( h(u) = \frac{f(u)}{g(u)} \), the derivative is given by \( h'(u) = \frac{f'(u)g(u) - f(u)g'(u)}{(g(u))^2} \). This rule is particularly useful when dealing with rational functions, like the innermost function of our example, \( \frac{u+1}{u-1} \).

To apply the quotient rule, identify the numerator and the denominator functions and differentiate them separately. Then plug these into the quotient formula. Remember that maintaining correct order and using subtraction carefully are crucial for accurate results. While it seems complex at first, practice will hone your skills in applying this rule seamlessly.

Exploring Trigonometric Functions

Trigonometric functions are another common element in calculus problems, and knowing how to differentiate them is important. Common trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). Each has its own differentiation rule. For example:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In our chain rule problem, the differentiation of \( \cos(v) \) becomes \( -\sin(v) \). Understanding these basic trig functions helps manage complex expressions by breaking them into smaller, manageable derivatives. The more familiar you become with these rules, the easier it becomes to tackle compound expressions involving trigonometric functions.