Problem 29 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=x^{5} \sec (1 / x) $$

Step-by-Step Solution

Answer

The derivative is \( 5x^4 \sec\left(\frac{1}{x}\right) - x^3 \sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right) \).

Step 1: Apply the Product Rule

To find the derivative \( \frac{d y}{d x} \) of the function \( y = x^5 \sec(\frac{1}{x}) \), recognize that this is a product of two functions: \( u = x^5 \) and \( v = \sec(\frac{1}{x}) \). The product rule states that \((uv)' = u'v + uv'\).

Step 2: Differentiate \( u = x^5 \)

Compute the derivative of \( u = x^5 \) using the power rule. \( \frac{d}{dx}(x^5) = 5x^4 \).

Step 3: Differentiate \( v = \sec(\frac{1}{x}) \) Using the Chain Rule

First, find the derivative of \( v = \sec(z) \) where \( z = \frac{1}{x} \). The derivative \( \sec(z) \) is \( \sec(z)\tan(z) \). Next, find \( \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \). Apply the chain rule: \( \frac{dv}{dx} = \frac{d}{dz}\sec(z) \cdot \frac{dz}{dx} = \sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \).

Step 4: Substitute into the Product Rule Formula

Combine the derivatives found in Steps 2 and 3 with the product rule formula: \( \frac{d}{dx}[x^5 \sec(\frac{1}{x})] = 5x^4 \sec\left(\frac{1}{x}\right) + x^5 \left(\sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)\right) \).

Step 5: Simplify the Expression

Simplify the expression from Step 4: \( 5x^4 \sec\left(\frac{1}{x}\right) - x^3 \sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right) \).

Key Concepts

The Product Rule

The product rule is an essential tool in calculus when dealing with the differentiation of products of two functions. In this example, we need it for the function, \(y = x^5 \sec(\frac{1}{x})\), which is a product of \(u = x^5\) and \(v = \sec(\frac{1}{x})\). To find the derivative of such a function, we apply the rule:

• First, differentiate the first function \(u\) while keeping \(v\) constant, giving us \(u'v\).
• Then, differentiate the second function \(v\) while keeping \(u\) constant, giving us \(uv'\).
• Finally, add these two derivatives together: \((uv)' = u'v + uv'\).

This method allows us to tackle complex expressions efficiently by breaking them down into manageable pieces. Here, the product rule helps handle the initial complexity of \(y\), paving the way for further steps.

The Chain Rule

The chain rule is a powerful technique used when differentiating composite functions. In our exercise, \(v = \sec(\frac{1}{x})\) involves a composition of functions: \(\sec(z)\) where \(z = \frac{1}{x}\). To tackle this, follow these steps:

• First, differentiate the outer function, \(\sec(z)\), which yields \(\sec(z)\tan(z)\).
• Then, find the derivative of the inner function \(\frac{1}{x}\), resulting in \(-\frac{1}{x^2}\).
• Multiply these two derivatives together to get the derivative of the composite function, \(\sec(z)\tan(z) \cdot (-\frac{1}{x^2})\).

Applying the chain rule in this way allows us to manage the layers of functions within a composite entity. It is crucial for handling expressions where a function is nested inside another, as in \(\sec(\frac{1}{x})\).

The Power Rule

One of the most straightforward and widely used techniques in differentiation, the power rule, allows us to quickly find derivatives of functions in the form \(x^n\). This rule states:

• The derivative of \(x^n\) is \(nx^{n-1}\).

In this problem, we apply the power rule to \(u = x^5\), giving:

• \(\frac{d}{dx}(x^5) = 5x^4\).

By applying the power rule, we quickly derive the part of the function \(x^5\), facilitating the use of the product rule for the entire expression. This method provides clarity and simplicity in deriving powers of \(x\), allowing us to focus on more complicated parts of the problem with confidence.