Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(f^{\prime}\left(\frac{\pi}{4}\right)\) if \(f(x)=\ln (\cos x)\)

Step-by-Step Solution

Answer

The derivative is \(-1\).

Step 1: Understand the Function and Problem

We are given the function \( f(x) = \ln(\cos x) \) and need to find the derivative of this function evaluated at \( x = \frac{\pi}{4} \). This means we need to compute \( f'(x) \), and then substitute \( x = \frac{\pi}{4} \) into it to find \( f^{\prime}\left(\frac{\pi}{4}\right) \).

Step 2: Find the Derivative of the Function

To find \( f'(x) \), we use the chain rule. Let \( u = \cos x \), so \( \ln(\cos x) = \ln(u) \). The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \), and the derivative of \( \cos x \) with respect to \( x \) is \( -\sin x \). So, applying the chain rule, we have: \[ f'(x) = \frac{d}{dx}[\ln(\cos x)] = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x}. \]

Step 3: Simplify the Derivative Using a Trigonometric Identity

The derivative \( f'(x) = -\frac{\sin x}{\cos x} \) simplifies to \( -\tan x \) because \( \tan x = \frac{\sin x}{\cos x} \). Thus, \[ f'(x) = -\tan x. \]

Step 4: Compute the Derivative at the Specific Point

Now evaluate \( f'(x) = -\tan x \) at \( x = \frac{\pi}{4} \). We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Therefore, \[ f^{\prime}\left(\frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1. \]

Key Concepts

Understanding the Chain Rule

The chain rule is an essential concept in calculus, especially when differentiating composite functions. A composite function involves one function inside another, like \( f(x) = \ln(\cos x) \). Here, we have an outer function \( \ln(u) \) and an inner function \( u = \cos x \). The chain rule allows us to differentiate these types of functions by multiplying the derivative of the outer function by the derivative of the inner function.

When applying the chain rule to \( f(x) = \ln(\cos x) \):

• Identify the outer function: \( \ln(u) \) with derivative \( \frac{1}{u} \).
• Identify the inner function: \( u = \cos x \) with derivative \( -\sin x \).
• Multiply these derivatives: \( \frac{1}{\cos x} \cdot (-\sin x) \).

This results in \( f'(x) = -\frac{\sin x}{\cos x} \), showing how the chain rule simplifies the process of differentiation with nested functions.

Using Trigonometric Identities

Trigonometric identities are very helpful in simplifying expressions, especially when they appear in derivatives. One common identity is \( \tan x = \frac{\sin x}{\cos x} \). These identities help us rewrite the expressions in a more manageable form.

When we have obtained \( f'(x) = -\frac{\sin x}{\cos x} \) using the chain rule, we can simplify this expression using the tan identity. The derivative becomes \(-\tan x\). Simplification not only makes computations more straightforward but can also provide insights into the behavior of the function.

Evaluating Derivatives at a Point

Once the derivative of a function has been found, you often need to evaluate it at a specific value of \( x \), known as evaluating the derivative at a point. This process involves substituting the given point into the derivative function.

In our exercise, we needed to find \( f'(\frac{\pi}{4}) \). We already simplified the derivative to \(-\tan x\). Plugging in \( x = \frac{\pi}{4} \) gives us:

• Calculate \( \tan(\frac{\pi}{4}) = 1 \) because the tangent of \( \frac{\pi}{4} \) is a known trigonometric value.
• Substitute into the simplified derivative: \( f'(\frac{\pi}{4}) = -1 \).

Evaluating derivatives at a point is particularly useful for understanding the instantaneous rate of change of a function at that specific value, and it helps in understanding the function's graph around that point.