Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ f^{\prime}\left(\frac{\pi}{4}\right) \text { if } f(x)=\ln (\cos x) $$

Step-by-Step Solution

Answer

The derivative at \( x = \frac{\pi}{4} \) is \( -1 \).

Step 1: Understanding the Function

We begin with the function given: \( f(x) = \ln(\cos x) \). In order to find its derivative at a specific value, we need to differentiate this function with respect to \( x \).

Step 2: Applying the Chain Rule

The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \cos x \). So, \( \frac{d}{dx}[\ln(\cos x)] = \frac{1}{\cos x} \cdot \frac{d}{dx}[\cos x] \).

Step 3: Derivative of Cosine

The derivative of \( \cos x \) is \( -\sin x \). So, when we apply this in the chain rule, we have \( \frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x} \).

Step 4: Simplifying the Expression

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

Step 5: Evaluating the Derivative at \( x = \frac{\pi}{4} \)

Now, substitute \( x = \frac{\pi}{4} \) into the derivative \( f'(x) = -\tan x \). We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Therefore, \( f'\left(\frac{\pi}{4}\right) = -1 \).

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. Let me simplify it for you: imagine you have two functions, say \( u(x) \) and \( y(u) \). The chain rule helps us differentiate a function that has another function inside it, like \( y(u(x)) \). Here’s what makes the chain rule so powerful: it tells us that

• the derivative of \( y \) with respect to \( x \) is the derivative of \( y \) with respect to \( u \) multiplied by the derivative of \( u \) with respect to \( x \).

In mathematical terms, if \( y = f(u) \) and \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) is given by\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}. \]In our given problem, your function is \( f(x) = \ln(\cos x) \). Here, the outer function is the natural logarithm, \( \ln \, u \), and the inner function is the trigonometric function, \( \cos x \). Applying the chain rule, we differentiate \( \ln u \) as \( \frac{1}{u} \) with \( u = \cos x \), and multiply it by the derivative of the inner function \( \cos x \) which is \( -\sin x \). Thus, the derivative of \( \ln(\cos x) \) is \( \frac{-\sin x}{\cos x} \).

Trigonometric Functions Differentiation

Differentiating trigonometric functions is an essential skill in calculus. When working with such functions, it’s crucial to remember the basic derivatives. For the primary trigonometric functions, we have:

• 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 exercise, the focus was on finding the derivative of \( \cos x \), which is \( -\sin x \). This result is used in applying the chain rule when dealing with the derivative of \( f(x) = \ln(\cos x) \). Knowing these key derivatives can make handling calculus problems much easier!
Once you've identified that the derivative of \( \cos x \) is \( -\sin x \), you can directly apply it within the chain rule to get the simplified form \( \frac{-\sin x}{\cos x} \), which turns into \( -\tan x \) using identity \( \tan x = \frac{\sin x}{\cos x} \).

Evaluating Derivatives at Specific Points

Evaluating derivatives at specific points is how we find the slope of the tangent line to a curve at a given point. In this problem, the specific point is \( x = \frac{\pi}{4} \). After determining the general derivative of \( f(x) = \ln(\cos x) \) to be \( -\tan x \), it’s time to evaluate it at \( x = \frac{\pi}{4} \).
Calculus often involves substituting a specific \( x \)-value into the derivative expression. At the angle \( \frac{\pi}{4} \), the trigonometric function \( \tan x \) equals 1. That’s due to the property of tangent when both sine and cosine of \( \frac{\pi}{4} \) are equal. So substituting into \( f'(x) = -\tan x \), we get:\[ f'\left(\frac{\pi}{4}\right) = -1 \times 1 = -1. \]By substituting, we gain insights on the behavior of the function at \( x = \frac{\pi}{4} \), knowing that the slope of the tangent line is \(-1\).