Problem 49 - Essential Calculus Early Transcendentals

Question

Let $$f(x)=c x+\ln (\cos x) .\( For what value of \)c\( is \)f^{\prime}(\pi / 4)=6 ?$$

Step-by-Step Solution

Answer

The value of \( c \) is 7.

Step 1: Identify the Derivative

First, we need to derive the function \( f(x) = cx + \ln(\cos x) \). The derivative, \( f'(x) \), is calculated using the sum rule for derivatives. We'll differentiate both components individually.

Step 2: Derivative of Components

For \( f(x) = cx + \ln(\cos x) \): - The derivative of \( cx \) with respect to \( x \) is \( c \).- The derivative of \( \ln(\cos x) \) is found using the chain rule: \(-\tan x \).Thus, the derivative \( f'(x) = c - \tan x \).

Step 3: Set the Derivative Equal to 6

Since we need \( f'(\pi/4) = 6 \), plug \( x = \pi/4 \) into the derived expression and set it equal to 6: \[ c - \tan(\pi/4) = 6 \]

Step 4: Calculate \( \tan(\pi/4) \)

The value of \( \tan(\pi/4) \) is 1. Substituting into the equation gives us: \[ c - 1 = 6 \]

Step 5: Solve for \( c \)

Add 1 to both sides of the equation \[ c - 1 = 6 \] to get: \[ c = 7 \]

Key Concepts

Derivative of a function

In calculus, the derivative of a function is a measure of how a function changes as its input changes. The derivative represents an instantaneous rate of change, given by the slope of the tangent line to the function graph at a point. To find it, use differentiation rules.

For example, in the function \( f(x) = cx + \ln(\cos x) \), the derivative, \( f'(x) \), is found by applying the sum rule. This rule states if you have two functions being added, \( u(x) + v(x) \), you can differentiate them separately and add the results:

• Derivative of \( cx \) is \( c \) since it's a linear function with slope \( c \).
• Derivative of \( \ln(\cos x) \) requires a bit more work, which leads us to our next section on trigonometric functions.

This forms the basis for calculating derivatives and forms a foundational concept in calculus, necessary for understanding changes and rates.

Trigonometric functions

Trigonometric functions such as sine, cosine, and tangent are vital in calculus, especially when dealing with derivatives. These functions are periodic and describe angles and relationships in circles. Understanding their derivatives is crucial.

In our function, \( f(x) = cx + \ln(\cos x) \), we encounter \( \ln(\cos x) \). When differentiating, finding \( \frac{d}{dx}[\ln(\cos x)] \) uses the chain rule (which we'll cover next).

• The derivative of \( \cos x \) is \(-\sin x \).
• For \( \ln(u) \), its derivative is \( \frac{1}{u} \cdot \frac{du}{dx} \).

Thus, \( \ln(\cos x) \) differentiates to \( -\tan x \), combining these elements since \( \tan x = \frac{\sin x}{\cos x} \).

Understanding these manipulations allows complex expressions to be simplified and used in calculus.

Chain rule

The chain rule is a powerful differentiation tool that allows us to handle composite functions. It breaks down a complex derivative into the product of two or more simpler derivatives.

The rule is stated as: if you have a composite function \( h(x) = f(g(x)) \), the chain rule tells you to multiply the derivative of the outer function by the derivative of the inner function:

• \( h'(x) = f'(g(x)) \cdot g'(x)\).

In \( f(x) = cx + \ln(\cos x) \), to differentiate \( \ln(\cos x) \), the chain rule is used:

- The outer function is \( \ln(u) \), where \( u = \cos x \), differentiating gives \( \frac{1}{u} \).
- The inner function \( \cos x \) differentiates to \(-\sin x \).

Multiply these to get \( \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \).

By applying the chain rule, we can tackle derivatives of more involved functions efficiently.