Problem 42 - Calculus Single Variable

Question

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos ^{2}(x), c=\pi / 4 $$

Step-by-Step Solution

Answer

The linearization of \(f(x) = \cos^2(x)\) at \(c = \pi/4\) is \(L(x) = \frac{\pi}{4} + \frac{1}{2} - x\).

Step 1: Find the Function Value

To find the linearization, we first need to evaluate the function at the given point. Calculate \(f(c)\) by substituting \(c = \pi / 4\) into \(f(x) = \cos^2(x)\). We have: \[ f\left( \frac{\pi}{4} \right) = \cos^2\left( \frac{\pi}{4} \right) = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}. \]

Step 2: Derive the Function

Next, find the derivative of the function, \(f'(x)\). Using the chain rule, where \(f(x) = (\cos(x))^2\): \[ f'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x) = -\sin(2x). \]

Step 3: Calculate the Derivative at c

Substitute \(c = \pi / 4\) into the derivative \(f'(x)\) to find \(f'(c)\): \[ f'\left( \frac{\pi}{4} \right) = -\sin\left( 2 \cdot \frac{\pi}{4} \right) = -\sin\left( \frac{\pi}{2} \right) = -1. \]

Step 4: Write the Linearization Equation

With \(f(c)\) and \(f'(c)\) calculated, plug these values into the linearization formula: \[ L(x) = f(c) + f'(c)\cdot(x-c). \]Substituting in the known values: \[ L(x) = \frac{1}{2} - 1 \cdot \left( x - \frac{\pi}{4} \right). \] Thus, \[ L(x) = \frac{1}{2} - (x - \frac{\pi}{4}) = \frac{1}{2} - x + \frac{\pi}{4}. \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. Composite functions are those where one function is nested within another. Consider a basic scenario: if you have a function within another function, like \( f(g(x)) \), then the chain rule helps us differentiate this effectively.

The chain rule states that the derivative of \( f(g(x)) \) is the derivative of the outer function \( f \) evaluated at \( g(x) \) times the derivative of the inner function \( g(x) \). It's expressed as:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \).

In the context of our exercise, we are tasked with differentiating \( f(x) = \cos^2(x) \). Here, we recognize \( \cos(x) \) as the inner function and \( x^2 \) as part of the outer function. This allows us to apply the chain rule, first taking the derivative of the outer part, which brings down the 2, leading to:

• \( 2 \cdot \cos(x) \cdot (-\sin(x)) \),
• resulting in: \( -2\cos(x)\sin(x) \).

The chain rule simplifies the process of taking derivatives of composite functions, making it a powerful tool in calculus.

Derivative Calculation

Calculating derivatives is a primary skill in calculus used to determine how a function changes at any given point. A derivative essentially measures how a function's output value shifts concerning a change in input value, \( x \).

When dealing with derivatives, it's crucial to recognize different rules and formulas involved. For basic polynomials, power rules apply, but for more complex functions, like those involving trigonometric identities, specialized techniques and rules like the chain rule come into play.

In the given exercise, after identifying \( f(x) = \cos^2(x) \), we use the chain rule to calculate its derivative. Differentiating \( \cos(x) \) results in \( -\sin(x) \), and with the multiplication principle, we end up with the derivative \( f'(x) = -2\cos(x)\sin(x) \), which can be further simplified.

• Simplified form: \( -\sin(2x) \), utilizing a trigonometric identity.

Finding derivatives with accuracy enables us to solve and analyze function behavior; for example, understanding how the function changes at \( c = \pi/4 \) in this case.

Trigonometric Functions

Trigonometric functions like \( \cos(x) \), \( \sin(x) \), and others play significant roles in many calculus problems. Understanding these functions is essential when dealing with problems involving oscillations and waves or circular motion.

In this exercise, \( f(x) = \cos^2(x) \) is expressed in terms of the cosine function. At \( c = \pi/4 \), we compute \( \cos(\pi/4) = \frac{1}{\sqrt{2}} \). Squaring this gives us \( f(c) = \frac{1}{2} \), showing its value at \( c = \pi/4 \).

Trigonometric identities also allow us to simplify expressions. Take the derivative \( -2\cos(x)\sin(x) \) from the chain rule application—it uses the double angle identity:

• \( \sin(2x) = 2\sin(x)\cos(x) \).

Thus, we transform our derivative expression into \( -\sin(2x) \). Mastering these trigonometric principles aids in making sense of results derived from complex functions, such as calculating the accurate linearization, \( L(x) \), in the problem.