Problem 70 - Calculus Single Variable

Question

find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=\sqrt{\sin ^{2}(3 x)+\cos ^{2}(3 x)} $$

Step-by-Step Solution

Answer

The derivative \( f'(x) \) is 0.

Step 1: Recognize Identity

The expression inside the square root is \( \sin^2(3x) + \cos^2(3x) \). Based on the Pythagorean trigonometric identity, we know that \( \sin^2(\theta) + \cos^2(\theta) = 1 \) for any angle \( \theta \). Applying this identity to our function, we simplify it to \( f(x) = \sqrt{1} \).

Step 2: Simplify the Function

Since \( \sqrt{1} = 1 \), the function \( f(x) = \sqrt{\sin^2(3x) + \cos^2(3x)} \) simplifies to \( f(x) = 1 \).

Step 3: Differentiate the Constant Function

The derivative of any constant function is zero. Therefore, if \( f(x) = 1 \), then the derivative, \( f'(x) \), is \( 0 \).

Key Concepts

Pythagorean Identity

The Pythagorean identity is foundational in trigonometry. It states that for any angle \( \theta \), the equation \( \sin^2(\theta) + \cos^2(\theta) = 1 \) always holds true. This identity is essentially a restatement of the Pythagorean Theorem, in a trigonometric form.

• It serves the purpose of simplifying complex trigonometric expressions.
• For example, in the given function \( f(x) = \sqrt{\sin^2(3x) + \cos^2(3x)} \), recognizing the identity allows for the expression within the square root to simplify to \( 1 \).
• This simplification is crucial in proceeding with the differentiation process.

Without the use of the Pythagorean identity, solving such trigonometric problems would be cumbersome, requiring more complex manipulations. When simplifying trigonometric expressions, always check if this reliable identity can be used to make your work easier.

Differentiation

Differentiation is a key process in calculus used to find the rate at which a function is changing at any given point. To differentiate a function, you calculate its derivative. The derivative of a function gives you the slope of the function's graph at any point.

• If a function describes the position of an object with respect to time, then its derivative will describe the object's velocity.
• For trigonometric functions, the process of differentiation often involves the use of special rules and identities, like the chain rule.
• In the example of the function \( f(x) = 1 \), there's no change in the function value concerning \( x \). This directly leads us to consider the nature of constant functions during differentiation.

Understanding differentiation involves recognizing when a function exhibits constant behavior and exploiting rules of calculus to simplify the problem. This allows for effective application to real-world problems involving rates of change.

Constant Function

Constant functions are deceptively simple but incredibly important in calculus. A constant function is one that always returns the same value, regardless of the input.

• In mathematical terms, if \( f(x) = c \), where \( c \) is a constant, then for every value of \( x \), \( f(x) \) will still be \( c \).
• The derivative of a constant function is \( 0 \), because the slope (or rate of change) of a constant function is always flat. There's no vertical change as \( x \) increases.
• For the function \( f(x) = 1 \), as simplified from the original trigonometric expression using the Pythagorean identity, its derivative \( f'(x) \) is \( 0 \).

Recognizing constant functions simplifies calculus problems significantly, often reducing complex expressions to basic solutions. This basic understanding of how constants behave in calculus is essential for mastering higher-level topics. It allows for a focus on the behavior of variables where they exist rather than constants.