Problem 23 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\sqrt{\cos (5 x)} $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = -\frac{5\sin(5x)}{2\sqrt{\cos(5x)}}\).

Step 1: Identify the Functions Involved

The given function is a composite function: \(f(x) = \sqrt{\cos(5x)}\). This can be rewritten as \(f(x) = (\cos(5x))^{1/2}\). The outer function is a power function, and the inner function is a cosine function within another linear function.

Step 2: Apply the Chain Rule

To differentiate a composite function, we use the chain rule. For a function \((g(x))^{n}\), the derivative is \(n(g(x))^{n-1} \cdot g'(x)\). Here, \(g(x) = \cos(5x)\) and the outer function is the square root (or power \(1/2\)).

Step 3: Differentiate the Outer Function

Differentiate the outer function \((\cos(5x))^{1/2}\) with respect to \(g(x) = \cos(5x)\). The result is \((1/2)(\cos(5x))^{-1/2}\).

Step 4: Differentiate the Inner Function

Now, differentiate the inner function \(g(x) = \cos(5x)\) with respect to \(x\). The derivative is \(-\sin(5x) \cdot 5\), which simplifies to \(-5\sin(5x)\).

Step 5: Combine Using Chain Rule

Combine the derivatives from Step 3 and Step 4 using the chain rule: \[ f'(x) = (1/2)(\cos(5x))^{-1/2} \cdot (-5\sin(5x)) \]Simplifying, we get:\[ f'(x) = -\frac{5\sin(5x)}{2\sqrt{\cos(5x)}} \]

Step 6: Simplify the Expression

The negative sign indicates the direction and magnitude of slope. Therefore, the simplified derivative is:\[ f'(x) = -\frac{5\sin(5x)}{2\sqrt{\cos(5x)}} \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, essential for differentiating composite functions. A composite function is a function formed by combining two or more functions. The chain rule helps you find the derivative when these functions are nested. Imagine peeling an onion; you perform differentiation at each layer until you reach the core.

To apply the chain rule:

• Identify the outer and inner functions.
• Differentially the outer function while keeping the inner untouched.
• Multiply the result by the derivative of the inner function.

In our example, we saw a function, \(f(x) = \sqrt{\cos(5x)}\). We identified the outer function as \((u^{1/2})\) and the inner function as \(u = \cos(5x)\). Differentiating step by step helps understand how each component contributes to the final derivative.

Composite Functions

Composite functions are those in which one function is applied to the results of another. They appear frequently in mathematics and calculus. Recognizing and understanding them is crucial for applying the chain rule.

Consider the function \(f(x) = \sqrt{\cos(5x)}\). It is a composite function because it combines a trigonometric function, cosine, with another function, a power function. Here is how it's broken down:

• The outer function: \(u^{1/2}\), where \(u = \cos(5x)\).
• The inner function: \(\cos(5x)\), which itself involves \(5x\).

By evaluating these components separately, you can systematically apply the chain rule to find the derivative effectively. The key is to clearly identify each component of the composite function.

Trigonometric Derivatives

Trigonometric derivatives are essential for calculating derivatives of functions that involve trigonometric components. They follow specific rules that make them unique but predictable.

Here are some of the basic trigonometric derivatives you need to remember:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• For composed functions like \(\cos(kx)\), apply the chain rule to account for the coefficient \(k\).

In our exercise, we dealt with \(\cos(5x)\). By differentiating this, we find \(-5\sin(5x)\) as the derivative due to the chain rule. The function's linearity, denoted by \(5x\), introduces a multiplication factor, which is essential to capture in differentiating.