Problem 19 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\cos ^{2}(3 \sqrt{x}) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\frac{3}{2}x^{-1/2}\sin(6\sqrt{x}) \).

Step 1: Identify Outer Function

Given the function \( f(x) = \cos^2(3\sqrt{x}) \), recognize that this is a composite function where \( u = \cos(3\sqrt{x}) \). The outer function is \( u^2 \), and its derivative with respect to \( u \) is \( 2u \).

Step 2: Apply Chain Rule to Outer Function

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. First, compute: \( \frac{d}{du}(u^2) = 2u \). Therefore, for our function, \( \frac{d}{dx}(\cos^2(3\sqrt{x})) = 2\cos(3\sqrt{x}) \cdot \frac{d}{dx}(\cos(3\sqrt{x})) \).

Step 3: Differentiate Inner Function's Outer Part

Now consider the function within \( \cos \), which is \( 3\sqrt{x} \). The derivative of \( \cos(v) \) where \( v = 3\sqrt{x} \) is \(-\sin(v)\cdot\frac{d}{dx}(v) \). Hence, \( \frac{d}{dx}(\cos(3\sqrt{x})) = -\sin(3\sqrt{x}) \cdot \frac{d}{dx}(3\sqrt{x}) \).

Step 4: Differentiate Inner-Inner Part

For the expression \( 3\sqrt{x} \), rewrite it as \( 3x^{1/2} \). Use power rule: \( \frac{d}{dx}(3x^{1/2}) = \frac{3}{2}x^{-1/2} \).

Step 5: Combine Derivatives

Substitute back into the expression from Step 3: \( \frac{d}{dx}(\cos(3\sqrt{x})) = -\sin(3\sqrt{x}) \cdot \frac{3}{2}x^{-1/2} \).

Step 6: Final Expression for f'(x)

Substitute the derivative from Step 5 into Step 2 to find the complete derivative: \( f'(x) = 2\cos(3\sqrt{x}) \cdot (-\sin(3\sqrt{x})) \cdot \frac{3}{2}x^{-1/2} \). Simplify this to obtain: \( f'(x) = -3\cos(3\sqrt{x})\sin(3\sqrt{x})x^{-1/2} \).

Step 7: Simplify Using Trigonometric Identity

Recognizing \( \cos(3\sqrt{x})\sin(3\sqrt{x}) \) as \( \frac{1}{2}\sin(6\sqrt{x}) \) using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we have: \( f'(x) = -\frac{3}{2}x^{-1/2}\sin(6\sqrt{x}) \).

Key Concepts

Chain Rule

The chain rule is a crucial tool in calculus for differentiating composite functions. When you have one function nested inside another, you need the chain rule to find the derivative efficiently.
To apply the chain rule, identify the 'outer' and 'inner' functions:

• The outer function is the one that applies to the entire expression—for example, the square or square root.
• The inner function is what's inside the outer function. In our exercise, this is the trigonometric function itself and its arguments.

In the example, the function is composed of an inner function being inserted into an outer one, \( \cos^2(3\sqrt{x}) \). Here, the outer is \( u^2 \) when you set \( u = \cos(3\sqrt{x}) \).
The chain rule suggests taking the derivative of the outer function \( 2u \), and then multiplying it by the derivative of the inner function. This approach breaks down the problem into manageable parts, simplifying the differentiation process.
This is compactly expressed as: \[(\text{outer derivative at inner}) \times (\text{inner derivative})\]

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. These identities are incredibly useful for simplifying expressions and solving trigonometry-related problems.
In this exercise, the identity we used was: \[\sin(2\theta) = 2\sin(\theta)\cos(\theta)\]
This identity is employed to simplify the product of sine and cosine functions. It transforms the expression \( \cos(3\sqrt{x})\sin(3\sqrt{x}) \) into \( \frac{1}{2}\sin(6\sqrt{x}) \). This simplification reduces the complexity of the derivative, making calculations more tractable.
Always remember that trigonometric identities are indispensable in calculus, not just for simplifying derivatives, but also for integrations and other operations involving trigonometric functions.

Composite Functions

Composite functions involve two or more functions combined such that the output of one function becomes the input of another. This nesting structure is common when transforming and manipulating functions in calculus.
In the core exercise, \(f(x)=\cos^2(3\sqrt{x}) \) is a composite function because it applies the trigonometric cosine function to \( 3\sqrt{x} \), then squares the result. Understanding composite functions is key to applying the chain rule effectively since you often need to differentiate each layer separately.
To successfully differentiate a composite function, accurately identify:

• The innermost function (innermost transformation)
• The sequence of operations outside of this to the outermost function

Each function layer has its own impact on the differentiation process, and recognizing these layers will facilitate correct application of calculus rules.
Composite functions show how far combinations in mathematics can go, offering vast possibilities in transforming and analyzing real-world problems.