Problem 57 - Calculus - AP Edition

Question

Find the indicated derivative. $$\frac{d}{d \omega}\left[a \cos ^{2} \pi \omega+b \sin ^{2} \pi \omega\right] \quad(a, b \text { constants })$$

Step-by-Step Solution

Answer

\( \pi(b-a)\sin(2\pi \omega) \)

Step 1: Understand the Function

We need to find the derivative of the function with respect to \( \omega \): \( a \cos^{2} \pi \omega + b \sin^{2} \pi \omega \). Notice that both terms involve the square of trig functions.

Step 2: Apply Derivative Properties to Trig Functions

Apply the chain rule as \( \cos^{2} \pi \omega \) and \( \sin^{2} \pi \omega \) can be written as compositions of functions. Recognize that the derivative of \( \cos^{2} x \) is \(-2 \cos x \sin x \) and the derivative of \( \sin^{2} x \) is \(2 \sin x \cos x \).

Step 3: Derivative of \( a \cos^{2} \pi \omega \)

Use the chain rule: Let \( u = \pi \omega \), then \( \frac{d}{d\omega}(a \cos^{2} \pi \omega) = a \cdot 2 \cdot (-\cos(\pi \omega)\sin(\pi \omega)) \cdot \pi \). This simplifies to \(-2a \pi \cos(\pi \omega)\sin(\pi \omega)\).

Step 4: Derivative of \( b \sin^{2} \pi \omega \)

Similarly, apply the chain rule: \( \frac{d}{d\omega}(b \sin^{2} \pi \omega) = b \cdot 2 \cdot \sin(\pi \omega)\cos(\pi \omega) \cdot \pi \). This simplifies to \(2b \pi \sin(\pi \omega)\cos(\pi \omega)\).

Step 5: Combine Derivatives

To find the overall derivative, combine the results: \(-2a \pi \cos(\pi \omega)\sin(\pi \omega) + 2b \pi \sin(\pi \omega)\cos(\pi \omega)\).

Step 6: Simplify Expression

Since both terms can be combined, factor out the common terms: \(2\pi \sin(\pi \omega)\cos(\pi \omega)(b - a)\). We know that \( \sin(2x) = 2\sin x \cos x \) so the expression simplifies to \( \pi(b-a)\sin(2\pi \omega) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus when it comes to differentiating composite functions. Essentially, a composite function is a function within another function. For example, if we have a function of the form \( f(g(x)) \), the Chain Rule guides us by telling us to first differentiate \( f \) with respect to \( g \), and then multiply by the derivative of \( g \) with respect to \( x \).

In the given exercise, we saw the function \( a \cos^{2} \pi \omega + b \sin^{2} \pi \omega \). The Chain Rule is applied when we have terms like \( \cos^{2} \pi \omega \). Here, the outer function is \( u^{2} \) where \( u = \cos \pi \omega \), and the inner function is \( \pi \omega \). So, the differentiation proceeds by using the Chain Rule: First, differentiate the outer function \( u^{2} \), and then multiply by the derivative of the inner function \( \cos \pi \omega \) with respect to \( \omega \). This is repeated similarly for \( \sin^{2} \pi \omega \).

Understanding the Chain Rule is crucial for tackling more complex differentiations, especially when they involve multiple layers of functions.

Trigonometric Functions

Trigonometric functions like sine and cosine are essential to many aspects of mathematics and modeling periodic phenomena. Differentiating these functions is a foundational skill that is often required in calculus.

In our task, we work with squared trigonometric functions, which themselves are compositions of functions. Knowing the derivatives of basic trigonometric functions is crucial:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).

When a trigonometric function is squared, additional steps are needed to find its derivative, often involving one of these basic derivatives as well as the application of the Chain Rule.

This exercise specially uses trigonometric identities during simplification. Recognizing identities like \( \sin(2x) = 2 \sin x \cos x \) is helpful for further simplifications after differentiation is completed.

Differentiation Techniques

Differentiation is at the core of calculus, focusing on finding how functions change at any point. This exercise demonstrates several differentiation techniques.

Key techniques involved include:

• The Power Rule: Used generally where an expression involves an exponent, for example, differentiating \( x^n \) results in \( nx^{n-1} \).
• Product Rule: While not directly used here, understanding that multiple terms can be treated individually and summed is useful.
• Chain Rule and Trigonometric Derivatives: As explained, these are pivotal in handling composite functions of trigonometry.

Combining these techniques effectively allows us to differentiate a wide array of functions smoothly. In this problem, after finding derivatives separately, combining and simplifying them using trigonometric identities leads to a clean, understandable result. Mastering these techniques offers powerful tools to tackle both simple and complex derivatives in calculus.