Problem 71 - Calculus Single Variable

Question

Find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=\sin ^{2}\left(\pi \sin \left(\pi x^{2}\right)\right) $$

Step-by-Step Solution

Answer

\(f'(x) = 2\pi^2 x \sin(2\pi \sin(\pi x^2)) \cos(\pi x^2)\)

Step 1: Understand the Function

The function given is composite and involves multiple nested functions: an outer sine squared function, an inner sine function multiplying by \( \pi \), and the innermost \( x^2 \). We need to use the chain rule to differentiate it step by step.

Step 2: Use the Chain Rule

The chain rule states that \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). For \( f(x) = \sin^2(u) \), where \( u = \pi \sin(v) \), and \( v = \pi x^2 \), we differentiate each part separately and multiply them together.

Step 3: Differentiate the Outer Function

Let \( u = \pi \sin(v) \), and \( f(u) = \sin^2(u) \). Then the derivative \( f'(u) \) is \( 2\sin(u)\cos(u) \) by using the derivative of \( \sin^2(u) \) which is \( 2\sin(u) \cos(u) \). Using the identity \( \sin(2u) = 2\sin(u)\cos(u) \), \( f'(u) = \sin(2u) \).

Step 4: Differentiate the Middle Function

For \( u = \pi \sin(v) \) where \( v = \pi x^2 \), the derivative \( \frac{du}{dv} = \pi \cos(v) \).

Step 5: Differentiate the Innermost Function

For \( v = \pi x^2 \), the derivative \( \frac{dv}{dx} = 2\pi x \).

Step 6: Combine the Derivatives Using the Chain Rule

The derivative of the entire function will be the product of Step 3, Step 4, and Step 5: \[ f'(x) = \sin(2\pi \sin(\pi x^2)) \cdot \pi \cos(\pi x^2) \cdot 2\pi x \] Simplifying gives:\[ f'(x) = 2\pi^2 x \sin(2\pi \sin(\pi x^2)) \cos(\pi x^2) \]

Key Concepts

Understanding the Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions, allowing us to break down complicated expressions into more manageable parts. Imagine nesting functions within other functions, like Russian dolls. Each layer (function) affects the one inside it. To understand it, think of a function \( f(g(x)) \), where \( g(x) \) is a function within \( f \). The chain rule formula is:

• \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).

This tells us that to find the derivative of a composite function, we need the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. When applied in steps, this rule transforms intimidating problems into simpler, stacked problems.
In our exercise, the function \( f(x) = \sin^2(\pi \sin(\pi x^2)) \) has multiple layers. The chain rule lets us peel back and differentiate each function sequentially.
This way, the chain rule becomes like a careful, systematic peeling back of nested layers to differentiate everything correctly.

Exploring Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) are crucial to understanding oscillating systems, waves, and circular motion. In calculus, they have distinct derivatives:

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

These formulas help us find the rate at which these functions change. In our function, \( f(u) = \sin^2(u) \), requires us to use the identity for the derivative: \( 2\sin(u) \cos(u) \), simplified using \( \sin(2u) = 2\sin(u)\cos(u) \).
This identity helps us tackle the doubled angles encountered in more complex trigonometric expressions.
Understanding these basic derivatives allows us to efficiently handle multiple layers of trigonometric functions occurring in our nested function without losing track of the process. It makes our work precise while handling oscillations and cycles in functions.

Navigating Composite Functions

Composite functions are like intricate machines made of several smaller, interconnected parts, each influencing the overall function. To break down these complex entities, consider a function \( f(x) \) being expressed as \( f(g(h(x))) \). Here, you have an innermost function \( h(x) \), a middle function \( g(x) \), and an outer function \( f(x) \).
The challenge is to differentiate such intertwined expressions correctly.
To embrace composite functions, you need to identify each layer and differentiate it individually. Start with the innermost function and work outward, applying the chain rule at each stage.
In our specific problem, the innermost function was \( v = \pi x^2 \), followed by the middle \( u = \pi \sin(v) \), leading to the outer \( \sin^2(u) \). This step-by-step method reveals just how interconnected these functions are, shedding light on their overall behavior.
Approaching composite functions requires patience and careful analysis, but the payoff is a clear understanding of how different mathematical operations combine to form richly layered functions.