Problem 230 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=\cos ^{3}(\pi x)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -3\pi \cos^2(\pi x) \sin(\pi x) \).

Step 1: Identify the Function and its Composition

The given function is \( y = \cos^3(\pi x) \). This is a composite function where the outer function is \( u^3 \) and the inner function is \( u = \cos(\pi x) \). We have to apply the chain rule to differentiate this.

Step 2: Differentiate the Outer Function

Differentiate the outer function with respect to the inner function \( u \). If \( u = \cos(\pi x) \), then \( y = u^3 \). The derivative \( \frac{d}{du}(u^3) = 3u^2 \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( u = \cos(\pi x) \) with respect to \( x \). The derivative of \( \cos(\pi x) \) is \( -\sin(\pi x) \), and using the chain rule, multiply by the derivative of \( \pi x \) which is \( \pi \). So, \( \frac{du}{dx} = -\pi \sin(\pi x) \).

Step 4: Apply the Chain Rule

Now apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substituting the derivatives found, we have \( \frac{dy}{dx} = 3u^2 \cdot (-\pi \sin(\pi x)) \).

Step 5: Substitute Back the Original Inner Function

Substitute back \( u = \cos(\pi x) \) into the derivative equation: \( \frac{dy}{dx} = 3(\cos(\pi x))^2 \cdot (-\pi \sin(\pi x)) \). Simplifying gives \( \frac{dy}{dx} = -3\pi \cos^2(\pi x) \sin(\pi x) \).

Key Concepts

Chain Rule

The chain rule is an essential calculus tool used when differentiating composite functions. It allows us to find the derivative of a function that is built from another function. We say a function is composite when it can be expressed as one function applied inside another, like nesting a digital picture inside a photo frame.

To use the chain rule, identify two functions: the outer function, which operates on the inner function. You first differentiate the outer function while keeping the inner one unchanged. Then, multiply this by the derivative of the inner function. In simple terms, you break the function into layers, differentiate each layer, and then multiply the results.

In our exercise, we see this with the function:

• Outer function: raising to the power of 3 (notated as \(u^3\))
• Inner function: \(u = \cos(\pi x)\)

Hence, the derivative requires these multiple steps, emphasizing the importance of using the chain rule systematically.

Composite Functions

Composite functions occur when one function is applied to the results of another. You can think of it like a nested box, where opening one reveals another inside. In mathematics, this is expressed as \(f(g(x))\).

Understanding composite functions is crucial before applying the derivative rules, as it identifies how many functions you are dealing with. In our example, \(y = \cos^3(\pi x)\), is composed of two parts:

• \(g(x) = \cos(\pi x)\)
• \(f(u) = u^3\)

Composite functions often appear in calculus problems, so becoming comfortable with their structure helps you apply rules like the chain rule more accurately. Recognizing them is the first step to solving them effectively.

Trigonometric Differentiation

Trigonometric differentiation focuses on differentiating functions that include trigonometric expressions like \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\). These functions are periodic, meaning they repeat after a certain interval, which adds a unique aspect when differentiating.

When handling the derivative of trigonometric functions, each has a standard derivative:

• \(\frac{d}{dx} \sin(x) = \cos(x)\)
• \(\frac{d}{dx} \cos(x) = -\sin(x)\)
• \(\frac{d}{dx} \tan(x) = \sec^2(x)\)

Applying these during differentiation, especially in composite functions, is crucial. For instance, in differentiating \(u = \cos(\pi x)\), you notice the derivative becomes \(-\pi \sin(\pi x)\). The extra \(\pi\) arises from the chain rule (due to \(\pi x\) being another function within \(\cos\)). Thus, understanding these derivatives helps tackle a wide array of calculus problems more easily.