Problem 33 - Calculus - AP Edition

Question

Find \(d y / d x\) $$y=\cos ^{3}(\sin 2 x)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -6[\cos(\sin(2x))]^2 \sin(\sin(2x)) \cos(2x) \).

Step 1: Identify the Outer Function

In the given equation, we have a composition of functions. The outermost function is the cube, i.e., \[ y = [ ext{something}]^3 \]. We'll differentiate this outer function first.

Step 2: Use the Chain Rule on the Outer Function

Apply the chain rule to the outer function. Differentiating \[ y = u^3 \] with respect to \( u \) gives \[ \frac{d}{du} (u^3) = 3u^2 \].Here \( u = \cos(\sin(2x)) \). The derivative becomes \[ \frac{dy}{du} = 3[\cos(\sin(2x))]^2 \].

Step 3: Identify the Middle Function

The middle function to differentiate is \( \cos(\sin(2x)) \). Identify \( f(v) = \cos(v) \) with \( v = \sin(2x) \).

Step 4: Use the Chain Rule on the Middle Function

Now differentiate the middle function \( \cos(v) \) with respect to \( v \). The derivative is \[ \frac{d}{dv} (\cos(v)) = -\sin(v) \].With \( v = \sin(2x) \), this gives \[ \frac{df}{dv} = -\sin(\sin(2x)) \].

Step 5: Identify and Differentiate the Inner Function

The innermost function is \( \sin(2x) \). The derivative of \( \sin(u) \) is \[ \frac{d}{du} (\sin(u)) = \cos(u) \].Furthermore, apply the chain rule again to \( 2x \). Differentiating gives \[ \frac{d}{dx} [2x] = 2 \].Combine these to get \[ \frac{dg}{dx} = 2 \cos(2x) \].

Step 6: Combine all Derivatives Using the Chain Rule

According to the chain rule, use the derivative results from each function: 1. For the outer function: \[ 3[\cos(\sin(2x))]^2 \] 2. For the middle function: \[ -\sin(\sin(2x)) \] 3. For the inner function: \[ 2 \cos(2x) \] The full derivative is: \[\frac{dy}{dx} = 3[\cos(\sin(2x))]^2 \cdot -\sin(\sin(2x)) \cdot 2 \cos(2x)\],which simplifies to \[\frac{dy}{dx} = -6[\cos(\sin(2x))]^2 \sin(\sin(2x)) \cos(2x). \]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. It allows us to take the derivative of a "chain" of functions by differentiating each part and then multiplying the results. When you have a function nested inside another function, the chain rule helps separate these layers.

Consider a function expressed as \( y = f(g(x)) \). To find the derivative of \( y \) with respect to \( x \), you first differentiate \( f \) with respect to \( g \), and then multiply by the derivative of \( g \) with respect to \( x \).

• Formula: \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)

In our exercise, we utilized the chain rule multiple times because of the layers of the function: the cube, cosine, and sine functions. This approach allows us to systematically differentiate each layer:

• Outer Layer: We started with the cube function \( y = u^3 \). Here, the chain rule was applied by differentiating the power with respect to \( u \), resulting in \( 3u^2 \).
• Middle Layer: Our inner \( u \) became \( \cos(v) \), requiring the derivative of cosine, resulting in \( -\sin(v) \).
• Innermost Layer: The variable \( v = \sin(2x) \) needed another application of the chain rule, as \( 2x \) demanded a derivative, producing \( 2 \cos(2x) \).

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in calculus, appearing frequently in differentiation problems. These functions have unique properties with well-defined derivatives, making their differentiation relatively straightforward.

Here's a quick refresher on derivatives of common trigonometric functions:

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

In our given problem, we worked with the function \( \cos(\sin(2x)) \). This required differentiation through nested trigonometric functions. First, understanding the chain rule applications:

• For \( \cos(\sin(2x)) \), the derivative required recognizing \( \cos \) as our middle function. The derivative of \( \cos(v) \) is \( -\sin(v) \).
• The next layer, \( \sin(2x) \), needed its own derivative calculations, using \( \cos(2x) \) as its output.

These steps highlight how we unwind complex trigonometric functions into simpler parts to apply differentiation rules effectively.

Implicit Differentiation

Implicit differentiation is a powerful technique used when a function is not explicitly solved for one variable in terms of another. Although it is not directly used in our example, understanding this concept helps clarify solving advanced differentiation problems.

We use implicit differentiation when dealing with equations like \( x^2 + y^2 = 1 \). Functions are mixed, so we differentiate both sides with respect to a chosen variable, usually \( x \). Treat all other variables as functions of \( x \) itself.

• The goal is to apply the chain rule to these implicit functions, solving for the derivative of interest.

For a quick guide:

• Differentiate all terms step by step, keeping track of \( dy/dx \) for terms involving \( y \).
• Collect all terms with \( dy/dx \) on one side of the equation.
• Solve for \( dy/dx \) by isolating it on one side.

With the explicit differentiation seen in the exercise above, knowing implicit techniques underscores how powerful the chain rule is across both explicit and implicit equations, giving you more tools to tackle even more complex differentiation problems.