Problem 31 - Calculus - AP Edition

Question

Find \(d y / d x\) $$y=\cos (\cos x)$$

Step-by-Step Solution

Answer

\(dy/dx = \sin(\cos(x)) \sin(x)\)

Step 1: Identify the Outer Function

The function given is a composition of two functions. The outer function is \( y = \cos(u) \), where \( u = \cos(x) \). First, we need to find the derivative of the outer function with respect to \( u \).

Step 2: Derivative of the Outer Function

The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \). Therefore, we have \( \frac{d}{du}(\cos u) = -\sin(u) \).

Step 3: Identify the Inner Function

The inner function is \( u = \cos(x) \). We need to find the derivative of this inner function with respect to \( x \).

Step 4: Derivative of the Inner Function

The derivative of \( \cos(x) \) with respect to \( x \) is \( -\sin(x) \). So, we have \( \frac{d}{dx}(\cos x) = -\sin(x) \).

Step 5: Apply the Chain Rule

According to the chain rule, the derivative of \( y = \cos(\cos(x)) \) with respect to \( x \) is the product of the derivatives found in Steps 2 and 4. Hence, \( \frac{dy}{dx} = \frac{d}{du}(\cos u) \cdot \frac{d}{dx}(u) \) which simplifies to \( \frac{dy}{dx} = -\sin(\cos(x)) \cdot (-\sin(x)) \).

Step 6: Simplify the Expression

Simplifying \(-\sin(\cos(x)) \cdot (-\sin(x))\) gives \( \sin(\cos(x)) \cdot \sin(x) \). Therefore, the derivative of \( y = \cos(\cos(x)) \) with respect to \( x \) is \( \sin(\cos(x)) \sin(x) \).

Key Concepts

Derivative of Trigonometric Functions

Taking the derivative of trigonometric functions is a fundamental skill in calculus. It's important as these functions appear frequently in both theoretical and applied math problems.
When we talk about the derivative of trigonometric functions, we are considering how these functions change as their inputs change. Common examples include derivatives of the sine, cosine, and tangent functions.

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

Understanding these basic derivatives allows us to tackle more complex problems, like those involving composite functions or the application of the chain rule.
Mastering these derivatives enables you to solve a wide variety of problems in calculus, which often extend to real-world applications, such as in physics and engineering.

Composite Functions

Composite functions occur when one function is nested within another. For example, in the function \( y = \cos(\cos(x)) \), the inner function \( u = \cos(x) \) is wrapped by the outer function \( y = \cos(u) \).
To tackle derivatives involving composite functions, we typically use the chain rule. The chain rule is a formula to compute the derivative of a composite function. It states:
If a function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Where \( \frac{dy}{du} \) is the derivative of the outer function with respect to the inner function, and \( \frac{du}{dx} \) is the derivative of the inner function with respect to \( x \).
Recognizing the presence of composite functions and applying the chain rule is key to finding solutions to more complex derivative problems.

Calculus Problem Solving

Calculus problem solving requires a step-by-step approach that combines various techniques depending on the problem at hand. Let's break down a typical approach to solving a derivative problem involving composite functions:

• Identify the functions: Determine which function is the outer and which is the inner. This will set the stage for applying the chain rule.
  
• Differentiate the inner function: Find the derivative of the inner function with respect to its variable.
  
• Differentiate the outer function: Determine the derivative of the outer function with respect to the inner function.
  
• Apply the chain rule: Use the results from the previous steps to find the overall derivative. This involves multiplying the derivative of the outer function by the derivative of the inner function.
  
• Simplify: As the final step, always simplify your expression to make it more readable and interpretable.
  

By systematically applying these steps, you can confidently solve calculus problems that involve composite functions and the chain rule. This methodical approach is crucial for accurately handling more complicated expressions in calculus.