Problem 31 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \sin(\cos(x)) \sin(x) \)

Step 1: Identify the Outer Function

The given function is a composition of functions. Identify the outer function, which is the cosine function: \[ y = oxed{ ext{cos}}( ext{cos}(x)) \]

Step 2: Identify the Inner Function

The inner function is the expression inside the outer function. Identify this as another cosine function: \[ u = ext{cos}(x) \] Thus, \[ y = ext{cos}(u) \]

Step 3: Differentiate the Outer Function with Respect to the Inner Function

Differentiate \(y = ext{cos}(u)\) with respect to \(u\). The derivative of \(\cos(u)\) is \(-\sin(u)\): \[ \frac{d y}{d u} = -\sin(u) \]

Step 4: Differentiate the Inner Function with Respect to \(x\)

Differentiate the inner function \(u = \text{cos}(x)\) with respect to \(x\). The derivative of \(\cos(x)\) is \(-\sin(x)\): \[ \frac{d u}{d x} = -\sin(x) \]

Step 5: Apply the Chain Rule

Use the chain rule to find \( \frac{d y}{d x} \) by multiplying the derivatives found in steps 3 and 4: \[ \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \] Substitute the values: \[ \frac{d y}{d x} = (-\sin(u)) \cdot (-\sin(x)) \]

Step 6: Simplify the Result

Substitute back \(u = \cos(x)\) into the expression and simplify: \[ \frac{d y}{d x} = \sin(\cos(x)) \cdot \sin(x) \] Thus, the derivative of \(y\) with respect to \(x\) is: \[ \frac{d y}{d x} = \sin(\cos(x)) \sin(x) \]

Key Concepts

Chain Rule

The chain rule is a powerful technique in calculus used to differentiate composite functions. A composite function involves one function inside another, like in the expression \( y = \cos(\cos x) \). The chain rule helps us find the derivative of such functions by taking into account both the outer and inner functions.

To apply the chain rule, we first identify the outer function and differentiate it with respect to the inner function. For instance, in \( y = \cos(\cos x) \), we recognize that \( \cos \) is the outer function and its derivative is \( -\sin \.\)

Next, we differentiate the inner function with respect to the variable \( x \). Here, \( \cos x \) is the inner function and its derivative is \( -\sin x \).

Finally, the chain rule states that the derivative of the composite function is the product of the derivative of the outer function and the derivative of the inner function. This is expressed as:

• \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \)

The chain rule allows us to efficiently solve problems involving complex nested functions.

Trigonometric Functions

Trigonometric functions such as cosine, sine, and tangent are fundamental in calculus and appear often in various functions. They are periodic, meaning they repeat their values in regular intervals. Understanding their derivatives is crucial when dealing with calculus problems.

Certain features make trigonometric functions unique:

• They have specific derivatives, like \( \frac{d}{dx} \cos(x) = -\sin(x) \).
• They are used to model waves and oscillations in real-world situations.

When differentiating trigonometric functions, it's essential to recall these basic derivatives. In our exercise where \( y = \cos(\cos x) \), both the outer and inner functions are trigonometric. By knowing the derivative \( \frac{d}{dx} \cos(x) = -\sin(x) \), you can effectively find the solution using the chain rule.

This knowledge not only aids in solving simple derivatives but also in tackling complex trigonometric expressions in calculus.

Composite Functions

Composite functions combine two or more functions into one, creating a more complex expression. They are written as \( y = f(g(x)) \), where \( g(x) \) and \( f(x) \) are individual functions.

The method of decomposing a function into its component functions is crucial. It helps us simplify the differentiation process using the chain rule.

Consider our example \( y = \cos(\cos x) \). Here, the inner function \( g(x) = \cos(x) \) and the outer function \( f(x) = \cos(x) \) combine to form a single composite function.

• First, separate the functions: identify \( u = \cos(x) \).
• Then, apply the outer function: \( y = \cos(u) \).

By breaking down a composite function, we can apply appropriate calculus rules effectively, solving complex problems systematically. Understanding this concept allows for precise manipulations when dealing with nested functions, especially in calculus.