Problem 27 - Calculus Single Variable

Question

Calculate the requested derivative. \(f^{(5)}(x)\) where \(f(x)=\sin (x)\)

Step-by-Step Solution

Answer

The 5th derivative of \(f(x) = \sin(x)\) is \(f^{(5)}(x) = \cos(x)\).

Step 1: Understanding the problem

We need to calculate the 5th derivative of the function \(f(x) = \sin(x)\), denoted as \(f^{(5)}(x)\). The five signifies that we need to differentiate the function five times.

Step 2: Calculate the first derivative

To find the first derivative of \(f(x) = \sin(x)\), we apply the derivative rule for sine: \(f'(x) = \cos(x)\).

Step 3: Calculate the second derivative

Now differentiate \(f'(x) = \cos(x)\): The derivative of \(\cos(x)\) is \(-\sin(x)\). So, \(f''(x) = -\sin(x)\).

Step 4: Calculate the third derivative

Differentiate \(f''(x) = -\sin(x)\): We get \(f'''(x) = -\cos(x)\) because the derivative of \(-\sin(x)\) is \(-\cos(x)\).

Step 5: Calculate the fourth derivative

Differentiate \(f'''(x) = -\cos(x)\): The derivative of \(-\cos(x)\) is \(\sin(x)\). So, \(f^{(4)}(x) = \sin(x)\).

Step 6: Calculate the fifth derivative

Differentiate \(f^{(4)}(x) = \sin(x)\) once more. As in Step 2, the derivative of \(\sin(x)\) is \(\cos(x)\). Hence, \(f^{(5)}(x) = \cos(x)\).

Key Concepts

Trigonometric Functions

Trigonometric functions are the backbone of many mathematical concepts, especially in calculus. These functions, such as \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\), represent relationships between angles and sides of triangles. In calculus, we focus on how these functions change, which involves determining their derivatives. Understanding the periodic nature of trigonometric functions is crucial. They repeat their values in regular intervals, a pattern known as periodicity.
For instance, \(\sin(x)\) is periodic with period \(2\pi\), meaning \(\sin(x + 2\pi) = \sin(x)\). This property significantly affects the behavior of their derivatives, often leading to cyclic patterns in higher-order derivatives.

Higher Order Derivatives

Higher order derivatives involve taking the derivative of a function multiple times. When you calculate higher order derivatives, like the fifth derivative in our original problem, the key is to recognize any repeating patterns that emerge.
For the function \(f(x) = \sin(x)\), the derivatives exhibit a cycle every four derivatives:

• 1st derivative: \(f'(x) = \cos(x)\)
• 2nd derivative: \(f''(x) = -\sin(x)\)
• 3rd derivative: \(f'''(x) = -\cos(x)\)
• 4th derivative: \(f^{(4)}(x) = \sin(x)\)

Notice that the fifth derivative starts the cycle again, \(f^{(5)}(x) = \cos(x)\). Each complete cycle mirrors the original function and its derivatives' relationships, showcasing the harmony found in trigonometric calculus.

Differentiation Rules

Differentiation rules are guiding principles that simplify the process of finding derivatives. For trigonometric functions, these rules help us determine how the functions change concerning one another. Let's focus on a key rule:
The derivative of \(\sin(x)\) is \(\cos(x)\) and vice versa with a negative sign: the derivative of \(\cos(x)\) is \(-\sin(x)\). This alternating nature makes trigonometric functions unique and often easier to differentiate repeatedly.

• General Rule: \(\frac{d}{dx} [\sin(x)] = \cos(x)\)
• General Rule: \(\frac{d}{dx} [\cos(x)] = -\sin(x)\)

When faced with a problem requiring multiple derivatives, it becomes efficient to remember these basic rules, because they form the base for calculating higher derivatives efficiently. Understanding these rules allows for quick determination of higher order derivatives without recalculating each step.