Problem 12 - Calculus for Biology and Medicine

Question

Find \(f^{(n)}(x)\) and \(f^{(n+1)}(x)\) if \(f(x)=x^{n}\).

Step-by-Step Solution

Answer

For \( f(x)=x^n \), \( f^{(n)}(x)=n! \) and \( f^{(n+1)}(x)=0 \).

Step 1: Understanding the Problem

We need to find the general expression for the first derivative, denoted as \( f'(x) \), of the function \( f(x) = x^n \). Then, we need to express the \( n^{th} \) and \( (n+1)^{th} \) derivatives, \( f^{(n)}(x) \) and \( f^{(n+1)}(x) \), respectively.

Step 2: Find the First Derivative

To find \( f'(x) \), apply the power rule for derivatives, which states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

Step 3: Identify Pattern in Successive Derivatives

The pattern in successive derivatives can be observed by continuing differentiation: - Second derivative \( f''(x) = n(n-1)x^{n-2} \).- Third derivative \( f'''(x) = n(n-1)(n-2)x^{n-3} \).This continues, forming a factorial pattern as \( n(n-1)(n-2)...(n-(k-1))x^{n-k} \) for the \( k^{th} \) derivative.

Step 4: Write the Formula for the nth Derivative

Based on the observed pattern, the \( n^{th} \) derivative of \( f(x) = x^n \) is given by:\[ f^{(n)}(x) = n! \cdot x^{n-n} = n! \cdot x^0 = n!\]This result holds for all non-negative integers \( n \).

Step 5: Find the (n+1)th Derivative

To find \( f^{(n+1)}(x) \), apply the derivative one more time to \( f^{(n)}(x) \):\[f^{(n+1)}(x) = \frac{d}{dx}(n!) = 0\]This holds true because the derivative of a constant (in this case, \( n! \)) is zero.

Key Concepts

Understanding the Power Rule in Calculus

In calculus, the power rule is a fundamental technique for differentiation. It's a simple formula used to find the derivative of a term with a variable raised to a power. The power rule states that if you have a function given by \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \). This means you multiply the original power by the coefficient (which is often 1 in basics) then reduce the power by one.

For example, consider the function \( f(x) = x^3 \). Using the power rule, the derivative is \( f'(x) = 3x^2 \). Here, the original power 3 is brought down as a coefficient and the new exponent becomes 2 (one less than the original 3).

The power rule is extremely useful because:

• It simplifies finding derivatives for polynomial functions.
• It's one of the first rules taught in calculus due to its simplicity and utility.
• Helps in understanding more complex derivative rules.

By mastering the power rule, you build a strong foundation for tackling more advanced calculus topics.

Exploring Factorial in Calculus

Factorials play a crucial role in calculus, particularly when dealing with derivatives and combinatorics within mathematics. A factorial, denoted by \( n! \), represents the product of all positive integers up to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are important when examining sequences and series or calculating permutations and combinations.

In differentiation, a factorial appears naturally when finding successive derivatives of polynomial functions. As seen in the solution, for a function \( f(x) = x^n \), the \( n^{th} \) derivative leads to the expression \( f^{(n)}(x) = n! \).

This is due to the repetitive multiplication of decreasing integers as you continue differentiating:

• The first derivative involves \( n \).
• The second derivative involves \( n(n-1) \).
• The third derivative involves \( n(n-1)(n-2) \).

The pattern concludes when the power is reduced to zero, resulting in a constant of \( n! \). Understanding factorials is essential as many advanced calculus problems rely on this compact and efficient notation.

The Concept of Successive Differentiation

Successive differentiation involves taking derivatives of a function multiple times. It is a process of repeated application of the differentiation rules on a given function, allowing us to find higher-order derivatives.

When tasked with finding \( f^{(n)}(x) \) for the function \( f(x) = x^n \), we observe a pattern that emerges with every derivative.

Here's how it unfolds:

• First derivative: \( nx^{n-1} \)
• Second derivative: \( n(n-1)x^{n-2} \)
• Third derivative: \( n(n-1)(n-2)x^{n-3} \)

This continues until we reach the \( n^{th} \) derivative which is simply \( n! \) because after differentiating \( n \) times, all variable terms will be canceled leaving only the factorial.

Finally, the \( (n+1)^{th} \) derivative becomes 0, as differentiating the constant \( n! \) results in zero.

Successive differentiation highlights the patterns formed in derivative sequences and helps us understand the behavior of polynomial functions deeply, making it a valuable tool in calculus.