Question

In this section, we showed that the rule \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) is valid for what values of \(n ?\)

Short answer

Answer: The derivative rule \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) is valid for all values of n.

Step-by-step solution

Recall the definition of the derivative

The derivative of a function f(x) with respect to x is given by: \[f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}\]

Apply the definition of the derivative to the function \(f(x) = x^n\)

In this case, \(f(x) = x^n\). Let's apply the definition of the derivative: \[f'(x) = \lim_{h \to 0}\frac{(x+h)^n - x^n}{h}\]

Use the binomial theorem to expand \((x+h)^n\)

According to the binomial theorem, we have: \[(x+h)^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}h^k\] Now, we can substitute this expansion into the expression for the derivative: \[f'(x) = \lim_{h \to 0}\frac{\sum_{k=0}^{n}\binom{n}{k}x^{n-k}h^k - x^n}{h}\]

Simplify the expression for the derivative

Let's split the summation and simplify: \[f'(x) = \lim_{h \to 0}\frac{\binom{n}{0}x^{n}h^0 + \sum_{k=1}^{n}\binom{n}{k}x^{n-k}h^k - x^n}{h}\] Notice that \(\binom{n}{0}x^{n}h^0 = x^n\) and \(h^0 = 1\). So the expression becomes: \[f'(x) = \lim_{h \to 0}\frac{\sum_{k=1}^{n}\binom{n}{k}x^{n-k}h^k}{h}\]

Evaluate the limit as \(h \to 0\)

When we take the limit as h approaches 0, all terms with a factor of h in the numerator will tend to 0. The only term that remains is the one with k=1: \[f'(x) = \left.\binom{n}{1}x^{n-1}h^0\right|_{h = 0} = nx^{n-1}\] Thus, the derivative rule \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) is valid for all values of n, since we didn't impose any restrictions on the values of n during our derivations.

Key concepts

Derivative Definition

In calculus, one of the foundational concepts is the derivative. A derivative measures how a function changes as its input changes. In a formal sense, the derivative of a function with respect to a variable gives the slope of the tangent line to the graph of the function at a particular point. This idea is captured by the limit definition of the derivative.Let's say we have a function \( f(x) \). The derivative \( f'(x) \) is expressed as:

• \( f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \)

This formula calculates the instantaneous rate of change of the function, providing insight into its behavior. The \( h \) represents a tiny increment, and as it approaches zero, the fraction approximates the slope at a particular point.Understanding this limit process helps us comprehend why derivatives reflect the instantaneous rate of change at a point, making it a versatile tool for solving various problems, such as finding maximum and minimum values of functions.
By grasping this concept, we lay the groundwork for further exploration into more complex calculus ideas.

Binomial Theorem

The Binomial Theorem is another fundamental idea in mathematics that simplifies the process of expanding expressions that are raised to a power. It provides a formula to expand binomials into a series of terms, making calculations more manageable.The theorem states that for any natural number \( n \),

• \((x + h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k \)

This expansion expresses \((x+h)^n\) as a sum of terms involving binomial coefficients \( \binom{n}{k} \), which are calculated as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Each term in the expansion represents contributions from the powers of \( x \) and \( h \).In calculus, the Binomial Theorem assists in the differentiation process by helping to expand expressions like \( (x+h)^n \), which appear in derivative calculations. This expansion allows us to simplify expressions, especially when applying limits. By mastering this theorem, students can handle complex algebraic manipulations required in various advanced mathematical tasks.

Limit Evaluation

Evaluating limits is an essential skill in calculus, as limits form the foundation of many core concepts, including derivatives and integrals. The process of evaluating a limit helps us understand the behavior of a function as its variables approach certain values.Consider a simple limit:

• \( \lim_{h \to 0} \frac{(x+h)^n - x^n}{h} \)

This expression is crucial for finding the derivative of \( x^n \). The key step in evaluating this limit is simplifying the expression, often achieved by expanding terms using techniques like the Binomial Theorem, and subsequently canceling out terms to isolate relevant parts.The essence of limit evaluation lies in carefully analyzing and simplifying the expression, ensuring all indeterminate forms are resolved before taking the limit itself. In most cases, terms involving \( h \), in the numerator will disappear as \( h \) approaches zero, highlighting the term of interest.Accuracy and attention to detail while solving limits can significantly affect the success of applying derivatives or integrals, making it a pivotal skill to master early in calculus study.