Question

Alternative proof of the Power Rule The Binomial Theorem states that for any positive integer \(n\) $$\begin{aligned} (a+b)^{n}=& a^{n}+n a^{n-1} b+\frac{n(n-1)}{2 \cdot 1} a^{n-2} b^{2} \\ &+\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} a^{n-3} b^{3}+\cdots+n a b^{n-1}+b^{n} \end{aligned}$$ Use this formula and the definition \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) to show that \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\) for any positive integer \(n\)

Short answer

Question: Provide an alternative proof of the Power Rule using the Binomial Theorem and the definition of a derivative using limits. Answer: Following a step-by-step process, we used the Binomial Theorem and the definition of a derivative to prove that the derivative of the function \(f(x) = x^n\) is given by the Power Rule: \(f'(x) = nx^{n-1}\).

Step-by-step solution

Define the function and its derivative

Let \(f(x) = x^n\), where \(n\) is a positive integer. To find its derivative \(f^{\prime}(x)\), we can use the definition of a derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Apply the Binomial Theorem

Now, we need to find an expression for \(f(x+h)\). Since \(f(x) = x^n\), we have \(f(x+h) = (x+h)^n\). Apply the Binomial Theorem to this expression, we get: $$ (x+h)^n = x^n + n x^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n$$

Simplify the derivative expression

Next, substitute the expression for \(f(x+h)\) into the definition of \(f'(x)\) and simplify the numerator: $$\begin{aligned} f'(x) &= \lim_{h\to 0} \frac{(x+h)^n-x^n}{h}\\ &= \lim_{h\to 0} \frac{(x^n + n x^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n )-x^n}{h} \end{aligned}$$ Now, cancel out the \(x^n\) terms and factor out an \(h\) from the remaining terms in the numerator: $$ f'(x) = \lim_{h\to 0} \frac{n x^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n}{h} = \lim_{h\to 0} (nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \cdots + h^{n-1}) $$

Evaluate the limit as h approaches 0

As \(h\) approaches 0, the terms with \(h\) in the expression go to 0: $$f'(x) = nx^{n-1} + 0 + 0 + \cdots + 0 = nx^{n-1}$$

Show that the resulting expression is the Power Rule

By following the steps above, we have shown that the derivative of \(f(x) = x^n\) with respect to \(x\) is: $$f'(x) = nx^{n-1}$$ This expression is indeed the Power Rule for differentiation. Thus, we have successfully provided an alternative proof of the Power Rule using the Binomial Theorem and the definition of the derivative.

Key concepts

Binomial Theorem

The Binomial Theorem is a powerful mathematical tool that expands expressions of the form \((a+b)^n\), where \(n\) is a positive integer. This theorem provides a way to express this expansion in terms of binomial coefficients, and it is written as: \[ (a+b)^n = a^n+n a^{n-1}b+ \frac{n(n-1)}{2}a^{n-2}b^2+ \cdots + b^n \] This formula simplifies the process of expanding powers of binomials into a series sum.

• "\(a^n\)" is the first term with \(b\) raised to the power zero.
• "\(n a^{n-1}b\)" is the second term, where \(b\) is raised to the first power, and so on.
• This pattern continues until the last term "\(b^n\)" where \(a\) is raised to the power zero.

With every term, the power of \(a\) decreases by one, and the power of \(b\) increases by one. Binomial coefficients, such as \(n,\ \frac{n(n-1)}{2}\), are calculated using combinations.

Derivative

The concept of a derivative is fundamental in calculus and is used to analyze the rate at which functions change. In simple terms, a derivative represents the instantaneous rate of change of a function with respect to one of its variables. The standard notation for a derivative of a function \(f(x)\) with respect to \(x\) is \(f'(x)\) or \(\frac{df}{dx}\). To find the derivative, you need to understand the definition \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

• "\(f(x + h)\)" is the function value at a slightly different point.
• "\(f(x)\)" is the original function value.
• "\(h\)" is a very small increment.

The formula represents the slope of the tangent to the curve at any point \(x\). As \(h\) approaches \(0\), you find the exact slope—the derivative.

Differentiation

Differentiation is the process through which you calculate the derivative of a function. It allows you to determine how a function changes at any given point, and it's a crucial operation in calculus. Differentiation can be applied to various kinds of functions, including polynomial, trigonometric, exponential, and more. The Power Rule, a basic tool for differentiation, simplifies finding the derivative of functions of the form \(x^n\). According to this rule: \[ \frac{d}{dx}x^n = nx^{n-1} \] This rule states that to differentiate \(x^n\), you multiply by the exponent \(n\) and reduce the exponent by one. Differentiation is valuable for finding:

• Rates of change across various fields.
• Velocity in physics or financial models.
• Optimization results to determine maxima or minima of functions.

Limit

The concept of a limit is fundamental to understanding calculus and derivatives. A limit explores the value that a function "approaches" as its input approaches a certain point. It is not necessarily the value the function reaches but the value it gets closer to as the process continues. Mathematically, if \(f(x)\) approaches a number \(L\) as \(x\) approaches \(c\), then you write: \[ \lim_{x \to c} f(x) = L \] Limits are crucial when you calculate derivatives or deal with indeterminate forms. In the context of derivatives, you need limits to find the exact slope of the tangent line at a point, transforming the "difference quotient" into a specific value.

• Limits help understand behavior near points where functions are not easily defined.
• They deal with infinitesimally small quantities, providing precise values.
• They form the foundation for further calculus operations such as integration.

Learning and applying the concept of limits is key to effective differentiation and more complex calculus concepts.