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Derivative of Power Function

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By Pimath, 5 June, 2025

In this page we will see how to compute the derivative of the power function using two equivalent forms of the difference quotient: one with the variable \(h\), where \(h\to 0\), and one with the variable \(x\), where \(x\to x_0\).

Let \(n\in\mathbb{N}\setminus\{0\}\), and consider the power function:

\[ f(x)=x^n \]

The two forms of the difference quotient are:

\[ \lim_{h\to 0}\frac{(x+h)^n-x^n}{h} \qquad , \qquad \lim_{x\to x_0}\frac{x^n-x_0^n}{x-x_0} \]


Table of Contents

  • Difference quotient for \( h\to 0 \)
  • Difference quotient for \( x\to x_0 \)

Difference quotient for \( h\to 0 \)

We compute the derivative of the power function using the definition of the difference quotient:

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

Substituting \(f(x)=x^n\), we obtain:

\[ f'(x) = \lim_{h\to 0} \frac{(x+h)^n-x^n}{h} \]

We now apply the binomial theorem:

\[ (x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n \]

Substituting the binomial expansion into the difference quotient:

\[ f'(x) = \lim_{h\to 0} \frac{ x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n - x^n }{h} \]

Simplifying the terms \(x^n\):

\[ f'(x) = \lim_{h\to 0} \frac{ nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n }{h} \]

Dividing each term by \(h\):

\[ f'(x) = \lim_{h\to 0} \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \frac{n(n-1)(n-2)}{6}x^{n-3}h^2 + \cdots + h^{n-1} \right) \]

Passing to the limit as \(h\to 0\), all terms containing positive powers of \(h\) tend to \(0\). Therefore:

\[ f'(x) = nx^{n-1} \]

We therefore conclude that:

\[ \frac{d}{dx}x^n = nx^{n-1} \qquad , \qquad \forall x\in\mathbb{R} \]


Difference quotient for \( x\to x_0 \)

Let us now compute the derivative of the power function in the form:

\[ f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} \]

Substituting \(f(x)=x^n\):

\[ f'(x_0) = \lim_{x\to x_0} \frac{x^n-x_0^n}{x-x_0} \]

The numerator is a difference of powers. We therefore use the factorization:

\[ x^n-x_0^n = (x-x_0) \left( x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1} \right) \]

Substituting into the difference quotient:

\[ f'(x_0) = \lim_{x\to x_0} \frac{ (x-x_0) \left( x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1} \right) }{x-x_0} \]

Simplifying the factor \(x-x_0\):

\[ f'(x_0) = \lim_{x\to x_0} \left( x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1} \right) \]

Passing to the limit as \(x\to x_0\), each term tends to \(x_0^{\,n-1}\). Since there are \(n\) terms equal to \(x_0^{\,n-1}\), we obtain:

\[ f'(x_0) = nx_0^{\,n-1} \]

In conclusion:

\[ f'(x) = nx^{n-1} \qquad , \qquad \forall x\in\mathbb{R} \]


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