Power Rule for Derivatives/Natural Number Index/Proof by Binomial Theorem

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Theorem

Let $n \in \N$.

Let $f: \R \to \R$ be the real function defined as $f \left({x}\right) = x^n$.


Then:

$f' \left({x}\right) = n x^{n-1}$

everywhere that $f \left({x}\right) = x^n$ is defined.


When $x = 0$ and $n = 0$, $f^{\prime} \left({x}\right)$ is undefined.


Proof

Let $f \left({x}\right) = x^n$ for $x \in \R, n \in \N$.

By the definition of the derivative:

$\displaystyle \dfrac \d {\d x} f \left({x}\right) = \lim_{h \mathop \to 0} \dfrac {f \left({x + h}\right) - f \left({x}\right)} h = \lim_{h \mathop \to 0} \dfrac{(x + h)^n - x^n} h$


Using the binomial theorem this simplifies to:

\(\displaystyle \) \(\) \(\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 0} x^n + {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n - x^n} h}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n - 1} x h^{n - 1} + {n \choose n} h^n} h}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \left({ {n \choose 1} x^{n - 1} + {n \choose 2} x^{n - 2} h^1 + \cdots + {n \choose n - 1} x h^{n - 2} + {n \choose n} h^{n - 1} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle {n \choose 1} x^{n - 1}\) evaluating the limit
\(\displaystyle \) \(=\) \(\displaystyle n x^{n - 1}\) Binomial Coefficient with One: $\dbinom r 1 = r$

$\blacksquare$