Power Rule for Derivatives/Natural Number Index/Proof by Induction

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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

We will use the notation $D f \left({x}\right) = f^{\prime} \left({x}\right)$ as it is convenient.


Let $n = 0$.

Then $\forall x \in \R: x^n = 1$.

Thus $f \left({x}\right)$ is the constant function $f_1 \left({x}\right)$ on $\R$.


Thus from Derivative of Constant, $D f \left({x}\right) = D \left({x^0}\right) = 0 x^{-1}$, except where $x = 0$.

So the result holds for $n = 0$.


Let $n = 1$.

Then:

$\forall x \in \R: f \left({x}\right) = x^n = x$

Then from Derivative of Identity Function:

$D \left({x}\right) = 1 = 1 \cdot x^{1-1}$

So the result holds for $n = 1$.


Now assume $D \left({x^k}\right) = k x^{k-1}$.

Then by the Product Rule for Derivatives:

$D \left({x^{k+1}}\right) = D \left({x^k x}\right) = x^k D \left({x}\right) + D \left({x^k}\right) x = x^k \cdot 1 + k x^{k-1} x = \left({k+1}\right) x^k$

The result follows by induction.

$\blacksquare$


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