proof of the power ruleProofOfPowerRule2002-02-24 21:19:192006-12-09 04:53:35Proofpower rule0\usepackage{amssymb}
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\newcommand{\D}[1]{\ensuremath{\mathrm{d}#1}}
\newcommand{\DDX}{\ensuremath{\frac{\D{}}{\D{x}}}}The power rule can be derived by repeated application of the product rule.
\subsubsection*{Proof for all positive integers $n$}
The power rule has been shown to hold for $n = 0$ and $n = 1$.
If the power rule is known to hold for some $k > 0$, then we have
\begin{eqnarray*}
\DDX x^{k+1}
& = & \DDX (x\cdot x^k) \\
& = & x\left(\DDX x^k\right) + x^k \\
& = & x\cdot(kx^{k-1}) + x^k \\
& = & kx^k + x^k \\
& = & (k + 1)x^k
\end{eqnarray*}
Thus the power rule holds for all positive integers $n$.
\subsubsection*{Proof for all positive rationals $n$}
Let $y = x^{p/q}$. We need to show
\begin{equation}
\frac{\D{y}}{\D{x}}(x^{p/q}) = \frac{p}{q}x^{p/q-1} \label{prat}
\end{equation}
The proof of this comes from implicit differentiation.
By definition, we have $y^q = x^p$. We now take the derivative with respect to $x$ on both sides of the equality.
\begin{eqnarray*}
\DDX y^q & = & \DDX x^p \\
\frac{\D{}}{\D{y}}(y^q)\frac{\D{y}}{\D{x}} & = & px^{p-1} \\
qy^{q-1}\frac{\D{y}}{\D{x}} & = & px^{p-1} \\
\frac{\D{y}}{\D{x}} & = & \frac{p}{q}\frac{x^{p-1}}{y^{q-1}} \\
&=& \frac{p}{q}x^{p-1}y^{1-q} \\
&=& \frac{p}{q}x^{p-1}x^{p(1-q)/q} \\
&=& \frac{p}{q}x^{p-1+p/q-p} \\
&=& \frac{p}{q}x^{p/q-1}
\end{eqnarray*}
\subsubsection*{Proof for all positive irrationals $n$}
For positive irrationals we claim continuity due to the fact that (\ref{prat})
holds for all positive rationals, and there are positive rationals that approach any positive irrational.
\subsubsection*{Proof for negative powers $n$}
We again employ implicit differentiation. Let $u = x$, and differentiate $u^n$ with respect to $x$ for some non-negative $n$. We must show
\begin{equation}
\frac{\D{u^{-n}}}{\D{x}} = -nu^{-n-1} \label{neg}
\end{equation}
By definition we have $u^nu^{-n} = 1$.
We begin by taking the derivative with respect to $x$ on both sides of the equality. By application of the product rule we get
\begin{eqnarray*}
\DDX(u^nu^{-n}) & = & 1 \\
u^n\frac{\D{u^{-n}}}{\D{x}} + u^{-n}\frac{\D{u^n}}{\D{x}} & = & 0\, \\
u^n\frac{\D{u^{-n}}}{\D{x}} + u^{-n}(nu^{n-1}) & = & 0\, \\
u^n\frac{\D{u^{-n}}}{\D{x}} & = & -nu^{-1} \\
\frac{\D{u^{-n}}}{\D{x}} & = & -nu^{-n-1}
\end{eqnarray*}