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% If you make a derivative work, please add a credits page and list me as the original author.
% -- James Sumners <james DOT sumners AT gmail DOT com>

\documentclass[11pt]{article}
\title{An Interesting Proof by Induction of Derivation}
\author{James Sumners}
\date{\today}

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%\parindent  15.mm % indent paragraph by this much
\parskip     2.mm   % space between paragraphs
\usepackage{amsmath, amsthm, amssymb} % Allows us to redefine theorem styles and easily write proofs.

\begin{document}
\maketitle

Problem:
\begin{em}
Use induction and the product rule to prove that if $f(x)$ is a differntiable function, then
\begin{eqnarray*}
(f(x)^{m})'=mf(x)^{m-1}f'(x),
\end{eqnarray*}
for all integers $m\ge1$.
\end{em}

\begin{proof}
\mbox{} % I like my proofs to start on the next line.

Consider the case where $m=1$. Then $(f(x)^{1})'=1f(x)^{1-1}f'(x) \rightarrow f'(x)=f'(x)$.

Now, assume $(f(x)^{m})'=mf(x)^{m-1}f'(x)$ for some integer $m>1$. Then the left hand side of the equation is $(f(x)^{m+1})'=(f(x)^{m}f(x))'=(f(x)^{m})'f(x)+f(x)^{m}f'(x)=[mf(x)^{m-1}f'(x)]f(x)+f(x)^{m}f'(x)=mf(x)^{m}f'(x)+f(x)^{m}f'(x)=(m+1)f(x)^{m}f'(x)$ for $m+1$.
Similarly, the right hand side of the equation is $(m+1)f(x)^{(m+1)-1}f'(x)=(m+1)f(x)^{m}f'(x)$ for $m+1$.

Therefore, if $f(x)$ is a differentiable function then $(f(x)^{m})'=mf(x)^{m-1}f'(x)$ for all integers $m\ge1$.
\mbox{}\newline
\end{proof}

\end{document}