The Quotient Rule - Proofs (For Green's MAT114 Class)

limit-definition.tex

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\iffalse
Josh Pruim wrote this. For Green's class. In October.
\fi
\noindent
The Limit Definition of a Derivative is as follows
\[
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.
\]
The function \textit{f} is defined below
\[
f(x) = \frac{g(x)}{h(x)}
\]
\begin{align*}
\text{First, I set up my definition}
\\
f'(x) &= \lim_{\Delta x \to 0} \frac{\frac{g(x + \Delta x)}{h(x + \Delta x)} - \frac{g(x)}{h(x)}}{\Delta x}.
\\
\text{Rearranging my equations}
\\
f'(x) &= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \cdot \frac{h(x+\Delta x) \cdot g(x) - h(x) \cdot g(x + \Delta x)}{g(x + \Delta x) \cdot g(x)}
\\
\text{Adding and subtracting } h(x) \cdot g(x)
\\
f'(x) &= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \cdot \frac{h(x+\Delta x) \cdot g(x) -h(x) \cdot g(x) + h(x) \cdot g(x) - h(x) \cdot g(x + \Delta x)}{g(x + \Delta x) \cdot g(x)}
\\
\text{To rewrite}
\\
f'(x) &= \lim_{\Delta x \to 0} \frac{1}{g(x + \Delta x) \cdot g(x)} \cdot \frac{h(x+\Delta x) \cdot g(x) -h(x) \cdot g(x) + h(x) \cdot g(x) - h(x) \cdot g(x + \Delta x)}{\Delta x}
\\
\text{Separate and Factor}
\\
f'(x) &= \lim_{\Delta x \to 0} \frac{1}{g(x + \Delta x) \cdot g(x)} \cdot g(x) \cdot \frac{h(x+\Delta x) -h(x)}{\Delta x} + h(x) \cdot \frac{ g(x) - g(x + \Delta x)}{\Delta x}
\\
\text{Using limits of parts}
\\
f'(x) &= \frac{1}{g(x) \cdot g(x)} \cdot g(x) \cdot h'(x) + h(x) \cdot g'(x)
\\
\text{Condense and rewrite}
\\
f'(x) &= \frac{g(x) \cdot h'(x) + h(x) \cdot g'(x)}{g(x)^2}
\\
\text{Thus giving us the Quotient Rule}
\end{align*}


\end{document}

product-rule.tex

\documentclass[10pt]{article}
\pagestyle{plain}
\usepackage{amsmath}
\usepackage[margin=0.25in]{geometry}
\usepackage{parselines} 
\begin{document}
\iffalse
Josh Pruim wrote this. For Green's class. In October.
\fi
\noindent
The function \textit{f} is defined below
\[
f(x) = \frac{g(x)}{h(x)}
\]
The Quotient Rule is as follows with respect to \textit{f(x)}
\[
f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}
\]
\begin{align*}
f(x) &= \frac{g(x)}{h(x)} = g(x) \cdot h(x)^{-1}
\\
\text{Apply Product Rule}
\\
f'(x) &= g'(x) \cdot h(x)^{-1} + g(x) \cdot (h(x)^{-1})'
\\
\text{Use Power Rule}
\\
f'(x) &= g'(x) \cdot h(x)^{-1} + g(x) \cdot (-1) \cdot h(x)^{-2} \cdot h'(x)
\\
\text{This can be rewritten as}
\\
f'(x) &= \frac{g'(x)}{h(x)} - \frac{g(x) \cdot h'(x)}{h(x)^2}
\\
\text{Multiply by 1 on the left}
\\
f'(x) &= (\frac{h(x)}{h(x)}) \cdot \frac{g'(x)}{h(x)} - \frac{g(x) \cdot h'(x)}{h(x)^2}
\\
\text{This results in}
\\
f'(x) &= \frac{g'(x) \cdot h(x)}{h(x)^2} - \frac{g(x) \cdot h'(x)}{h(x)^2}
\\
\text{Consolidate over a common denominator}
\\
f'(x) &= \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h(x)^2}
\end{align*}

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