m8.tex

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\title{MATH 4338 Main Problem 8}
\date{}
\author{Andy Lu}

\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\newtheorem{theorem*}{Question}

\begin{document}
  \maketitle
  \begin{proof}
    Verify
    $$g'(x_0) = \frac{1}{f'[g(x_0)]}$$
    
    Computing $f'(x)$:
    \begin{align*}
       \indent \text{Note: Computing }a'(x) \text{ and } b'(x): & & \\
      a'(x) = 3x^{3-1} = 3x^2 & & \text{[by Theorem 4.7]} \\
      b'(x) = 1x^{1-1} + 0 = 1 & & \text{[by Theorem 4.3, 4.7, and 4.1]} \\
      %f(x) = (x-1)^3 \, , \, I = {x:1<x<\infty} & & \\
    \end{align*}
    \begin{align*}
      f'(x) = a'(b(x)) \cdot b'(x) & & \text{[by the chain rule and}\\
      & & \text{ where } a(x) = x^3 \text{ and } b(x) = x-1 ]\\
       = (3(x-1)^2)(1) & & \text{[substituting in } a'(x) \,, b(x)\,, b'(x)] \\
       = 3(x-1)^2 & & \\
    \end{align*}
    Computing $g'(x)$:
    \begin{align*}
      x =& (y-1)^3 \\
      \sqrt[3]{x}=&(y-1) \\
      \sqrt[3]{x} + 1 =& y \\
      \sqrt[3]{x} + 1 =& g(x) \\
    \end{align*}
    Let $c(x) = \sqrt[3]{x}$ and $d(x) = 1$. Then by Theorem 4.9,4.7, and 4.1, 
    $c'(x) = 3^{-1}(x)^{3^{-1} - 1}(1) = \frac{x^{-2/3}}{3} = 
    \frac{1}{3x^{2/3}}$. Also by Theorem 4.1, $d'(x) = 0$. Thus,
    \begin{align*}
      g'(x) = c'(x) + d'(x) && \text{[by Theorem 4.3]} \\
      g'(x) = \frac{1}{3x^{2/3}} + 0 && \text{[by substitution]} \\
      g'(x) = \frac{1}{3x^{2/3}} && \\
    \end{align*}
    Now we verify this is Formula 4.13 of the Inverse Differentiation Theorem.
    \begin{align*}
      \frac{1}{f'[g(x_0)]} &= &&\\
      &= \frac{1}{3(g(x_0)-1)^2}  && \text{[by substitution]}\\
      &= \frac{1}{3(\sqrt[3]{x_0} + 1-1)^2}  && \text{[by substitution]}\\
      &= \frac{1}{3(\sqrt[3]{x_0})^2}  && \text{[combine like terms]}\\
      &= \frac{1}{3((x_0)^{1/3})^2}  && \text{[square root property]}\\
      &= \frac{1}{3(x_0)^{2/3}}  && \text{[exponent property]}\\
      &= g(x_0) && \text{[definition of } g'(x) ]
    \end{align*}
    Thus, Formula 4.13 of the Inverse Differentiation Theorem is verified.
  \end{proof}
\end{document}