documentclass[a4paper,11pt]{book}
usepackage[T1]{fontenc}
usepackage[utf8]{inputenc}
usepackage{lmodern}
usepackage{classicthesis}
usepackage{hyperref}
usepackage{graphicx}
usepackage[english]{babel}
usepackage{amsmath,mathtools}
usepackage{amsfonts}
usepackage{color}
usepackage{url}
usepackage{enumitem}
usepackage{multicol}
usepackage{tabu}
usepackage{amsthm}
usepackage{fourier-orns}
usepackage{framed}
usepackage{tcolorbox}
usepackage[toc,page]{appendix}
usepackage[all]{xy}

begin{document}

begin{enumerate}
item For the function $y=x,$ find the derivative, $y'.$
[y'=lim_{hto 0}frac{(x+h)-x}{h}=lim_{hto 0}frac{h}{h}=lim_{hto 0}1=1.]
item For the function $y=x^2,$ find the derivative, $y'.$
begin{align*}
y'=lim_{hto 0}frac{(x+h)^2-x^2}{h}&=lim_{hto 0}frac{x^2+2xh+h^2-x^2}{h}\
&=lim_{hto 0}frac{2xh+h^2}{h}=lim_{hto 0}(2x+h)=2x.
end{align*}
    begin{center}
        begin{framed}
For the next two examples, we will use the angle sum formula for sine, which is
[sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta.]
        end{framed}
    end{center}
item For the function $y=sin x,$ find the derivative, $y'.$
begin{align*}
y'=lim_{hto 0}frac{sin (x+h)-sin x}{h}&=lim_{hto 0}frac{sin xcos h+cos xsin h-sin x}{h}\
&=lim_{hto 0}frac{sin x(cos h-1)+cos xsin h}{h}\
&=left[lim_{hto 0}(sin x)left(frac{cos h-1}{h}right)right]left[lim_{hto 0}(cos x)left(frac{sin h}{h}right)right]\
&=(sin x)cdot 0+(cos x)cdot 1\
&=cos x,
end{align*}
where we have used $displaystylelim_{hto 0}frac{sin h}{h}=1$ and $displaystylelim_{hto 0}frac{cos h-1}{h}=0$ from section (1.3).
item Find $dfrac{d}{dx}tan x.$
begin{align*}
dfrac{d}{dx}tan x=displaystylelim_{hto 0}dfrac{tan(x+h)-tan x}{h}&=displaystylelim_{hto 0}left(dfrac{sin(x+h)}{hcos(x+h)}-dfrac{sin x}{hcos x}right)\
&=displaystylelim_{hto 0}left(dfrac{sin(x+h)cos x-cos(x+h)sin x}{hcos(x+h)cos x}right)\
&=displaystylelim_{hto 0}left(dfrac{sin[(x+h)-x]}{hcos(x+h)cos x}right)\
&=displaystylelim_{hto 0}left[left(dfrac{sin h}{h}right)left(dfrac{1}{cos(x+h)cos x}right)right]\
&=dfrac{1}{cos^2(x)}=sec^2 x.
end{align*}
end{enumerate}

end{document}

documentclass[a4paper,11pt]{book}
usepackage[T1]{fontenc}
usepackage[utf8]{inputenc}
usepackage{lmodern}
usepackage{classicthesis}
usepackage{hyperref}
usepackage{graphicx}
usepackage[english]{babel}
usepackage{amsmath,mathtools}
usepackage{amsfonts}
usepackage{color}
usepackage{url}
usepackage{enumitem}
usepackage{multicol}
usepackage{tabu}
usepackage{amsthm}
usepackage{fourier-orns}
usepackage{framed}
usepackage[many]{tcolorbox}
usepackage[toc,page]{appendix}
usepackage[all]{xy}
usepackage[margin=2cm,showframe]{geometry}

begin{document}

begin{enumerate}
item For the function $y=x,$ find the derivative, $y'.$
[y'=lim_{hto 0}frac{(x+h)-x}{h}=lim_{hto 0}frac{h}{h}=lim_{hto 0}1=1.]
item For the function $y=x^2,$ find the derivative, $y'.$
begin{align*}
y'=lim_{hto 0}frac{(x+h)^2-x^2}{h}&=lim_{hto 0}frac{x^2+2xh+h^2-x^2}{h}\
&=lim_{hto 0}frac{2xh+h^2}{h}=lim_{hto 0}(2x+h)=2x.
end{align*}
      begin{center}leavevmode
        begin{framed}
For the next two examples, we will use the angle sum formula for sine, which is
[sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta.]
        end{framed}
        end{center}
        makebox[linewidth]{%
          begin{tcolorbox}[arc=0pt,outer arc=0pt,colback=white,width=.6textwidth]
      For the next two examples, we will use the angle sum formula for sine, which is
      [sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta.]
        end{tcolorbox}}
item For the function $y=sin x,$ find the derivative, $y'.$
begin{align*}
y'=lim_{hto 0}frac{sin (x+h)-sin x}{h}&=lim_{hto 0}frac{sin xcos h+cos xsin h-sin x}{h}\
&=lim_{hto 0}frac{sin x(cos h-1)+cos xsin h}{h}\
&=left[lim_{hto 0}(sin x)left(frac{cos h-1}{h}right)right]left[lim_{hto 0}(cos x)left(frac{sin h}{h}right)right]\
&=(sin x)cdot 0+(cos x)cdot 1\
&=cos x,
end{align*}
where we have used $displaystylelim_{hto 0}frac{sin h}{h}=1$ and $displaystylelim_{hto 0}frac{cos h-1}{h}=0$ from section (1.3).
item Find $dfrac{d}{dx}tan x.$
begin{align*}
dfrac{d}{dx}tan x=displaystylelim_{hto 0}dfrac{tan(x+h)-tan x}{h}&=displaystylelim_{hto 0}left(dfrac{sin(x+h)}{hcos(x+h)}-dfrac{sin x}{hcos x}right)\
&=displaystylelim_{hto 0}left(dfrac{sin(x+h)cos x-cos(x+h)sin x}{hcos(x+h)cos x}right)\
&=displaystylelim_{hto 0}left(dfrac{sin[(x+h)-x]}{hcos(x+h)cos x}right)\
&=displaystylelim_{hto 0}left[left(dfrac{sin h}{h}right)left(dfrac{1}{cos(x+h)cos x}right)right]\
&=dfrac{1}{cos^2(x)}=sec^2 x.
end{align*}
end{enumerate}

end{document}