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\begin_body

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%MODULE: Continuity
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%USER DEFINED COMMANDS
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\begin_inset FormulaMacro
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%INPUT THE INSTRUCTIONS PAGE 
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%BUILD  TITLE PAGE 
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\begin_layout Title
Implicit Differentiation
\end_layout

\begin_layout Subtitle
There are cases where we would like to compute the derivative for a 
\emph on
relation
\emph default
 that is not a function, or that cannot easily be written in the form 
\begin_inset Formula $\ec y=f(x)$
\end_inset

.
 In these circumstances, it can still make sense to compute a derivative---all
 we need is that the limit of the slopes of the secant lines exists---but
 since we cannot isolate 
\begin_inset Formula $\ec y$
\end_inset

 from 
\begin_inset Formula $\ec x$
\end_inset

, we need a more general method than the ones used heretofore.
 We call this method 
\emph on

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 implicit differentiation
\emph default
.
 
\end_layout

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%1
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\end_layout

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{
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Defining the problem
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}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
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\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

 How can we find the derivative at any point of the following relations
 
\emph on
quickly
\emph default
? 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec x^{2}+y^{2}=1$
\end_inset

 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec x^{2}e^{xy}=x+y$
\end_inset

 
\end_layout

\begin_layout Itemize
\begin_inset Formula $\ec\arcsin x$
\end_inset

, 
\begin_inset Formula $\ec\arctan x$
\end_inset

, 
\begin_inset Formula $\ec\arcsec x$
\end_inset

 
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset

 
\begin_inset ERT
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\begin_layout Plain Layout

%2
\end_layout

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout BeginFrame
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

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SAGE worksheets
\begin_inset ERT
status collapsed

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}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
You will need a blank SAGE worksheet for this module.
\end_layout

\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Implicit Differentiation
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout OverlayArea
\begin_inset ERT
status open

\begin_layout Plain Layout

{
\backslash
textwidth}{6in}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
If we look back at the first example,
\begin_inset Formula 
\[
{\ec x^{2}+y^{2}=1},
\]

\end_inset

 we notice that it is possible to solve for 
\begin_inset Formula $\ec y$
\end_inset

 explicitly:
\begin_inset Formula 
\[
{\ec y=\pm\sqrt{1-x^{2}}}.
\]

\end_inset


\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
This is not a function, but we can choose either the positive or negative
 branch to compute the derivative at a given point.
\end_layout

\begin_layout Standard
\noindent
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/circle.png
	lyxscale 20
	scale 15

\end_inset

 
\begin_inset space \qquad{}
\end_inset


\begin_inset Graphics
	filename figures/circle_two_branches.png
	lyxscale 20
	scale 15

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2->
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset VSpace -6ex
\end_inset


\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/circle.png
	lyxscale 20
	scale 15

\end_inset

 
\begin_inset space \qquad{}
\end_inset


\begin_inset Graphics
	filename figures/circle_two_branches.png
	lyxscale 20
	scale 15

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout

\end_layout

\end_inset


\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
On the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 upper (positive) branch
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 we have
\begin_inset Formula 
\begin{align*}
\ec y & {\ec=\sqrt{1-x^{2}}=(1-x^{2})^{1/2}}.
\end{align*}

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
On the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

upper (positive) branch
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 we have
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\sqrt{1-x^{2}}=(1-x^{2})^{1/2}\\
\ec y' & {\ec=\frac{1}{2}(1-x^{2})^{-1/2}\cdot(-2x)}.
\end{align*}

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Don't forget the Chain Rule.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<4>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
On the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 lower (negative) branch
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 we have
\begin_inset Formula 
\begin{align*}
\ec y & {\ec=-\sqrt{1-x^{2}}=-(1-x^{2})^{1/2}}.
\end{align*}

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<5>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
On the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

lower (negative) branch
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

 we have
\begin_inset Formula 
\begin{align*}
\dims y & \dims=-\sqrt{1-x^{2}}=-(1-x^{2})^{1/2}\\
\ec y' & {\ec=-\frac{1}{2}(1-x^{2})^{-1/2}\cdot(-2x)}.
\end{align*}

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Don't forget the Chain Rule.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
Either way, we found
\begin_inset Formula 
\[
{\ec y'=\mp\frac{x}{\sqrt{1-x^{2}}}}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Implicit Differentiation
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
Even though 
\begin_inset Formula $\ec y$
\end_inset

 is not a function of 
\begin_inset Formula $\ec x$
\end_inset

 in the circle 
\begin_inset Formula $\ec C$
\end_inset

, we can find functions 
\begin_inset Formula $\ec y_{1}$
\end_inset

 and 
\begin_inset Formula $\ec y_{2}$
\end_inset

 such that 
\begin_inset Formula $\ec C$
\end_inset

 is the union of the points described by two functions.
 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/circle.png
	lyxscale 20
	scale 20

\end_inset

 
\begin_inset space \qquad{}
\end_inset


\begin_inset Graphics
	filename figures/circle_two_branches.png
	lyxscale 20
	scale 20

\end_inset

 
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
Thus the relation 
\begin_inset Formula $\ec x^{2}+y^{2}=1$
\end_inset

 defines 
\begin_inset Formula $\ec y$
\end_inset

 as a function of 
\begin_inset Formula $\ec x$
\end_inset

 
\emph on
locally
\emph default
.
 We can treat 
\begin_inset Formula $\ec y$
\end_inset

 as a function that depends on 
\begin_inset Formula $\ec x$
\end_inset

, and differentiate using the Chain Rule.
\end_layout

\begin_layout BeginFrame
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset

First Example
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}
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\end_inset


\end_layout

\begin_layout OverlayArea
\begin_inset ERT
status open

\begin_layout Plain Layout

{
\backslash
textwidth}{6in}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
We demonstrate this using the first example.
\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\ec x^{2}+y^{2} & {\ec=1}.
\end{align*}

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}+y^{2} & \dims=1\\
\ec2x+2y\cdot{\hilite y'} & {\ec=0}.
\end{align*}

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Since 
\begin_inset Formula $y$
\end_inset

 is a function of 
\begin_inset Formula $x$
\end_inset

, the Chain Rule tells us that the derivative of 
\begin_inset Formula $y^{2}$
\end_inset

 is 
\begin_inset Formula $2y\cdot{y'}$
\end_inset

.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}+y^{2} & \dims=1\\
\dims2x+2y\cdot y' & \dims=0\\
\ec y' & {\ec=-\frac{x}{y}}.
\end{align*}

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Solve for 
\begin_inset Formula $y'$
\end_inset

.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<4>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}+y^{2} & \dims=1\\
\dims2x+2y\cdot y' & \dims=0\\
\dims y' & \dims=-\frac{x}{y}.
\end{align*}

\end_inset

 We illustrate this by computing the slopes of lines tangent to the circle
 at various points.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<5>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims y' & \dims=-\frac{x}{y}.
\end{align*}

\end_inset

 We illustrate this by computing the slopes of lines tangent to the circle
 at various points.
\end_layout

\begin_layout Standard
The point 
\begin_inset Formula $\hilite(0,1)$
\end_inset

 is on the curve, since 
\begin_inset Formula $\ec0^{2}+1^{2}=1$
\end_inset

.
 At this point, 
\begin_inset Formula 
\[
{\ec y'=-0/1=0}.
\]

\end_inset

 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/tanline_at_top.png
	lyxscale 20
	scale 20

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<6>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims y' & \dims=-\frac{x}{y}.
\end{align*}

\end_inset

 We illustrate this by computing the slopes of lines tangent to the circle
 at various points.
\end_layout

\begin_layout Standard
The point 
\begin_inset Formula $\hilite(\sqrt{2}/2,\sqrt{2}/2)$
\end_inset

 is on the curve, since 
\begin_inset Formula $\ec\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}=1$
\end_inset

.
 At this point, 
\begin_inset Formula 
\[
{\ec{y'}=-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=-1}.
\]

\end_inset

 
\begin_inset VSpace -6ex
\end_inset


\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
 
\end_layout

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/tanline_at_upper_right.png
	lyxscale 20
	scale 20

\end_inset


\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<7>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims y' & \dims=-\frac{x}{y}.
\end{align*}

\end_inset

 We illustrate this by computing the slopes of lines tangent to the circle
 at various points.
\end_layout

\begin_layout Standard
The point 
\begin_inset Formula $\hilite(1,0)$
\end_inset

 is on the curve, since 
\begin_inset Formula $\ec1^{2}+0^{2}=1$
\end_inset

.
 At this point, 
\begin_inset Formula 
\[
{\ec{y'}=-\frac{1}{0}},\mathrm{\ corresponding\ to\ a\ vertical\ line}.
\]

\end_inset

 
\begin_inset Float figure
placement h!
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figures/tanline_at_right.png
	lyxscale 20
	scale 20

\end_inset

 
\end_layout

\end_inset


\end_layout

\end_deeper
\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Implicit Plots
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\noindent
It isn't always easy to view the graph of a relation.
 If you can't separate 
\begin_inset Formula $\ec y$
\end_inset

 and 
\begin_inset Formula $\ec x$
\end_inset

, you can still plot it in SAGE using the 
\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 implicit_plot()
\family default
 command.
 For example, the graphs of the circles were generated using these commands:
\emph on
 (Try it!)
\end_layout

\begin_layout Standard
\begin_inset Box Boxed
position "c"
hor_pos "c"
has_inner_box 1
inner_pos "c"
use_parbox 1
use_makebox 0
width "8.75in"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\family typewriter
\size scriptsize
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 circleplot = implicit_plot(x‸2+y‸2-1,(x,-2,2),(y,-2,2),cmap='jet',plot_points=5
00)
\family default

\begin_inset Newline newline
\end_inset


\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 circleplot.show(aspect_ratio=1)
\family default

\begin_inset Newline newline
\end_inset


\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 upperplot = implicit_plot(x‸2+y‸2-1,(x,-2,2),(y,0,2),cmap='jet',plot_points=500
)
\family default

\begin_inset Newline newline
\end_inset


\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 lowerplot = implicit_plot(x‸2+y‸2-1,(x,-2,2),(y,-2,0),cmap='hsv',plot_points=50
0)
\family default

\begin_inset Newline newline
\end_inset


\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 (upperplot+lowerplot).show(aspect_ratio=1)
\end_layout

\end_inset


\end_layout

\begin_layout Standard
The first argument is an expression 
\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 exp
\family default
 which corresponds to the equation 
\begin_inset Formula $\ec\mathtt{\ttcolor exp}=0$
\end_inset

.
 We want to plot 
\begin_inset Formula $\ec x^{2}+y^{2}=1$
\end_inset

, which corresponds to 
\begin_inset Formula $\ec x^{2}+y^{2}-1=0$
\end_inset

, so 
\begin_inset Formula $\ec\mathtt{\ttcolor exp}=x^{2}+y^{2}-1$
\end_inset

.
 The second and third arguments are a range of 
\begin_inset Formula $x$
\end_inset

- and 
\begin_inset Formula $y$
\end_inset

-values over which to graph 
\begin_inset Formula $\ec{\ttcolor\texttt{exp}}=0$
\end_inset

.
 The third argument corresponds to a color.
 The usual colors don't apply for implicit plots; stick with 
\family typewriter

\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
ttcolor
\end_layout

\end_inset

 'jet'
\family default
 (blue), 
\family typewriter

\begin_inset ERT
status collapsed

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 'hsv'
\family default
 (red), and 
\family typewriter

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 'summer'
\family default
 (green).
 The final argument ensures an accurate graph; if the number is too small,
 the graph will be wrong.
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Second Example
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\begin_deeper
\begin_layout Standard
\noindent
We can also differentiate the second example, where it is not possible to
 isolate 
\begin_inset Formula $\ec y$
\end_inset

.
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<1>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\ec x^{2}e^{xy} & {\ec=x+y}.
\end{align*}

\end_inset


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<2>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}e^{xy} & {\dims=x+y}\\
\ec2xe^{xy}+x^{2}e^{xy}\cdot\left(y+x{\hilite y'}\right) & {\ec=1+{\hilite y'}}.
\end{align*}

\end_inset

 
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Use the Product Rule and the Chain Rule.
 We only write 
\begin_inset Formula $y'$
\end_inset

 when we take the derivative of 
\begin_inset Formula $y$
\end_inset

.
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<3>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}e^{xy} & {\dims=x+y}\\
\dims2xe^{xy}+x^{2}e^{xy}\cdot\left(y+xy'\right) & \dims=1+y'\\
\ec2xe^{xy}+x^{2}ye^{xy}+x^{3}e^{xy}\cdot{\hilite y'} & {\ec=1+{\hilite y'}}.
\end{align*}

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Distribution.
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<4>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}e^{xy} & {\dims=x+y}\\
\dims2xe^{xy}+x^{2}e^{xy}\cdot\left(y+xy'\right) & \dims=1+y'\\
2xe^{xy}+x^{2}ye^{xy}+x^{3}e^{xy}\cdot{y'} & =1+{y'}\\
\ec\left(x^{3}e^{xy}-1\right){\hilite y'} & {\ec=1-e^{xy}\left(x^{2}+2x\right)}.
\end{align*}

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Everything with 
\begin_inset Formula $y'$
\end_inset

 on one side;
\begin_inset Newline newline
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everything without it to the other.
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<5>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims x^{2}e^{xy} & {\dims=x+y}\\
\dims2xe^{xy}+x^{2}e^{xy}\cdot\left(y+xy'\right) & \dims=1+y'\\
2xe^{xy}+x^{2}ye^{xy}+x^{3}e^{xy}\cdot{y'} & =1+{y'}\\
\left(x^{3}e^{xy}-1\right){y'} & {=1-e^{xy}\left(x^{2}+2x\right)}\\
\ec{y'} & {\ec=\frac{1-e^{xy}\left(x^{2}+2x\right)}{x^{3}e^{xy}-1}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\end_inset

Solve for 
\begin_inset Formula $y'$
\end_inset

.
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Third Example
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\begin_deeper
\begin_layout Standard
\noindent
The third example asks us to find the derivatives of the inverse functions
 of 
\begin_inset Formula $\ec\sin x$
\end_inset

, 
\begin_inset Formula $\ec\tan x$
\end_inset

, and 
\begin_inset Formula $\ec\sec x$
\end_inset

.
 The derivatives of these functions have an interesting form.
 Implicit differentation allows us to find these derivatives.
 Start with 
\begin_inset Formula $\ec\arcsin x$
\end_inset

:
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\ec y & {\ec=\arcsin x}.
\end{align*}

\end_inset


\end_layout

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\begin_layout Pause

\end_layout

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<2>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=\arcsin x}\\
\ec\sin y & {\ec=x}.
\end{align*}

\end_inset

 
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Properties of inverse functions.
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<3>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims\sin y & {\dims=x}\\
\ec\cos y\cdot y' & {\ec=1}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\end_inset

Differentiated both sides with respect to 
\begin_inset Formula $x$
\end_inset

.
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<4>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims\cos y\cdot y' & {\dims=1}\\
\ec y' & {\ec=\frac{1}{\cos y}}.
\end{align*}

\end_inset

 
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\end_inset

Solve for 
\begin_inset Formula $y'$
\end_inset

.
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<5>
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\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims y' & {\dims=\frac{1}{\cos y}}\\
\ec y' & {\ec=\frac{1}{\cos\left(\arcsin x\right)}}.
\end{align*}

\end_inset

 
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Substitute 
\begin_inset Formula $y=\arcsin x$
\end_inset

.
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<6>
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\begin_deeper
\begin_layout Standard
\noindent
 
\begin_inset Formula 
\begin{align*}
\dims y' & {\dims=\frac{1}{\cos\left(\arcsin x\right)}}\\
\ec y' & {\ec=\frac{1}{\sqrt{1-x^{2}}}}.
\end{align*}

\end_inset

 
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\end_inset

This is a property learned in precalculus; we explain it briefly.
 We want to know 
\begin_inset Formula $\cos\alpha$
\end_inset

 where 
\begin_inset Formula $\alpha=\arcsin x$
\end_inset

.
 Thus 
\begin_inset Formula $x=\sin\alpha$
\end_inset

, so we can draw a right triangle with hypotenuse of length 
\begin_inset Formula $1$
\end_inset

 that has an angle of measure 
\begin_inset Formula $\alpha$
\end_inset

 and a leg opposite 
\begin_inset Formula $\alpha$
\end_inset

 of length 
\begin_inset Formula $x$
\end_inset

.
 The length of the side adjacent to 
\begin_inset Formula $\alpha$
\end_inset

 will then be 
\begin_inset Formula $\sqrt{1-x^{2}}$
\end_inset

.
 
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Third Example
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\begin_layout Standard
\noindent
We have the following.
\end_layout

\begin_layout Theorem
\noindent
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[
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\end_inset

(Derivatives of inverse trig functions)
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]
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arcsin x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{\sqrt{1-x^{2}}}$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arctan x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{{1+x^{2}}}$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arcsec x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{x\sqrt{x^{2}-1}}$
\end_inset

.
 
\end_layout

\end_deeper
\begin_layout Standard
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\end_inset

You should check on your own that the other two are correct.
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Conclusion
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\end_layout

\begin_layout Itemize

\emph on
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hilite
\end_layout

\end_inset

 Implicit differentiation
\emph default
 allows us to compute the derivative in cases where 
\begin_inset Formula $\ec y$
\end_inset

 may not be a function of 
\begin_inset Formula $\ec x$
\end_inset

.
 
\end_layout

\begin_layout Itemize
Implicit differentation uses the chain rule: whenever you have to take the
 derivative of 
\begin_inset Formula $\ec y$
\end_inset

, you multiply by 
\begin_inset Formula $\hilite y'$
\end_inset

.
 
\end_layout

\begin_layout Itemize
Implicit differentiation allowed us to determine the derivatives of three
 new functions:
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\begin_deeper
\begin_layout Theorem
\begin_inset ERT
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\begin_layout Plain Layout

[
\end_layout

\end_inset

(Derivatives of inverse trig functions)
\begin_inset ERT
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\begin_layout Plain Layout

]
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arcsin x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{\sqrt{1-x^{2}}}$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arctan x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{{1+x^{2}}}$
\end_inset

.
 
\end_layout

\begin_layout Itemize
If 
\begin_inset Formula $\ec f(x)=\arcsec x$
\end_inset

, then 
\begin_inset Formula $\ec f'(x)=\frac{1}{x\sqrt{x^{2}-1}}$
\end_inset

.
 
\end_layout

\end_deeper
\begin_layout Standard
You should remember these derivatives.
 
\end_layout

\end_deeper
\begin_layout EndFrame

\end_layout

\end_body
\end_document
