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%MODULE 1: Intro to SAGE
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\begin_layout Title
Derivatives of logarithmic functions
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\begin_layout Subtitle
In this module we determine the derivative of the logarithmic function 
\begin_inset Formula $\ec\ln x$
\end_inset

.
 We also use it to find the derivatives of more difficult functions.
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{
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Defining the problem
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\backslash
begin{question}
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 What are the derivatives of the following functions? 
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\begin_layout Itemize
\begin_inset Formula $\ec y=\ln x$
\end_inset

 
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\begin_layout Itemize
\begin_inset Formula $\ec y=x^{\sqrt{x}}$
\end_inset

 
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{
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The derivative of 
\begin_inset Formula $\ln x$
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}
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\begin_deeper
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\noindent
We start with the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

.
 Before showing explicitly what its derivative is, let's take an educated
 guess using the SAGElets 
\begin_inset Quotes eld
\end_inset

Tangent Line Guesser
\begin_inset Quotes erd
\end_inset

 and 
\begin_inset Quotes eld
\end_inset

Function Factory
\begin_inset Quotes erd
\end_inset

.
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\begin_layout Pause

\end_layout

\begin_layout Standard
Using the Tangent Line Guesser, estimate the value of the derivative of
 
\begin_inset Formula $\ec\ln x$
\end_inset

 at the following values of 
\begin_inset Formula $\ec x$
\end_inset

.
 
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{
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\backslash
ec
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\begin_inset Tabular
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1 
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2 
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5 
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10 
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<3>
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\begin_deeper
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(SAGE may report an error after you change the value of 
\begin_inset Formula $\ec f$
\end_inset

 in the SAGElet to 
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{
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\backslash
ttcolor
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\end_inset

 ln(x)
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}
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\end_inset

.
 This is expected, because 
\begin_inset Formula $\ec a$
\end_inset

 is initially 0 and 
\begin_inset Formula $\ec\ln0$
\end_inset

 is undefined.
 Change the value of 
\begin_inset Formula $\ec a$
\end_inset

 and the error should disappear.)
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<4>
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Once you have some values, plug them into the Function Factory.
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<5->
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\begin_deeper
\begin_layout Standard
Once you have some values, plug them into the Function Factory.
 Your function should look like the one below.
 You should recognize it; modify 
\begin_inset Formula $\ec f$
\end_inset

 to match it.
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<5>
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	filename figures/points_for_deriv.png
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\begin_layout Standard
\noindent
This looks like 
\begin_inset Formula $\ec1/x$
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.
 But is it?
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{
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The derivative of 
\begin_inset Formula $\ln x$
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<1>
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\begin_deeper
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\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\ec y & {\ec=\ln x}.
\end{align*}

\end_inset


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\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\ec e^{y} & {\ec=e^{\ln x}}.
\end{align*}

\end_inset

 
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\backslash
begin{explanation}
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The exponential function is one-to-one.
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\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\ec e^{y} & {\ec=x}.
\end{align*}

\end_inset

 
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\backslash
begin{explanation}
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\end_inset

The exponential function and the logarithm are inverse functions.
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<4>
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\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\ec e^{y}\cdot y' & {\ec=1}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

Took the derivative of both sides 
\emph on
with respect to 
\begin_inset Formula $x$
\end_inset


\emph default
;
\begin_inset Newline newline
\end_inset

 implicit differentation tells us that the derivative of 
\begin_inset Formula $e^{y}$
\end_inset


\begin_inset Newline newline
\end_inset

 with respect to 
\begin_inset Formula $x$
\end_inset

 is 
\begin_inset Formula $e^{y}\cdot y'$
\end_inset

.
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\end_inset


\end_layout

\end_deeper
\begin_layout Pause

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\begin_inset ERT
status open

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<5>
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\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\ec y' & {\ec=e^{-y}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

Multiplied both sides by 
\begin_inset Formula $e^{-y}$
\end_inset

.
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\end_layout

\end_deeper
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<6>
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\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\ec y' & {\ec=e^{-\ln x}}.
\end{align*}

\end_inset

 
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\backslash
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\end_inset

Substituted 
\begin_inset Formula $y=\ln x$
\end_inset

---see first line.
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end{explanation}
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<7>
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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\ec y' & {\ec=e^{\ln x^{-1}}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

Property of logarithms.
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end{explanation}
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\end_deeper
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<8>
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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\dims y' & \dims=e^{\ln x^{-1}}\\
\ec y' & {\ec=x^{-1}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

The exponential function and the logarithm are inverses.
\begin_inset ERT
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end{explanation}
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\end_inset


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<9>
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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
To find the derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

, we will use implicit differentiation.
 
\begin_inset Formula 
\begin{align*}
\dims y & \dims=\ln x\\
\dims e^{y} & \dims=e^{\ln x}\\
\dims e^{y} & \dims=x\\
\dims e^{y}\cdot y' & \dims=1\\
\dims y' & \dims=e^{-y}\\
\dims y' & \dims=e^{-\ln x}\\
\dims y' & \dims=e^{\ln x^{-1}}\\
\dims y' & \dims=x^{-1}\\
\ec y' & {\ec=\frac{1}{x}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

Property of exponents.
\begin_inset ERT
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\end_deeper
\end_deeper
\begin_layout BeginFrame
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{
\end_layout

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The derivative of 
\begin_inset Formula $\ln x$
\end_inset


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}
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\end_layout

\begin_layout Standard
\noindent
The preceding argument confirms our initial guess.
\end_layout

\begin_layout Theorem
The derivative of 
\begin_inset Formula $\ec\ln x$
\end_inset

 is 
\begin_inset Formula $\ec1/x$
\end_inset

.
\end_layout

\begin_layout BeginFrame
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{
\end_layout

\end_inset

Logarithmic differentiation
\begin_inset ERT
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}
\end_layout

\end_inset


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{
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\begin_deeper
\begin_layout Standard
\noindent
Once we know the derivative of the logarithm, we can use it to find derivatives
 of functions where 
\emph on
the variable is in both the base and the exponent
\emph default
.
 We call this technique 
\begin_inset ERT
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{
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\backslash
hilite
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 logarithmic differentiation
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}
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.
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<2>
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\begin_deeper
\begin_layout Standard
The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\ec y & {\ec=x^{\sqrt{x}}}.
\end{align*}

\end_inset


\end_layout

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\begin_layout Pause

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<3>
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\end_layout

\begin_deeper
\begin_layout Standard
The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\ec\ln y & {\ec=\ln x^{\sqrt{x}}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
\end_layout

\end_inset

Took the logarithm of both sides.
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end{explanation}
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<4>
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\end_layout

\begin_deeper
\begin_layout Standard
The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\ec\ln y & {\ec=\sqrt{x}\ln x}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\backslash
begin{explanation}
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\end_inset

Property of logarithms.
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end{explanation}
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The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\ec\frac{1}{y}\cdot y' & {\ec=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}.
\end{align*}

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Took the derivative of both sides.
 By implicit differentiation and the derivative of the logarithm, 
\begin_inset Formula $(\ln y)'=\frac{1}{y}\cdot y'$
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.
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 The derivative of the right hand side uses the product rule.
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\begin_deeper
\begin_layout Standard
The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\ln x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\dims\frac{1}{y}\cdot y' & {\dims=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}\\
\ec y' & {\ec=y\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}.
\end{align*}

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Multiplied 
\begin_inset Formula $y$
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 to both sides.
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<7>
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\begin_deeper
\begin_layout Standard
The second example illustrates the technique.
 We take the derivative of 
\begin_inset Formula 
\begin{align*}
\dims y & {\dims=x^{\sqrt{x}}}\\
\dims\ln y & {\dims=\sqrt{x}\ln x}\\
\dims\frac{1}{y}\cdot y' & {\dims=\frac{1}{2\sqrt{x}}\ln x+\sqrt{x}\cdot\frac{1}{x}}\\
\dims y' & {\dims=y\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}\\
\ec y' & {\ec=x^{\sqrt{x}}\left(\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\right)}.
\end{align*}

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Substitued 
\begin_inset Formula $y=x^{\sqrt{x}}$
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---see first line.
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Derivatives of other logarithms
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\noindent
For the derivative of a different logarithmic base, we can reason from the
 
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 base conversion rule
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 for logarithms that 
\begin_inset Formula 
\[
{\ec\left(\log_{a}x\right)'=\left(\frac{\ln x}{\ln a}\right)'=\frac{1}{\ln a}\left(\ln x\right)'=\frac{1}{\ln a}\frac{1}{x}=\frac{1}{x\ln a}}.
\]

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 Thus
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\begin_layout Theorem
\noindent
 For any 
\begin_inset Formula $\ec a\in\mathbb{R}$
\end_inset

 where 
\begin_inset Formula $\ec a>0$
\end_inset

 and 
\begin_inset Formula $\ec a\neq1$
\end_inset

, the derivative of 
\begin_inset Formula $\ec\log_{a}x$
\end_inset

 is 
\begin_inset Formula $\ec\frac{1}{x\ln a}$
\end_inset

.
 
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Conclusion
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\begin_layout Itemize
We determined the derivative of 
\begin_inset Formula $\ec\ln x$
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 both geometrically and precisely, using implicit differentiation.
 
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\begin_layout Itemize
The derivative 
\begin_inset Formula $\ec\ln x$
\end_inset

 allows us to compute the derivatives of functions with variables in both
 the base and the exponent using 
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{
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 logarithmic differentiation
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.
 
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\end_body
\end_document
