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Contents Index Dark Mode Prev Up Next \(\DeclareMathOperator{\erf}{erf}
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Section 8.11 The Derivative of the Natural Logarithm
Since the
natural logarithm is the inverse of the
natural exponential we can find its derivative using the same trick that we used to find the derivatives of the inverse trig functions. Start with the identity from
Logarithm Property #2 :
\begin{equation}
e^{\ln(x)} =x \tag{8.27}
\end{equation}
and make this easier on our eyes with the substitution
\(\ln(x)=y\text{.}\) Next we differentiate
equation (8.27) and see that
\begin{equation*}
e^y\dx{y} = \dx{x}\text{.}
\end{equation*}
Thus
\begin{equation*}
\dx{y}
=\frac{1}{e^y}\dx{x}\text{.}
\end{equation*}
To get back to the original variable \(x\) we substitute \(y=\ln(x)\) giving
\begin{equation*}
\dx{y}
=\frac{1}{e^{\ln(x)}}\dx{x}\text{,}
\end{equation*}
and since \(e^{\ln(x)}=x\) we have
\begin{equation*}
\dx{y} =\frac{1}{x}\dx{x}\text{.}
\end{equation*}
\begin{equation*}
\dx{y}
=\frac{1}{x}\dx{x}
\end{equation*}
\begin{equation}
\dx{(\ln(x))} =x^{-1} \dx{x}.\tag{8.28}
\end{equation}
equation (8.28) is the last differentiation rule you need to memorize. Memorize it.
As always the difficulty here is not remembering what the differentiation rule is, the difficulty is in learning to use it in tandem with all of the other rules. So practice is required. Lots of practice.
Drill 8.11.1 .
For each of the following find \(\dx{y}\text{,}\) and \(\dfdx{y}{x}\text{.}\)
\(\displaystyle y=\ln\left(\frac{7-x}{5+x}\right)\)
\(\displaystyle y=\dfrac{\ln(x)}{e^x}\)
\(\displaystyle y=\tan(\ln(x))\)
\(\displaystyle y=\ln(\tan(x))\)
\(\displaystyle y=\inverse\tan(\ln(x))\)
\(\displaystyle y=x\ln(x)-x\)
\(\displaystyle y=\ln(\ln(\sec(x)))\)
\(\displaystyle y=\ln(\sin(x^2+1))\)
\(\displaystyle y=\sqrt{x}\cdot\ln\left(\sqrt{x}\right)\)
\(\displaystyle y=\ln(2x+2)\)
\(\displaystyle \ln(y)=\sin(e^x)\)
\(\displaystyle \ln(x+y)=x\inverse\tan(y)\)
\(\displaystyle \ln(x+y)=e^{x+y}\)
Problem 8.11.2 .
For each of the following find
\(\dx{y}\text{,}\) and
\(\dfdx{y}{x}\text{.}\)
(a)
\(y=\ln\left(\frac{x-1}{x+1}\right)\)
Hint .
Use the properties of logarithms to make these drills easier on the eyes.
(b)
\(y=\ln(\sec(x)\tan(x))\)
(c)
\(y=\ln \left( \frac{(2x+1)^2(3x-2)}{x^2+1} \right)\)
(d)
\(y=\ln\left(\frac{e^{x^2}\sin(x)}{\sqrt{x^3-7x+4}}\right)\)
(e)
\(y=\ln\left(\left[\frac{(2x+3)(4x+5)^{20}}{x^{\frac12}-\ln(x)}\right]^{\frac43}\right)\)
Problem 8.11.3 .
Recall that in
Problem 6.8.5 we showed the curvature of the graph of a function
\(y=y(x)\) is given by
\begin{equation*}
\kappa=\frac{\abs{\dfdxn{y}{x}{2}}}{\left(1+\left(\dfdx{y}{x}\right)^2\right)^{\frac32}}.
\end{equation*}
(a)
Show that the curvature of the graph of \(y=e^x\) is given by
\begin{equation*}
\kappa_e=\frac{e^x}{(1+e^{2x})^{\frac32}}\text{.}
\end{equation*}
Use the graph of \(\kappa_e=\frac{e^x}{(1+e^{2x})^{\frac32}}\) (or use Newton’s Method) to approximate the coordinates of the point \((x,y)\text{,}\) where the curvature of the graph of the natural exponential function is the greatest.
(b)
Show that the curvature of the graph of the \(y=\ln(x)\) is given by
\begin{equation*}
\kappa_l=\frac{x}{(1+x^2)^{\frac32}}\text{.}
\end{equation*}
Use the graph of \(\kappa_l=\frac{x}{(1+x^2)^{\frac32}}\) (or use Newton’s Method) to approximate the coordinates of the point \((x,y)\text{,}\) where the curvature of the graph of the natural logarithm is the greatest.
(c) Graph the natural logarithm and the natural exponential functions on the same set of axes and identify the points where the curvature is greatest in parts (a) and (b)? How are these points related? Does this surprise you? Explain.
DIGRESSION: A Curious Fact. From the
Power Rule 4.3.37 guarantees that the derivative of any monomial
\((x^\alpha)\) is another monomial (
\(\alpha x^{\alpha -1}\) ). But what happens if we try to go the other direction?
If \(\dfdx{y}{x}=x^2\) then clearly \(y=\frac{x^3}{3}\text{.}\) In general if
\begin{equation*}
\dfdx{y}{x}=x^\alpha
\end{equation*}
then
\begin{align*}
y= \frac{x^{\alpha+1}}{\alpha+1}\amp{}\amp{} \text{ for }\amp{}\amp{} \alpha\ne -1.
\end{align*}
Aside: Comment.
Actually, this should be
\(y=\frac{x^{\alpha+1 }}{\alpha +1}\) plus an arbitrary constant. Do you see why? Since the constant is irrelevant to our current point, we ignore it.
But if
\(\alpha=-1\) running the Power Rule backwards would give us
\(y= \frac{x^{-1+1}}{-1+1}\) which is meaningless since the denominator is
\(-1+1=0.\) When
\(\alpha=-1\) we get is
\(y=\ln(x).\)
It seems very strange that there would be this one, single exception doesn’t it? Or, to come at the question a little differently, maybe this says that the logarithm function is somehow related to polynomials? What do you think?
