4.1: Logs and Derivatives
- Page ID
- 530
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definition of the Natural Logarithm
Recall that
\[ \int x^n\, dx = \dfrac{x^{n+1}}{n+1} +C. \nonumber \]
What is
\[ \int x^{-1} dx \text{?} \nonumber \]
For \(x > 0\) we define
\[ \int_1^x \dfrac{1}{t} dt = \ln x. \nonumber \]
Properties of \(\ln x\):
- \(\ln 1 = 0\)
- \(\ln (ab) = \ln a + \ln b\)
- \(\ln(a^n) = n \ln a\)
- \(\ln \left( \dfrac{a}{b} \right) = \ln a - \ln b\)
Proof of (3)
\[\dfrac{d}{dx} \ln x^n = \dfrac{d}{dx} \int_1^{x^n} \dfrac{1}{t} dt \nonumber \]
\[ = \dfrac{1}{x^n} \left( n\,x^{n-1}\right) = n \left(\dfrac{1}{x} \right) \nonumber \]
\[ = n \dfrac{d}{dx} \int _1^x \dfrac{1}{t} dt = n\ln x. \nonumber \]
So that
\[ \ln x^n \nonumber \]
and
\[ n \ln x \nonumber \]
have the same derivative. Hence
\[ \ln x^n = n \ln x + C. \nonumber \]
Plugging in \(x = 1\) we have that \(C = 0\).
Definition of \(e\)
Let \(e\) be such that
\[ \ln e = 1 \nonumber \]
ie.
\[ \int_1^e \dfrac{1}{t} dt = 1. \nonumber \]
Find the derivative of
\[ \ln (x^2 + 1). \nonumber \]
Solution
We use the chain rule with \( y = \ln u\) so \(u = x^2 + 1 \),
\[ y' = (2x)(1/u) = \dfrac{2x}{x^2+1}. \nonumber \]
Find the derivatives of the following functions:
- \(\ln (\ln x)\)
- \( \dfrac{\ln x}{x} \)
- \((\ln x)^2 \)
- \(\ln (\sec x)\)
- \(\ln (\csc x) \)
- Show that \( y= 3 \ln x - 4 \) is a solution of the differential equation \(xy'' + y' = 0\).
- Find the relative extrema of \( x \ln x\).
- Find the equation of the tangent line to \(y = 3x^2 - \ln x \) at \((1,3)\).
- Find \(\dfrac{dy}{dx}\) for \( \ln (xy) + 2x^2 = 30 \).
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.