Understand the natural logarithm and the mathematical constant e using integrals
Identify how to differentiate the natural logarithm function
Perform integrations where the natural logarithm is involved
Understand how to find derivatives and integrals of exponential functions
Convert logarithmic and exponential expressions to base e forms
The Natural Logarithm as an Integral
The Main Idea
Definition of Natural Logarithm: [latex]\ln x = \int_1^x \frac{1}{t} dt[/latex] for [latex]x > 0[/latex]
Key Properties:
[latex]\frac{d}{dx} \ln x = \frac{1}{x}[/latex]
[latex]\int \frac{1}{u} du = \ln |u| + C[/latex]
[latex]\ln 1 = 0[/latex]
[latex]\ln(ab) = \ln a + \ln b[/latex]
[latex]\ln(\frac{a}{b}) = \ln a - \ln b[/latex]
[latex]\ln(a^r) = r \ln a[/latex] (for rational r)
Graphical Interpretation:
Area under [latex]y = \frac{1}{t}[/latex] from [latex]1[/latex] to [latex]x (x > 1)[/latex]
Negative area under [latex]y = \frac{1}{t}[/latex] from [latex]x[/latex] to [latex]1[/latex] ([latex]0 < x < 1[/latex])
Problem-Solving Strategy
Recognize logarithmic forms in integrals
Use logarithm properties to simplify expressions
Apply chain rule for derivatives of logarithmic functions
Use u-substitution for integrals involving [latex]\frac{1}{u}[/latex]
Calculate the following derivatives:
[latex]\frac{d}{dx}\text{ln}(2{x}^{2}+x)[/latex]
[latex]\frac{d}{dx}{(\text{ln}({x}^{3}))}^{2}[/latex]
Show Solution
[latex]\frac{d}{dx}\text{ln}(2{x}^{2}+x)=\frac{4x+1}{2{x}^{2}+x}[/latex]
[latex]\frac{d}{dx}{(\text{ln}({x}^{3}))}^{2}=\frac{6\text{ln}({x}^{3})}{x}[/latex]
Watch the following video to see the worked solution to this example.
VIDEO
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You can view the transcript for this segmented clip of “6.7 Try It Problems” here (opens in new window) .
Calculate the integral [latex]\displaystyle\int \frac{x}{{x}^{2}+4}dx.[/latex]
Show Solution
Using [latex]u[/latex]-substitution, let [latex]u={x}^{2}+4.[/latex] Then [latex]du=2xdx[/latex] and we have
[latex]\displaystyle\int \frac{x}{{x}^{2}+4}dx=\frac{1}{2}\displaystyle\int \frac{1}{u}du\frac{1}{2}\text{ln}|u|+C=\frac{1}{2}\text{ln}|{x}^{2}+4|+C=\frac{1}{2}\text{ln}({x}^{2}+4)+C.[/latex]
Use properties of logarithms to simplify the following expression into a single logarithm:
[latex]\text{ln}8-\text{ln}2-\text{ln}(\frac{1}{4}).[/latex]
Show Solution
[latex]4\text{ln}2[/latex]
Evaluate [latex]\int \frac{x^3 + 2x}{x^4 + 4x^2 + 4} dx[/latex]
Show Answer
Recognize this as [latex]\int \frac{1}{u} \cdot \frac{du}{dx} dx[/latex]
Let [latex]u = x^4 + 4x^2 + 4 = (x^2 + 2)^2[/latex]
[latex]\frac{du}{dx} = 4x^3 + 8x = 4x(x^2 + 2)[/latex]
Rewrite integral:
[latex]\frac{1}{4} \int \frac{1}{(x^2 + 2)^2} \cdot 4x(x^2 + 2) dx = \frac{1}{4} \int \frac{1}{x^2 + 2} dx[/latex]
Final answer:
[latex]\frac{1}{4} \ln|x^2 + 2| + C[/latex]
Properties of the Exponential Function
The Main Idea
Definition of [latex]e[/latex]:
[latex]e[/latex] is the unique number such that [latex]\ln e = 1[/latex]
Approximately [latex]2.71828182846...[/latex]
Exponential Function:
Defined as the inverse of [latex]\ln x[/latex]
[latex]e^x = \exp(x)[/latex] for all real [latex]x[/latex]
Key Properties:
[latex]e^{\ln x} = x[/latex] for [latex]x > 0[/latex]
[latex]\ln(e^x) = x[/latex] for all [latex]x[/latex]
[latex]e^p e^q = e^{p+q}[/latex]
[latex]\frac{e^p}{e^q} = e^{p-q}[/latex]
[latex]{({e}^{p})}^{r}={e}^{pr}[/latex] (r rational)
Derivatives and Integrals:
[latex]\frac{d}{dx}e^x = e^x[/latex]
[latex]\int e^x dx = e^x + C[/latex]
Evaluate the following derivatives:
[latex]\frac{d}{dx}(\frac{{e}^{{x}^{2}}}{{e}^{5x}})[/latex]
[latex]\frac{d}{dt}{({e}^{2t})}^{3}[/latex]
Show Solution
[latex]\frac{d}{dx}(\frac{{e}^{{x}^{2}}}{{e}^{5x}})={e}^{{x}^{2}-5x}(2x-5)[/latex]
[latex]\frac{d}{dt}{({e}^{2t})}^{3}=6{e}^{6t}[/latex]
Watch the following video to see the worked solution to the above Try It.
VIDEO
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “6.7 Try It Problems” here (opens in new window) .
Evaluate the following integral: [latex]\displaystyle\int 2x{e}^{\text{−}{x}^{2}}dx.[/latex]
Show Solution
Using [latex]u[/latex]-substitution, let [latex]u=\text{−}{x}^{2}.[/latex] Then [latex]du=-2xdx,[/latex] and we have
[latex]\displaystyle\int 2x{e}^{\text{−}{x}^{2}}dx=\text{−}\displaystyle\int {e}^{u}du=\text{−}{e}^{u}+C=\text{−}{e}^{\text{−}{x}^{2}}+C.[/latex]
Find [latex]\frac{d}{dx}(e^{2x^3 + \sin x})[/latex]
Show Answer
Recognize this as a composition of functions
Apply the chain rule:
[latex]\frac{d}{dx}(e^{2x^3 + \sin x}) = e^{2x^3 + \sin x} \cdot \frac{d}{dx}(2x^3 + \sin x)[/latex]
Evaluate the inner derivative:
[latex]\frac{d}{dx}(2x^3 + \sin x) = 6x^2 + \cos x[/latex]
Combine:
[latex]\frac{d}{dx}(e^{2x^3 + \sin x}) = e^{2x^3 + \sin x}(6x^2 + \cos x)[/latex]
General Logarithmic and Exponential Functions
The Main Idea
Definition of General Exponential Functions:
For [latex]a > 0[/latex] and any real [latex]x[/latex]: [latex]y = a^x = e^{x \ln a}[/latex]
Derivatives of General Exponential Functions:
[latex]\frac{d}{dx} a^x = a^x \ln a[/latex]
Integrals of General Exponential Functions:
[latex]\int a^x dx = \frac{1}{\ln a} a^x + C[/latex]
General Logarithm Functions:
Inverse of [latex]a^x[/latex] when [latex]a \neq 1[/latex]
[latex]y = \log_a x[/latex] if and only if [latex]x = a^y[/latex]
Relationship to Natural Logarithm:
[latex]\log_a x = \frac{\ln x}{\ln a}[/latex]
Derivatives of General Logarithm Functions:
[latex]\frac{d}{dx} \log_a x = \frac{1}{x \ln a}[/latex]
Evaluate the following derivatives:
[latex]\frac{d}{dt}{4}^{{t}^{4}}[/latex]
[latex]\frac{d}{dx}{\text{log}}_{3}(\sqrt{{x}^{2}+1})[/latex]
Show Solution
[latex]\frac{d}{dt}{4}^{{t}^{4}}={4}^{{t}^{4}}(\text{ln}4)(4{t}^{3})[/latex]
[latex]\frac{d}{dx}{\text{log}}_{3}(\sqrt{{x}^{2}+1})=\frac{x}{(\text{ln}3)({x}^{2}+1)}[/latex]
Watch the following video to see the worked solution to the above Try It.
VIDEO
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “6.7 Try It Problems” here (opens in new window) .
Evaluate the following integral: [latex]\displaystyle\int \frac{3}{{2}^{3x}}dx.[/latex]
Show Solution
Use [latex]u\text{-substitution}[/latex] and let [latex]u=-3x.[/latex] Then [latex]du=-3dx[/latex] and we have
[latex]\displaystyle\int \frac{3}{{2}^{3x}}dx=\displaystyle\int 3·{2}^{-3x}dx=\text{−}\displaystyle\int {2}^{u}du=-\frac{1}{\text{ln}2}{2}^{u}+C=-\frac{1}{\text{ln}2}{2}^{-3x}+C.[/latex]
Evaluate the following derivative:
[latex]\frac{d}{dx} (3^x \cdot \log_2(x^2 + 1))[/latex]
Show Answer
Use the product rule:
[latex]\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'[/latex]
Let [latex]u = 3^x[/latex] and [latex]v = \log_2(x^2 + 1)[/latex]
Calculate [latex]u'[/latex]:
[latex]u' = 3^x \ln 3[/latex]
Calculate [latex]v'[/latex]:
[latex]v' = \frac{1}{(x^2 + 1) \ln 2} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{(x^2 + 1) \ln 2}[/latex]
Apply the product rule:
[latex]\begin{array}{rcl}