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Let’s now apply this definition to calculate a differentiation formula for ${a}^{x}.$ We have
The corresponding integration formula follows immediately.
Let $a>0.$ Then,
and
If $a\ne 1,$ then the function ${a}^{x}$ is one-to-one and has a well-defined inverse. Its inverse is denoted by ${\text{log}}_{a}x.$ Then,
Note that general logarithm functions can be written in terms of the natural logarithm. Let $y={\text{log}}_{a}x.$ Then, $x={a}^{y}.$ Taking the natural logarithm of both sides of this second equation, we get
Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base $a.$ Again, let $y={\text{log}}_{a}x.$ Then,
Let $a>0.$ Then,
Evaluate the following derivatives:
We need to apply the chain rule as necessary.
Evaluate the following derivatives:
Evaluate the following integral: $\int \frac{3}{{2}^{3x}}}dx.$
Use $u\text{-substitution}$ and let $u=\mathrm{-3}x.$ Then $du=\mathrm{-3}dx$ and we have
Evaluate the following integral: $\int {x}^{2}{2}^{{x}^{3}}dx}.$
$\int {x}^{2}{2}^{{x}^{3}}dx}=\frac{1}{3\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}2}{2}^{{x}^{3}}+C$
For the following exercises, find the derivative $\frac{dy}{dx}.$
$y=\text{ln}\left(2x+1\right)$
$y=\frac{1}{\text{ln}\phantom{\rule{0.2em}{0ex}}x}$
$-\frac{1}{x{(\text{ln}\phantom{\rule{0.2em}{0ex}}x)}^{2}}$
For the following exercises, find the indefinite integral.
$\int \frac{dt}{3t}$
For the following exercises, find the derivative $dy\text{/}dx.$ (You can use a calculator to plot the function and the derivative to confirm that it is correct.)
[T] $y=\frac{\text{ln}\left(x\right)}{x}$
[T] $y=x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)$
$\text{ln}\left(x\right)+1$
[T] $y={\text{log}}_{10}x$
[T] $y=\text{ln}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x\right)$
$\text{cot}\left(x\right)$
[T] $y=\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)$
[T] $y=7\phantom{\rule{0.2em}{0ex}}\text{ln}\left(4x\right)$
$\frac{7}{x}$
[T] $y=\text{ln}\left({\left(4x\right)}^{7}\right)$
[T] $y=\text{ln}\left(\text{tan}\phantom{\rule{0.2em}{0ex}}x\right)$
$\text{csc}\left(x\right)\text{sec}\phantom{\rule{0.2em}{0ex}}x$
[T] $y=\text{ln}\left(\text{tan}\left(3x\right)\right)$
[T] $y=\text{ln}\left({\text{cos}}^{2}x\right)$
$\mathrm{-2}\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}x$
For the following exercises, find the definite or indefinite integral.
$\int}_{0}^{1}\frac{dx}{3+x$
$\int}_{0}^{1}\frac{dt}{3+2t$
$\frac{1}{2}\text{ln}\left(\frac{5}{3}\right)$
$\int}_{0}^{2}\frac{x\phantom{\rule{0.2em}{0ex}}dx}{{x}^{2}+1$
$\int}_{0}^{2}\frac{{x}^{3}dx}{{x}^{2}+1$
$2-\frac{1}{2}\text{ln}\left(5\right)$
$\int}_{2}^{e}\frac{dx}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x$
$\int}_{2}^{e}\frac{dx}{{\left(x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)\right)}^{2}$
$\frac{1}{\text{ln}\left(2\right)}-1$
$\int \frac{\text{cos}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx}{\text{sin}\phantom{\rule{0.2em}{0ex}}x}$
${\int}_{0}^{\pi \text{/}4}\text{tan}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx$
$\frac{1}{2}\text{ln}\left(2\right)$
$\int \text{cot}\left(3x\right)dx$
$\int \frac{{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}^{2}dx}{x}$
$\frac{1}{3}{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}^{3}$
For the following exercises, compute $dy\text{/}dx$ by differentiating $\text{ln}\phantom{\rule{0.2em}{0ex}}y.$
$y=\sqrt{{x}^{2}+1}$
$y=\sqrt{{x}^{2}+1}\sqrt{{x}^{2}-1}$
$\frac{2{x}^{3}}{\sqrt{{x}^{2}+1}\sqrt{{x}^{2}-1}}$
$y={e}^{\text{sin}\phantom{\rule{0.2em}{0ex}}x}$
$y={x}^{\mathrm{-1}\text{/}x}$
${x}^{\mathrm{-2}-(1\text{/}x)}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x-1\right)$
$y={e}^{\left(ex\right)}$
$y={x}^{\left(ex\right)}$
$y=\sqrt{x}\phantom{\rule{0.2em}{0ex}}\sqrt[3]{x}\phantom{\rule{0.2em}{0ex}}\sqrt[6]{x}$
$1$
$y={x}^{\mathrm{-1}\text{/}\text{ln}\phantom{\rule{0.2em}{0ex}}x}$
$y={e}^{\text{\u2212}\text{ln}\phantom{\rule{0.2em}{0ex}}x}$
$-\frac{1}{{x}^{2}}$
For the following exercises, evaluate by any method.
$\int}_{5}^{10}\frac{dt}{t}-{\displaystyle {\int}_{5x}^{10x}\frac{dt}{t}$
$\int}_{1}^{{e}^{\pi}}\frac{dx}{x}}+{\displaystyle {\int}_{\mathrm{-2}}^{\mathrm{-1}}\frac{dx}{x$
$\pi -\text{ln}\left(2\right)$
$\frac{d}{dx}{\displaystyle {\int}_{x}^{1}\frac{dt}{t}}$
$\frac{d}{dx}{\displaystyle {\int}_{x}^{{x}^{2}}\frac{dt}{t}}$
$\frac{1}{x}$
$\frac{d}{dx}\text{ln}\left(\text{sec}\phantom{\rule{0.2em}{0ex}}x+\text{tan}\phantom{\rule{0.2em}{0ex}}x\right)$
For the following exercises, use the function $\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ If you are unable to find intersection points analytically, use a calculator.
Find the area of the region enclosed by $x=1$ and $y=5$ above $y=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$
${e}^{5}-6\phantom{\rule{0.2em}{0ex}}{\text{units}}^{2}$
[T] Find the arc length of $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ from $x=1$ to $x=2.$
Find the area between $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ and the x -axis from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2.$
$\text{ln}\left(4\right)-1\phantom{\rule{0.2em}{0ex}}{\text{units}}^{2}$
Find the volume of the shape created when rotating this curve from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=2$ around the x -axis, as pictured here.
[T] Find the surface area of the shape created when rotating the curve in the previous exercise from $x=1$ to $x=2$ around the x -axis.
$2.8656$
If you are unable to find intersection points analytically in the following exercises, use a calculator.
Find the area of the hyperbolic quarter-circle enclosed by $x=2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=2$ above $y=1\text{/}x.$
[T] Find the arc length of $y=1\text{/}x$ from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4.$
$3.1502$
Find the area under $y=1\text{/}x$ and above the x -axis from $x=1\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=4.$
For the following exercises, verify the derivatives and antiderivatives.
$\frac{d}{dx}\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)=\frac{1}{\sqrt{1+{x}^{2}}}$
$\frac{d}{dx}\text{ln}\left(\frac{x-a}{x+a}\right)=\frac{2a}{\left({x}^{2}-{a}^{2}\right)}$
$\frac{d}{dx}\text{ln}\left(\frac{1+\sqrt{1-{x}^{2}}}{x}\right)=-\frac{1}{x\sqrt{1-{x}^{2}}}$
$\frac{d}{dx}\text{ln}\left(x+\sqrt{{x}^{2}-{a}^{2}}\right)=\frac{1}{\sqrt{{x}^{2}-{a}^{2}}}$
$\int \frac{dx}{x\phantom{\rule{0.2em}{0ex}}\text{ln}\left(x\right)\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)}}=\text{ln}\left(\text{ln}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}x\right)\right)+C$
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