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## #lim _(x -> oo) ((x)/(2+x))^(x-2)# ?

Andrea S.
Featured 1 month ago

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = \frac{1}{e} ^ 2$

#### Explanation:

Write the expression as:

${\left(\frac{x}{2 + x}\right)}^{x - 2} = \frac{1}{{\left(\frac{x}{2 + x}\right)}^{- \left(x - 2\right)}} = {\left(\frac{x + 2}{x}\right)}^{2 - x} = {\left(1 + \frac{2}{x}\right)}^{2 - x}$

Now take the logarithm of this expression:

$\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right) = \left(2 - x\right) \ln \left(1 + \frac{2}{x}\right) = 2 \ln \left(1 + \frac{2}{x}\right) - x \ln \left(1 + \frac{2}{x}\right)$

We can see that:

${\lim}_{x \to \infty} 2 \ln \left(1 + \frac{2}{x}\right) = 2 \ln \left(1\right) = 0$

so we can ignore this term.

Focusing on the other addendum we have:

$x \ln \left(1 + \frac{2}{x}\right) = 2 \ln \frac{1 + \frac{2}{x}}{\frac{2}{x}}$

Substituting $y = \frac{2}{x}$ we have ${\lim}_{x \to \infty} y \left(x\right) = 0$ so that:

${\lim}_{x \to \infty} \ln \frac{1 + \frac{2}{x}}{\frac{2}{x}} = {\lim}_{y \to 0} \ln \frac{1 + y}{y} = 1$

(you can find the explanation here )

Putting this together we can see that:

${\lim}_{x \to \infty} \ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right) = {\lim}_{x \to \infty} 2 \ln \left(1 + \frac{2}{x}\right) - x \ln \left(1 + \frac{2}{x}\right) = 0 - 2 \cdot 1 = - 2$

Now note that:

${e}^{\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)} = {\left(\frac{x}{2 + x}\right)}^{x - 2}$

so that:

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = {\lim}_{x \to \infty} \left({e}^{\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)}\right)$

and as ${e}^{x}$ is continuous in all of $\mathbb{R}$:

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = {e}^{{\lim}_{x \to \infty} \left(\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)\right)} = {e}^{- 2} = \frac{1}{e} ^ 2$

## How do you find the area of one petal of #r=6sin2theta#?

Steve
Featured 1 month ago

$\frac{9 \pi}{2}$

#### Explanation:

Area in polar coordinates is given by:

$A = {\int}_{\alpha}^{\beta} \frac{1}{2} {r}^{2} \setminus d \theta$

The first step is to plot the polar curve to establish the appropriate range of $\theta$

From the graph we can see that for the petal in Q1 then $\theta \in \left[0 , \frac{\pi}{2}\right]$

Hence,

$A = {\int}_{0}^{\frac{\pi}{2}} \frac{1}{2} {\left(6 \sin 2 \theta\right)}^{2} \setminus d \theta$
$\setminus \setminus \setminus = 18 \setminus {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} 2 \theta \setminus d \theta$
$\setminus \setminus \setminus = 18 \setminus {\int}_{0}^{\frac{\pi}{2}} \frac{1}{2} \left(1 - \cos 4 \theta\right) \setminus d \theta$
$\setminus \setminus \setminus = 9 \setminus {\int}_{0}^{\frac{\pi}{2}} 1 - \cos 4 \theta \setminus d \theta$
$\setminus \setminus \setminus = 9 \setminus {\left[\theta - \frac{1}{4} \sin 4 \theta\right]}_{0}^{\frac{\pi}{2}}$
$\setminus \setminus \setminus = 9 \setminus \left\{\left(\frac{\pi}{2} - \frac{1}{4} \sin \left(2 \pi\right)\right) - \left(0\right)\right\}$
$\setminus \setminus \setminus = \frac{9 \pi}{2}$

## How do you solve #\int \frac { e ^ { x } - 1} { e ^ { x } + 1} d x#?

HSBC244
Featured 1 month ago

The integral equals $x - 2 \ln | {e}^{x} / \left({e}^{x} + 1\right) | + C$

#### Explanation:

By partial fractions:

$\frac{A}{1} + \frac{B}{{e}^{x} + 1} = {e}^{x} - 2 + 1$

$A \left({e}^{x} + 1\right) + B = {e}^{x} - 1$

$A {e}^{x} + A + B = {e}^{x} - 1$

$A {e}^{x} + \left(A + B\right) = {e}^{x} - 1$

We can write a system of equations.

$\left\{\begin{matrix}A = 1 \\ A + B = - 1\end{matrix}\right.$

Solving, we get $A = 1$ and $B = - 2$. The integral becomes:

$\implies \int 1 - \frac{2}{{e}^{x} + 1} \mathrm{dx}$

We can separate.

$\implies \int 1 \mathrm{dx} - 2 \int \frac{1}{{e}^{x} + 1} \mathrm{dx}$

Let $u = {e}^{x}$. Then $\mathrm{du} = {e}^{x} \mathrm{dx}$. Then $\mathrm{dx} = \frac{\mathrm{du}}{e} ^ x = \frac{\mathrm{du}}{u}$

$\implies \int 1 \mathrm{dx} - 2 \int \frac{1}{\left(u + 1\right) u} \mathrm{du}$

We're going to use partial fractions again.

$\frac{A}{u + 1} + \frac{B}{u} = 1$

$A u + B u + B = 1$

Write a system of equations

$\left\{\begin{matrix}A + B = 0 \\ B = 1\end{matrix}\right.$

Solve to get $A = - 1$ and $B = 1$.

The expression becomes:

$\implies \int 1 \mathrm{dx} - 2 \left(\int \frac{1}{u} \mathrm{du} - \int \frac{1}{u + 1} \mathrm{du}\right)$

$\implies \int 1 \mathrm{dx} - 2 \int \frac{1}{u} \mathrm{du} + 2 \int \frac{1}{u + 1} \mathrm{du}$

$\implies x - 2 \ln | u | + 2 \ln | u + 1 | + C$

$\implies x - 2 \left(\ln | u | - \ln | u + 1 |\right) + C$

The logarithms can be combined.

$\implies x - 2 \ln | \frac{u}{u + 1} | + C$

$\implies x - 2 \ln | {e}^{x} / \left({e}^{x} + 1\right) | + C$

Hopefully this helps!

## How do you find the derivative of #-ln(x-(x^2+1)^(1/2))#?

Rory H.
Featured 1 month ago

The derivative of $- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$ with respect to x is $\frac{1}{\sqrt{{x}^{2} + 1}} .$ To solve this use the chain rule carefully. See the explanation for details.

#### Explanation:

Starting with:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right)$

use the chain rule to get:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right)$

$= - \frac{1}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} \cdot \frac{d}{\mathrm{dx}} \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$

use the chain rule once again on the remaining derivative:

$= - \frac{1}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} \cdot \left(1 - \frac{1}{2} {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \left(2 x\right)\right)$

Simplify:

$= \frac{\frac{x}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right) - 1}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

Note that $1 = {\left({x}^{2} + 1\right)}^{\frac{1}{2}} / {\left({x}^{2} + 1\right)}^{\frac{1}{2}}$, then substitute this for 1:

$= \frac{\frac{x}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right) - {\left({x}^{2} + 1\right)}^{\frac{1}{2}} / {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= \frac{\frac{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= \left(\frac{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)\right) \div i \mathrm{de} \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$

$= \left(\frac{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)\right) \cdot \frac{1}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= \frac{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} \cdot \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= 1 \cdot \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{1}{2}}} = \frac{1}{\sqrt{\left({x}^{2} + 1\right)}} .$

Finally:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right) = \frac{1}{\sqrt{\left({x}^{2} + 1\right)}} .$

If you have any questions about the use of the chain rule or any other part of this solution, then please ask.

Rory.

## Can someone help out with the question below?

Stefan V.
Featured 3 weeks ago

$P ' \left(16\right) = \frac{75}{32}$

#### Explanation:

The idea here is that you can find the rate of change of the pollution with respect to time by taking the first derivative of your function $P \left(t\right)$.

So this is pretty much an exercise in finding the derivative of the function

$P \left(t\right) = {\left({t}^{\frac{1}{4}} + 3\right)}^{3}$

which can be calculated by using the chain rule and the power rule. Keep in mind that you have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{d}{\mathrm{dt}} \left[u {\left(t\right)}^{n}\right] = n \cdot u {\left(t\right)}^{n - 1} \cdot \frac{d}{\mathrm{dt}} \left[u \left(t\right)\right]}}}$

$\left\{\begin{matrix}u \left(t\right) = {t}^{\frac{1}{4}} + 3 \\ n = 3\end{matrix}\right.$

This means that the derivate of $P \left(t\right)$ will be

${\overbrace{\frac{d}{d} \left(\mathrm{dt}\right) \left[P \left(t\right)\right]}}^{\textcolor{b l u e}{= {P}^{'} \left(t\right)}} = 3 \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2} \cdot \frac{d}{\mathrm{dt}} \left({t}^{\frac{1}{4}} + 3\right)$

${P}^{'} \left(t\right) = 3 \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2} \cdot \frac{1}{4} \cdot {t}^{\left(\frac{1}{4} - 1\right)}$

${P}^{'} \left(t\right) = \frac{3}{4} \cdot {t}^{- \frac{3}{4}} \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2}$

Now all you have to do is plug in $16$ for $t$ to find

${P}^{'} \left(t\right) = \frac{3}{4} \cdot {16}^{- \frac{3}{4}} \cdot {\left({16}^{\frac{1}{4}} + 3\right)}^{2}$

Since you know that

$16 = {2}^{4}$

you can rewrite the equation as

${P}^{'} \left(16\right) = \frac{3}{4} \cdot {2}^{\left[4 \cdot \left(- \frac{3}{4}\right)\right]} \cdot {\left[{2}^{\left(4 \cdot \frac{1}{4}\right)} + 3\right]}^{2}$

${P}^{'} \left(16\right) = \frac{3}{4} \cdot {2}^{- 3} \cdot {\left(2 + 3\right)}^{2}$

${P}^{'} \left(16\right) = \frac{3}{4} \cdot \frac{1}{8} \cdot 25$

${P}^{'} \left(16\right) = \frac{75}{32}$

And there you have it -- the rate at which the pollution changes after $16$ years.

## How do you differentiate #y=sqrt(4+3x)#?

Johnson Z.
Featured 3 weeks ago

$y = \frac{3}{2 {\left(4 + 3 x\right)}^{\frac{1}{2}}}$

#### Explanation:

Recall the formula for chain rule:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ or $\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} f ' \left(x\right) = g ' \left[h \left(x\right)\right] h ' \left(x\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and the formula for power rule:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

To start, recognize the inside and outside functions of $y = \textcolor{g r e e n}{\sqrt{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{4 + 3 x}}}$.

Inside function: $y = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{4 + 3 x}$
Outside function: $y = \textcolor{g r e e n}{\sqrt{a}}$

How to Differentiate Using Chain Rule

$1$. Take the derivative of the outside function, $y = \sqrt{a}$, but replace the $a$ with the inside function, $4 + 3 x$.

$2$. Multiply by the derivative of the inside function, $4 + 3 x$.

Applying Chain Rule
1. The derivative of the outside function, $y = \sqrt{a}$, would be $\frac{1}{2} {a}^{- \frac{1}{2}}$, using the power rule. However, $a$ needs to be replaced by the inside function, so it becomes $\frac{1}{2} {\left(4 + 3 x\right)}^{- \frac{1}{2}}$.

$y = \sqrt{a}$

$\textcolor{red}{\downarrow}$

$y = \frac{1}{2} {a}^{- \frac{1}{2}}$

$\textcolor{red}{\downarrow}$

$y = \frac{1}{2} {\left(4 + 3 x\right)}^{- \frac{1}{2}}$

2.$\textcolor{w h i t e}{i}$Then we need to multiply by the derivative of the inside function, $4 + 3 x$, which becomes $3$ using the power rule.

$y = \frac{1}{2} {\left(4 + 3 x\right)}^{- \frac{1}{2}}$

$\textcolor{red}{\downarrow}$

$y = \frac{1}{2} {\left(4 + 3 x\right)}^{- \frac{1}{2}} \left(3\right)$

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} y = \frac{3}{2 {\left(4 + 3 x\right)}^{\frac{1}{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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