## Featured Answers

22
Active contributors today

## How do you find the derivative of #y=ln((x-1)/(x+1))^(1/3)#?

Steve
Featured 2 months ago

#### Answer:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3 {x}^{2} - 3}$

#### Explanation:

$y = \ln {\left(\frac{x - 1}{x - 1}\right)}^{\frac{1}{3}}$
$\therefore y = \frac{1}{3} \ln \left(\frac{x - 1}{x + 1}\right)$
$\therefore 3 y = \ln \left(\frac{x - 1}{x + 1}\right)$
$\therefore {e}^{3 y} = \left(\frac{x - 1}{x + 1}\right)$ ..... [1]

Differentiating the LHS implicity and the RHS using the quotient rule gives:

$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x + 1\right) \left(1\right) - \left(x - 1\right) \left(1\right)}{x + 1} ^ 2$
$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 1 - x + 1}{x + 1} ^ 2$
$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x + 1} ^ 2$

$\therefore 3 \left(\frac{x - 1}{x + 1}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x + 1} ^ 2$ (using [1])
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{x + 1} ^ 2 \cdot \frac{x + 1}{x - 1}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{x + 1} \cdot \frac{1}{x - 1}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{{x}^{2} - 1}$

Hence,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3 {x}^{2} - 3}$

## Use Newton's method to approximate the given number correct to eight decimal places?

Henry W.
Featured 1 month ago

#### Answer:

$1.04910303 \left(8 d . p .\right)$

#### Explanation:

Let $x = {95}^{\frac{1}{95}}$

${x}^{95} = 95$

${x}^{95} - 95 = 0$

Let $f \left(x\right) = {x}^{95} - 95$

Then $f ' \left(x\right) = 95 {x}^{94}$

Now we simply plug into the formula with a starting point.

${x}_{0} = 1.2$

${x}_{1} = {x}_{0} - \frac{f \left({x}_{0}\right)}{f ' \left({x}_{0}\right)}$

$= 1.2 - \frac{{1.2}^{95} - 95}{95 {\left(1.2\right)}^{94}}$

$= 1.18736845$

Second iteration:

${x}_{2} = {x}_{1} - \frac{f \left({x}_{1}\right)}{f ' \left({x}_{1}\right)}$

$= 1.17486993$

And so on, starting with $1.2$ it took me 15 iterations to get to an approximation with 8 constant d.p. Answer should be $1.04910303 \left(8 d . p .\right)$

## How do I find the time, 𝑇(𝑥), that is required to travel from A to C via P?

Steve
Featured 2 weeks ago

#### Answer:

$D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

$T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$

From this we get the minimum time as $1.65$ (hours) which corresponds to a distance of $7$ (km).

#### Explanation:

A) The distance $D \left(x\right)$ from A to C via P as a function of $x$.

By Pythagoras;

$\setminus \setminus \setminus \setminus \setminus A {P}^{2} = A {B}^{2} + B {P}^{2}$
$\therefore A {P}^{2} = {3}^{2} + {x}^{2}$
$\therefore A {P}^{2} = 9 + {x}^{2}$
$\therefore \setminus \setminus A P = \sqrt{9 + {x}^{2}}$ (must be the +ve root)

Then;

$\setminus \setminus \setminus \setminus \setminus D \left(x\right) = A P + P C$
$\therefore D \left(x\right) = \sqrt{9 + {x}^{2}} + \left(6 - x\right)$

B) Find the time, $T \left(x\right)$, that is required to travel from A to C via P.

Using $\text{speed" = "distance"/"time}$:

Along AP the speed is $4$ km/hour, provided $x > 0$ (otherwise we are going along AB at $5$ km/hour) and so:

$\text{ } 4 = \frac{\sqrt{9 + {x}^{2}}}{t} _ 1$
$\therefore {t}_{1} = \frac{\sqrt{9 + {x}^{2}}}{4}$

Along PC (or AB) the speed is $5$ km/hour, and so:

$\text{ } 5 = \frac{6 - x}{t} _ 2$
$\therefore {t}_{2} = \frac{6 - x}{5}$

And so, the total time is given by:

$\setminus \setminus \setminus \setminus \setminus T \left(x\right) = {t}_{1} + {t}_{2}$
$\therefore T \left(x\right) = \frac{\sqrt{9 + {x}^{2}}}{4} + \frac{6 - x}{5}$ for ($x > 0$)

C) Stevie's Quest

In order for Stevie to complete the quest we need to find a critical point of T(x):

Differentiating wrt $x$ we get:

$\text{ } T ' \left(x\right) = \frac{1}{4} \cdot \frac{1}{2} {\left(9 + {x}^{2}\right)}^{- \frac{1}{2}} \cdot 2 x + \frac{1}{5} \left(- 1\right)$
$\therefore T ' \left(x\right) = \frac{x}{4 \sqrt{\left(9 + {x}^{2}\right)}} - \frac{1}{5}$

At a critical point, $T ' \left(x\right) = 0$

$\implies \frac{x}{4 \sqrt{9 + {x}^{2}}} - \frac{1}{5} = 0$
$\therefore 5 x - 4 \sqrt{9 + {x}^{2}} = 0$
$\therefore 5 x = 4 \sqrt{9 + {x}^{2}}$
$\therefore 25 {x}^{2} = 16 \left(9 + {x}^{2}\right)$
$\therefore 25 {x}^{2} = 144 + 16 {x}^{2}$
$\therefore 9 {x}^{2} = 144$
$\therefore {x}^{2} = \frac{144}{9}$
$\therefore {x}^{2} = 16$
$\therefore x = 4$ (must be the +ve root)

When $x = 4$ we have:

$\therefore D \left(4\right) = \sqrt{9 + 16} + \left(6 - 4\right) = 7$
$\therefore T \left(4\right) = \frac{\sqrt{9 + 16}}{4} + \frac{6 - 4}{5} = \frac{33}{20} = 1.65$

We can confirm visually that this corresponds to a minimum by looking at the graph:

graph{sqrt(9 + x^2)/4 + (6-x)/5 [-15, 15, -1, 10]}

## How do you determine whether the function satisfies the hypotheses of the Mean Value Theorem for #f(x)=x^(1/3)# on the interval [-5,4]?

Jim H
Featured 1 week ago

#### Answer:

It doesn't. But it does satisfy the conclusion. See below.

#### Explanation:

There are two hypotheses for MVT.

The function must be continuous on $\left[- 5 , 4\right]$?

The function must be differentiable on $\left(- 5 , 4\right)$?

The first is true (satisfied) because ${x}^{\frac{1}{3}}$ is the 3rd root function and a root function is continuous on its domain. In this case the domain is $\mathbb{R}$, so it includes $\left[- 5 , 4\right]$.

This function fails to be differentiable at $0$, so the second hypothesis is not satisfied on $\left(- 5 , 4\right)$.

This function does not satisfy the hypotheses of the Mean Value Theorem on this interval.

Bonus material

This function DOES satisfy the conclusion of the MVT on this interval. We cannot use the Mean Value Theorem to conclude that there is a $c$ in $\left(- 5 , 4\right)$ such that $f ' \left(c\right) = \frac{f \left(4\right) - f \left(- 5\right)}{4 - \left(- 5\right)}$.

We can, however solve $f ' \left(c\right) = \frac{f \left(4\right) - f \left(- 5\right)}{4 - \left(- 5\right)}$ algebraically. We find that there are two solutions in the interval.

The algebra is be tedious to type, but the graph makes this plausible.
It shows $f \left(x\right) = {x}^{\frac{1}{3}}$ and the secant line joining the points with $x = - 5$ and $x = 4$.

graph{(y-x^(1/3)) (y-4^{1/3)-((4^(1/3)+5^(1/3))/9)(x-4))= 0 [-6.06, 5.04, -2.864, 2.685]}

## How do you integrate #int 2/(y^4sqrt(y^2-25)# using trig substitutions?

HSBC244
Featured 6 days ago

#### Answer:

$\frac{2 \sqrt{{y}^{2} - 25}}{625 y} - \frac{2 {\left({y}^{2} - 25\right)}^{\frac{3}{2}}}{1875 {y}^{3}} + C$

#### Explanation:

Our goal here is to get rid of the #√# by making use of a pythagorean identity. Let $y = 5 \sec \theta$. Then $\mathrm{dy} = 5 \sec \theta \tan \theta d \theta$.

$= \int \frac{2}{{\left(5 \sec \theta\right)}^{4} \sqrt{{\left(5 \sec \theta\right)}^{2} - 25}} \cdot 5 \sec \theta \tan \theta d \theta$

Apply the identity ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$= \int \frac{2}{625 {\sec}^{4} \theta \sqrt{25 {\tan}^{2} \theta}} \cdot 5 \sec \theta \tan \theta d \theta$

$= \int \frac{2}{625 {\sec}^{4} \left(5 \tan \theta\right)} \cdot 5 \sec \theta \tan \theta d \theta$

$= \int \frac{2}{625 {\sec}^{3} \theta} d \theta$

Rewrite using $\cos x = \frac{1}{\sec} x$

$= \int \frac{2}{625} {\cos}^{3} \theta d \theta$

$= \int \frac{2}{625} {\cos}^{2} \theta \left(\cos \theta\right) d \theta$

Rearrange using the pythagorean identity ${\cos}^{2} x + {\sin}^{2} x = 1$:

$= \int \frac{2}{625} \left(1 - {\sin}^{2} \theta\right) \cos \theta d \theta$

Let $u = \sin \theta$. Then $\mathrm{du} = \cos \theta d \theta$ and $d \theta = \frac{\mathrm{du}}{\cos} \theta$.

$= \int \frac{2}{625} \left(1 - {u}^{2}\right) \cos \theta \cdot \frac{\mathrm{du}}{\cos} \theta$

$= \int \frac{2}{625} \left(1 - {u}^{2}\right) \mathrm{du}$

Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ where $n \ne - 1$.

$= \frac{2}{625} \left(u - \frac{1}{3} {u}^{3}\right) + C$

$= \frac{2}{625} \sin \theta - \frac{2}{1875} {\sin}^{3} \theta + C$

Draw a triangle to represent $\frac{y}{5} = \sec \theta$.

$= \frac{2}{625} \frac{\sqrt{{y}^{2} - 25}}{y} - \frac{2}{1875} {\left(\frac{\sqrt{{y}^{2} - 25}}{y}\right)}^{3} + C$

$= \frac{2 \sqrt{{y}^{2} - 25}}{625 y} - \frac{2 {\left({y}^{2} - 25\right)}^{\frac{3}{2}}}{1875 {y}^{3}} + C$

Hopefully this helps!

## How do you integrate #(x^4+x^3+x^2+1)/(x^2+x-2)# using partial fractions?

Douglas K.
Featured 2 days ago

#### Answer:

Because the order of the numerator is greater than the denominator some reduction must be done before expanding the partial fractions. Please see the explanation.

#### Explanation:

Given: $\frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Begin reducing by adding 0 to the numerator in the form of $- 2 {x}^{2} + 2 {x}^{2}$

$\frac{{x}^{4} + {x}^{3} - 2 {x}^{2} + 2 {x}^{2} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Break into two fractions:

$\frac{{x}^{4} + {x}^{3} - 2 {x}^{2}}{{x}^{2} + x - 2} + \frac{2 {x}^{2} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Factor ${x}^{2}$ from the numerator of the first term:

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

The first term becomes ${x}^{2}$ and we combine like terms in the numerator of the second fraction:

${x}^{2} + \frac{3 {x}^{2} + 1}{{x}^{2} + x - 2}$

Add 0 to the numerator of the second term in the form of $3 x - 6 - 3 x + 6$:

${x}^{2} + \frac{3 {x}^{2} + 3 x - 6 - 3 x + 6 + 1}{{x}^{2} + x - 2}$

Break the second term into two fractions:

${x}^{2} + \frac{3 {x}^{2} + 3 x - 6}{{x}^{2} + x - 2} + \frac{- 3 x + 6 + 1}{{x}^{2} + x - 2}$

Remove a factor of 3 from the first numerator and combine like terms in the last fraction:

${x}^{2} + \frac{3 \left(\cancel{{x}^{2} + x - 2}\right)}{\cancel{{x}^{2} + x - 2}} + \frac{7 - 3 x}{{x}^{2} + x - 2}$

The second term becomes 3:

${x}^{2} + 3 + \frac{7 - 3 x}{{x}^{2} + x - 2}$

Partial Fraction Expansion of the last term:

$\frac{7 - 3 x}{{x}^{2} + x - 2} = \frac{A}{x + 2} + \frac{B}{x - 1}$

$7 - 3 x = A \left(x - 1\right) + B \left(x + 2\right)$

Make B disappear by letting x = -2

$7 - 3 \left(- 2\right) = A \left(- 2 - 1\right) + B \left(- 2 + 2\right)$

$A$ = -13/3

Make A disappear by Letting x = 1:

$7 - 3 \left(1\right) = A \left(1 - 1\right) + B \left(1 + 2\right)$

$B = \frac{4}{3}$

Check

$\frac{- \frac{13}{3}}{x + 2} + \frac{\frac{4}{3}}{x - 1}$

$\frac{- \frac{13}{3}}{x + 2} \frac{x - 1}{x - 1} + \frac{\frac{4}{3}}{x - 1} \frac{x + 2}{x + 2}$

$\frac{\left(- \frac{13}{3}\right) \left(x - 1\right) + \left(\frac{4}{3}\right) \left(x + 2\right)}{\left(x + 2\right) \left(x - 1\right)}$

$\frac{7 - 3 x}{\left(x + 2\right) \left(x - 1\right)}$

This checks

Returning to the main problem:

$\int \frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2} \mathrm{dx} = \int {x}^{2} \mathrm{dx} + 3 \int \mathrm{dx} - \frac{13}{3} \int \frac{1}{x + 2} \mathrm{dx} + \frac{4}{3} \int \frac{1}{x - 1} \mathrm{dx}$

$\int \frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2} \mathrm{dx} = \frac{1}{3} {x}^{3} + 3 x - \frac{13}{3} \ln \left(x + 2\right) + \frac{4}{3} \ln \left(x - 1\right) + C$

##### Questions
Ask a question Filters
Loading...
• · 10 minutes ago · in Chain Rule
• 34 minutes ago · in Chain Rule
• · 59 minutes ago · in Quotient Rule
• An hour ago · in Chain Rule
• · 2 hours ago · in Integration by Parts
• 4 hours ago · in Power Rule
• 5 hours ago · in Chain Rule
This filter has no results, see all questions.
×
##### Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions