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3

## How did earlier mathematicians calculate limits so accurately?

Steve
Featured 1 week ago

With the exception of using L'Hospital's Rule and Taylor Series (which both rely on Calculus) they would use the same techniques that we currently use.

Most notably is that the limit you have used as an example, which can easily be derived using basic geometry.

Example

This is a technique used by Archimedes circa 250BC where he used regular polygons inside and outside to obtain approximations for $\pi$. I quite like this example as it shows the brilliance of the early mathematicians centuries before modern mathematics.

The above figure shows a regular $12$-sided polygon inscribed in a circle of radius $1$ unit, centre $O$, $A B$ is one of the sides of the polygon. $C$ is the midpoint of $A B$. Archimedes used the fact that the circumference of the circle is greater that the perimeter of this polygon.

Now,

${360}^{o} \div 12 = {30}^{o} \implies \hat{A O B} = {30}^{o}$
$C$ is the midpoint of $A B \implies A B = 2 B C$
$\therefore C O B = {15}^{o}$

By trigonometry we have:

$\sin \hat{C O B} = \frac{B C}{1}$
But $A B = 2 B C$ and $\hat{C O B} = {15}^{o} \implies \sin {15}^{'} = \frac{1}{2} A B$
$\therefore A B = 2 \sin {15}^{o}$

Using $\cos 2 A \equiv 1 - 2 {\sin}^{2} A$ we have:

$\cos {30}^{o} = 1 - 2 {\sin}^{2} {15}^{o}$
$\therefore \frac{\sqrt{3}}{2} = 1 - 2 {\sin}^{2} {15}^{o}$
$\therefore 2 {\sin}^{2} {15}^{o} = 1 - \frac{\sqrt{3}}{2}$
$\therefore {\sin}^{2} {15}^{o} = \frac{1 - \frac{\sqrt{3}}{2}}{2}$
$\therefore {\sin}^{2} {15}^{o} = \frac{2 - \sqrt{3}}{4}$
 :. sin 15^o = 1/2sqrt(2-sqrt(3)

Then he calculated the perimeter of the polygon using:

$P = 12 A B$
$\setminus \setminus \setminus = 24 \sin {15}^{o}$
 \ \ \ = 12 sqrt(2-sqrt(3)

Then as above:

Circumference of circle $>$ perimeter of polygon
 :. (2pi)(1) gt 12 sqrt(2-sqrt(3)
 :. pi gt 6 sqrt(2-sqrt(3)  ..... [A]

Next Archimedes considered a 12-sided polygon that lies outside a circle of radius $1$ unit, which touches each side of the polygon. $F$ is the midpoint of $D E$. Archimedes used the fact that the circumference of the circle is less that the perimeter of this polygon.

As before,

$\hat{D O E} = {30}^{o} \implies \hat{F O E} = {15}^{o}$
And, $D E = 2 F E$

By trigonometry:

$\tan \hat{F O E} = \frac{F E}{1}$
But $D E = 2 F E$ and $\hat{F O E} = {15}^{o} \implies \tan {15}^{o} = \frac{1}{2} \setminus D E$

Using $\tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ we have:

$\tan {30}^{o} = \frac{2 \tan {15}^{o}}{1 - {\tan}^{2} {15}^{o}}$
$\therefore \frac{1}{\sqrt{3}} = \frac{2 t}{1 - {t}^{2}}$, where $t = \tan {15}^{o}$
$\therefore 2 \sqrt{3} t = 1 - {t}^{2}$
$\therefore {t}^{2} + 2 \sqrt{3} t - 1 = 0$

This is quadratic in $t \left(= \tan {15}^{o}\right)$, and so using the quadratic formula:

$t = \frac{- 2 \sqrt{3} \pm \sqrt{2 {\sqrt{}}^{2} - 4 \left(1\right) \left(- 1\right)}}{2}$
$\setminus \setminus = - \sqrt{3} \pm 2$
But as $t = \tan {15}^{o} > 0 \implies \tan {15}^{o} = 2 - \sqrt{3}$

And so the perimeter of this polygon is:

$P = 12 D E$
$\setminus \setminus \setminus = 12 \tan {15}^{o}$
$\setminus \setminus \setminus = 24 \left(2 - \sqrt{3}\right)$

This time:

Circumference of circle $<$ perimeter of polygon
$\therefore \left(2 \pi\right) \left(1\right) < 24 \left(2 - \sqrt{3}\right)$
$\therefore \pi < 12 \left(2 - \sqrt{3}\right)$ ..... [B}

Combining the results [A] and [B} Archimedes then showed that:

$6 \sqrt{2 - \sqrt{3}} < \pi < 12 \left(2 - \sqrt{3}\right)$
$\therefore 3.10582 \ldots < \pi < 3.21539 \ldots$
$\therefore 3.106 < \pi < 3.215$

And of course we know that:

$3.141592653589793238462 \ldots$

Archimedes used this same technique on a $24$-sided polygon, and finally $48$- and $96$-sided polygons. Those early Greek philosophers sure did now their stuff !

As we take larger and larger $n$-sided polygons, this is the equivalent of the limiting process. Remember this was several hundred years before Isaac Newton, and Leibniz started using limits of infinitesimals for early Calculus in the 17th Century.

3

## What is int arctan(x^2) dx?

Eric Sia
Featured 2 weeks ago

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

#### Explanation:

Let

$I = \int {\tan}^{- 1} \left({x}^{2}\right) \mathrm{dx}$

Apply integration by parts:

$f \left(x\right) = {\tan}^{- 1} \left({x}^{2}\right)$, $f ' \left(x\right) = \frac{2 x}{{x}^{4} + 1}$
$g ' \left(x\right) = 1$, $g \left(x\right) = x$

Hence

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \int \frac{2 {x}^{2}}{{x}^{4} + 1} \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \int \frac{2 {x}^{2}}{{x}^{4} + 2 {x}^{2} + 1 - 2 {x}^{2}} \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \int \frac{2 {x}^{2}}{{\left({x}^{2} + 1\right)}^{2} - 2 {x}^{2}} \mathrm{dx}$

Apply the difference of squares ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$:

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \int \frac{2 {x}^{2}}{\left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right)} \mathrm{dx}$

Apply partial fraction decomposition:

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{1}{\sqrt{2}} \int \left(\frac{x}{{x}^{2} - \sqrt{2} x + 1} - \frac{x}{{x}^{2} + \sqrt{2} x + 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \int \left(\frac{2 x}{{x}^{2} - \sqrt{2} x + 1} - \frac{2 x}{{x}^{2} + \sqrt{2} x + 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \int \left(\frac{2 x - \sqrt{2} + \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} - \frac{2 x + \sqrt{2} - \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1}\right) \mathrm{dx}$

Integration is distributive $\int \left(a + b\right) \mathrm{dx} = \int a \mathrm{dx} + \int b \mathrm{dx}$:

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \int \left(\frac{2 x - \sqrt{2}}{{x}^{2} - \sqrt{2} x + 1} - \frac{2 x + \sqrt{2}}{{x}^{2} + \sqrt{2} x + 1}\right) \mathrm{dx} - \frac{1}{2} \int \left(\frac{1}{{x}^{2} - \sqrt{2} x + 1} + \frac{1}{{x}^{2} + \sqrt{2} x + 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \left(\ln | {x}^{2} - \sqrt{2} x + 1 | - \ln | {x}^{2} + \sqrt{2} x + 1 |\right) - \int \left(\frac{1}{2 {x}^{2} - 2 \sqrt{2} x + 2} + \frac{1}{2 {x}^{2} + 2 \sqrt{2} x + 2}\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \ln | \frac{{x}^{2} - \sqrt{2} x + 1}{{x}^{2} + \sqrt{2} x + 1} | - \int \left(\frac{1}{2 {x}^{2} - 2 \sqrt{2} x + 2} + \frac{1}{2 {x}^{2} + 2 \sqrt{2} x + 2}\right) \mathrm{dx}$

Complete the square in the denominators:

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \ln | \frac{{x}^{2} - \sqrt{2} x + 1}{{x}^{2} + \sqrt{2} x + 1} | - \int \left(\frac{1}{{\left(\sqrt{2} x - 1\right)}^{2} + 1} + \frac{1}{{\left(\sqrt{2} x + 1\right)}^{2} + 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \ln | \frac{{x}^{2} - \sqrt{2} x + 1}{{x}^{2} + \sqrt{2} x + 1} | - \frac{1}{\sqrt{2}} \left({\tan}^{- 1} \left(\sqrt{2} x - 1\right) + {\tan}^{- 1} \left(\sqrt{2} x + 1\right)\right) + C$

Apply the trigonometric angle-addition formula $\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \cdot \tan b}$:

$I = x {\tan}^{- 1} \left({x}^{2}\right) - \frac{\sqrt{2}}{4} \ln | \frac{{x}^{2} - \sqrt{2} x + 1}{{x}^{2} + \sqrt{2} x + 1} | - \frac{1}{\sqrt{2}} {\tan}^{- 1} \left(\frac{\sqrt{2} x}{1 - {x}^{2}}\right) + C$

3

## How do you solve y''+3y'-4y=100cos(2x)?

Steve
Featured 2 weeks ago

$y = A {e}^{x} + B {e}^{- 4 x} - 8 \cos \left(2 x\right) + 6 \sin \left(2 x\right)$

#### Explanation:

We have:

$y ' ' + 3 y ' - 4 y = 100 \cos \left(2 x\right)$ ..... [A]

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, ${y}_{p}$ of the non-homogeneous equation.

Complimentary Function

The homogeneous equation associated with [A] is

$y ' ' + 3 y ' - 4 y = 0$

And it's associated Auxiliary equation is:

${m}^{2} + 3 m - 4 = 0$

Which has two real and distinct solutions $m = - 4 , 1$

Thus the solution of the homogeneous equation is:

${y}_{c} = A {e}^{x} + B {e}^{- 4 x}$

Particular Solution

With this particular equation [A], a probably solution is of the form:

$y = a \cos \left(2 x\right) + b \sin \left(2 x\right)$

Where $a$ and $b$ are constants to be determined by substitution

Let s assume the above solution works, in which case be differentiating wrt $x$ we have:

$y ' \setminus \setminus = - 2 a \sin \left(2 x\right) + 2 b \cos \left(2 x\right)$
$y ' ' = - 4 a \cos \left(2 x\right) - 4 b \sin \left(2 x\right)$

Substituting into the initial Differential Equation $\left[A\right]$ we get:

$- 4 a \cos \left(2 x\right) - 4 b \sin \left(2 x\right) + 3 \left\{- 2 a \sin \left(2 x\right) + 2 b \cos \left(2 x\right)\right\} - 4 \left\{a \cos \left(2 x\right) + b \sin \left(2 x\right)\right\} = 100 \cos \left(2 x\right)$

$\therefore - 4 a \cos \left(2 x\right) - 4 b \sin \left(2 x\right) - 6 a \sin \left(2 x\right) + 6 b \cos \left(2 x\right) - 4 a \cos \left(2 x\right) - 4 b \sin \left(2 x\right) = 100 \cos \left(2 x\right)$

Equating coefficients of $\cos \left(2 x\right)$ and $\sin \left(2 x\right)$ we get:

$\cos \left(2 x\right) : - 4 a + 6 b - 4 a = 100 \implies - 8 a + 6 b = 100$
$\sin \left(2 x\right) : - 4 b - 6 a - 4 b = 0 \setminus \setminus \setminus \setminus \setminus \implies - 6 a - 8 b = 0$

Solving simultaneously we get:

$a = - 8$ and $b = 6$

And so we form the Particular solution:

${y}_{p} = - 8 \cos \left(2 x\right) + 6 \sin \left(2 x\right)$

Which then leads to the GS of [A}

$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A {e}^{x} + B {e}^{- 4 x} - 8 \cos \left(2 x\right) + 6 \sin \left(2 x\right)$

3

## How to find formula for nth derivative of f(x)=e^(2x)(x^2-3x+2)?

Andrea S.
Featured 2 weeks ago

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} \left({x}^{2} - 3 x + 2\right)\right) = {2}^{n} {e}^{2 x} \left({x}^{2} + x \left(n - 3\right) + \frac{{n}^{2} - 7 n + 8}{2}\right)$

#### Explanation:

Pose:

$P \left(x\right) = {x}^{2} - 3 x + 2$

The first derivative can be evaluated with the product rule:

$\frac{d}{\mathrm{dx}} \left({e}^{2 x} \left({x}^{2} - 3 x + 2\right)\right) = {e}^{2 x} \left(2 x - 3\right) + 2 {e}^{2 x} \left({x}^{2} - 3 x + 2\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{2 x} \left({x}^{2} - 3 x + 2\right)\right) = {e}^{2 x} \left(2 x - 3 + 2 {x}^{2} - 6 x + 4\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{2 x} \left({x}^{2} - 3 x + 2\right)\right) = {e}^{2 x} \left(2 {x}^{2} - 4 x + 1\right)$

For $n \ge 2$, using the product rule for the $n$-th derivative:

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} P \left(x\right)\right) = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {d}^{n - k} / {\mathrm{dx}}^{n - k} \left({e}^{2 x}\right)$

Note now that, as $P \left(x\right)$ is a polynomial of second degree, its derivative from the third order onward are identically null:

$\frac{d}{\mathrm{dx}} P \left(x\right) = 2 x - 3$

${d}^{2} / {\mathrm{dx}}^{2} P \left(x\right) = 2$

${d}^{k} / {\mathrm{dx}}^{k} P \left(x\right) = 0$ for $k > 2$

So, for $n \ge 2$ in the sum we can anyway ignore the terms for $k > 2$

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} P \left(x\right)\right) = {\sum}_{k = 0}^{2} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {d}^{n - k} / {\mathrm{dx}}^{n - k} \left({e}^{2 x}\right)$

On the other hand:

${d}^{j} / {\mathrm{dx}}^{j} \left({e}^{2 x}\right) = {2}^{j} {e}^{2 x}$

so:

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} P \left(x\right)\right) = {\sum}_{k = 0}^{2} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {2}^{n - k} {e}^{2 x}$

and as ${e}^{2 x}$ is common to all terms:

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} P \left(x\right)\right) = {e}^{2 x} {\sum}_{k = 0}^{2} {2}^{n - k} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right)$

There are only three terms in the sum and we can write them explicitly:

$k = 0 \implies {2}^{n - k} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) = {2}^{n} \left(\begin{matrix}n \\ 0\end{matrix}\right) P \left(x\right) = {2}^{n} \left({x}^{2} - 3 x + 2\right)$

$k = 1 \implies {2}^{n - k} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) = {2}^{n - 1} \left(\begin{matrix}n \\ 1\end{matrix}\right) \frac{\mathrm{dP} \left(x\right)}{\mathrm{dx}} = {2}^{n - 1} n \left(2 x - 3\right) = {2}^{n} \left(n x - \frac{3 n}{2}\right)$

$k = 2 \implies {2}^{n - k} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) = {2}^{n - 2} \left(\begin{matrix}n \\ 2\end{matrix}\right) \frac{{d}^{2} P \left(x\right)}{\mathrm{dx}} ^ 2 = {2}^{n - 2} \frac{n \left(n - 1\right)}{2} \times 2 = {2}^{n} \frac{n \left(n - 1\right)}{4}$

Then:

${\sum}_{k = 0}^{2} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {2}^{n - k} = {2}^{n} \left({x}^{2} - 3 x + 2 + n x - \frac{3 n}{2} + \frac{n \left(n - 1\right)}{4}\right)$

simplifying:

${\sum}_{k = 0}^{2} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {2}^{n - k} = {2}^{n} \left({x}^{2} + x \left(n - 3\right) + \frac{8 - 6 n + {n}^{2} - n}{4}\right)$

${\sum}_{k = 0}^{2} \left(\begin{matrix}n \\ k\end{matrix}\right) {d}^{k} / {\mathrm{dx}}^{k} \left(P \left(x\right)\right) {2}^{n - k} = {2}^{n} \left({x}^{2} + x \left(n - 3\right) + \frac{{n}^{2} - 7 n + 8}{4}\right)$

and in conclusion:

${d}^{n} / {\mathrm{dx}}^{n} \left({e}^{2 x} \left({x}^{2} - 3 x + 2\right)\right) = {2}^{n} {e}^{2 x} \left({x}^{2} + x \left(n - 3\right) + \frac{{n}^{2} - 7 n + 8}{2}\right)$

By direct inspection we can see the formula gives the right derivative also for $n = 1$ and for $n = 0$ (the function itself).

3

## How to find formula for nth derivative of f(x)=e^(2x)(x^2-3x+2)?

Steve
Featured 2 weeks ago

${f}^{\left(n\right)} \left(x\right) = \left\{4 {x}^{2} + 4 n x + {n}^{2} - 7 n - 12 x + 8\right\} \setminus {2}^{n - 2} \setminus {e}^{2 x}$

#### Explanation:

We have:

$f \left(x\right) = {e}^{2 x} \left({x}^{2} - 3 x + 2\right)$

Although we could leave the function "as is" and use the product rule to differentiate, it makes more sense, to multiply out and analyze each component separately:

$f \left(x\right) = {x}^{2} {e}^{2 x} - 3 x {e}^{2 x} + 2 {e}^{2 x}$
$\text{ } = {f}_{1} \left(x\right) - 3 {f}_{2} \left(x\right) + 2 {f}_{3} \left(x\right)$

say, where:

${f}_{1} \left(x\right) = {x}^{2} {e}^{2 x}$
${f}_{2} \left(x\right) = x {e}^{2 x}$
${f}_{3} \left(x\right) = {e}^{2 x}$

Then trivially, the $n t h$ derivative of $f \left(x\right)$ is:

${f}^{\left(n\right)} \left(x\right) = {f}_{1}^{\left(n\right)} \left(x\right) - 3 {f}_{2}^{\left(n\right)} \left(x\right) + 2 {f}_{3}^{\left(n\right)} \left(x\right)$ ..... [A]

$n t h$ derivative of ${f}_{3} \left(x\right) = {e}^{2 x}$

Differentiating wrt $x$ we have

${f}_{3}^{\left(1\right)} \left(x\right) = 2 {e}^{2 x} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {2}^{1} {e}^{2 x}$
${f}_{3}^{\left(2\right)} \left(x\right) = 2 \cdot 2 {e}^{2 x} \setminus \setminus \setminus \setminus \setminus = {2}^{2} {e}^{2 x}$
${f}_{3}^{\left(3\right)} \left(x\right) = 2 \cdot 2 \cdot 2 {e}^{2 x} = {2}^{3} {e}^{2 x}$

${f}_{3}^{\left(n\right)} \left(x\right) = {2}^{n} \setminus {e}^{2 x}$ ..... [B]

$n t h$ derivative of ${f}_{2} \left(x\right) = x {e}^{2 x}$

We have:

${f}_{2} \left(x\right) = x {f}_{3} \left(x\right)$

Differentiating wrt $x$, and applying the product rule, we have

${f}_{2}^{\left(1\right)} \left(x\right) = \left(x\right) \left({f}_{3}^{\left(1\right)} \left(x\right)\right) + \left(1\right) \left({f}_{3} \left(x\right)\right)$
$\text{ } = x \setminus {f}_{3}^{\left(1\right)} \left(x\right) + {f}_{3} \left(x\right)$

${f}_{2}^{\left(2\right)} \left(x\right) = \left(x\right) \left({f}_{3}^{\left(2\right)} \left(x\right)\right) + \left(1\right) \left({f}_{3}^{\left(1\right)} \left(x\right)\right) + {f}_{3}^{\left(1\right)} \left(x\right)$
$\text{ } = x \setminus {f}_{3}^{\left(2\right)} \left(x\right) + 2 \setminus {f}_{3}^{\left(1\right)} \left(x\right)$

${f}_{2}^{\left(3\right)} \left(x\right) = \left(x\right) \left({f}_{3}^{\left(3\right)} \left(x\right)\right) + \left(1\right) \left({f}_{3}^{\left(2\right)} \left(x\right)\right) + 2 \setminus {f}_{3}^{\left(2\right)} \left(x\right)$
$\text{ } = x {f}_{3}^{\left(3\right)} \left(x\right) + 3 \setminus {f}_{3}^{\left(2\right)} \left(x\right)$

${f}_{2}^{\left(n\right)} \left(x\right) = x {f}_{3}^{\left(n\right)} \left(x\right) + n \setminus {f}_{3}^{\left(n - 1\right)} \left(x\right)$ ..... [C}

And using [B] this becomes:

${f}_{2}^{\left(n\right)} \left(x\right) = x \setminus {2}^{n} \setminus {e}^{2 x} + n \setminus {2}^{n - 1} \setminus {e}^{2 x}$
$\text{ } = x \setminus 2 \setminus {2}^{n - 1} \setminus {e}^{2 x} + n \setminus {2}^{n - 1} \setminus {e}^{2 x}$
$\text{ } = \left(2 x + n\right) \setminus {2}^{n - 1} \setminus {e}^{2 x}$ ..... [D]

$n t h$ derivative of ${f}_{1} \left(x\right) = {x}^{2} {e}^{2 x}$

We have:

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

Using the above result [C] we conclude that:

${f}_{1}^{\left(n\right)} \left(x\right) = x {f}_{2}^{\left(n\right)} \left(x\right) + n \setminus {f}_{2}^{\left(n - 1\right)} \left(x\right)$

Using the result [D] this becomes:

${f}_{1}^{\left(n\right)} \left(x\right) = x \left(2 x + n\right) \setminus {2}^{n - 1} \setminus {e}^{2 x} + n \left(2 x + n - 1\right) \setminus {2}^{n - 1 - 1} \setminus {e}^{2 x}$
$\text{ } = x \left(2 x + n\right) \setminus {2}^{n - 1} \setminus {e}^{2 x} + n \left(2 x + n - 1\right) \setminus {2}^{n - 2} \setminus {e}^{2 x}$
$\text{ } = 2 x \left(2 x + n\right) \setminus {2}^{n - 2} \setminus {e}^{2 x} + n \left(2 x + n - 1\right) \setminus {2}^{n - 2} \setminus {e}^{2 x}$
$\text{ } = \left(2 x \left(2 x + n\right) + n \left(2 x + n - 1\right)\right) \setminus {2}^{n - 2} \setminus {e}^{2 x}$
$\text{ } = \left(4 {x}^{2} + 2 n x + 2 n x + {n}^{2} - n\right) \setminus {2}^{n - 2} \setminus {e}^{2 x}$
$\text{ } = \left(4 {x}^{2} + 4 n x + {n}^{2} - n\right) \setminus {2}^{n - 2} \setminus {e}^{2 x}$

Combining the results:

Recall from [A] we have:

${f}^{\left(n\right)} \left(x\right) = {f}_{1}^{\left(n\right)} \left(x\right) - 3 {f}_{2}^{\left(n\right)} \left(x\right) + 2 {f}_{3}^{\left(n\right)} \left(x\right)$

Combining the above results we get:

${f}^{\left(n\right)} \left(x\right) = \left(4 {x}^{2} + 4 n x + {n}^{2} - n\right) \setminus {2}^{n - 2} \setminus {e}^{2 x} - 3 \left(2 x + n\right) \setminus {2}^{n - 1} \setminus {e}^{2 x} + 2 \left({2}^{n} \setminus {e}^{2 x}\right)$

$\text{ } = \left(4 {x}^{2} + 4 n x + {n}^{2} - n\right) {2}^{n - 2} \setminus {e}^{2 x} - 3 \left(2 x + n\right) \setminus 2 \setminus {2}^{n - 2} \setminus {e}^{2 x} + 2 \left(2 \setminus 2 \setminus {2}^{n - 2} \setminus {e}^{2 x}\right)$

 " " = {(4x^2+4nx +n^2-n ) -6 (2x+n) + 8} \ 2^(n-2) \ e^(2x)) }

$\text{ } = \left\{4 {x}^{2} + 4 n x + {n}^{2} - n - 12 x - 6 n + 8\right\} \setminus {2}^{n - 2} \setminus {e}^{2 x}$

$\text{ } = \left\{4 {x}^{2} + 4 n x + {n}^{2} - 7 n - 12 x + 8\right\} \setminus {2}^{n - 2} \setminus {e}^{2 x}$

2

## Answer the differential equation (dx/x^2+x) + (dy/y^2+y)=0 ; y(2)=1?

Ratnaker Mehta
Featured 2 weeks ago

$\left(1\right) : \frac{x y}{\left(x + 1\right) \left(y + 1\right)} = c , \text{ is the GS.}$

$\left(2\right) : 2 x y = x + y + 1 , \text{ is the PS.}$

#### Explanation:

The given Diff. Eqn. $\frac{\mathrm{dx}}{{x}^{2} + x} + \frac{\mathrm{dy}}{{y}^{2} + y} = 0 ,$ with

Initial Condition (IC) $y \left(2\right) = 1.$

It is a Separable Variable Type Diff. Eqn., &, to obtain its

General Solution (GS), we integrate term-wise.

$\therefore \int \frac{\mathrm{dx}}{x \left(x + 1\right)} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \frac{\left(x + 1\right) - x}{x \left(x + 1\right)} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \left\{\frac{x + 1}{x \left(x + 1\right)} - \frac{x}{x \left(x + 1\right)}\right\} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \left\{\frac{1}{x} - \frac{1}{x + 1}\right\} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \left\{\ln x - \ln \left(x + 1\right)\right\} + \left\{\ln y - \ln \left(y + 1\right)\right\} = \ln c .$

$\therefore \ln \left\{\frac{x}{x + 1}\right\} + \ln \left\{\frac{y}{y + 1}\right\} = \ln c .$

$\therefore \ln \left\{\frac{x y}{\left(x + 1\right) \left(y + 1\right)}\right\} = \ln c .$

$\therefore \frac{x y}{\left(x + 1\right) \left(y + 1\right)} = c , \text{ is the GS.}$

To find its Particular Solution (PS), we use the given IC, that,

$y \left(2\right) = 1 , i . e . , w h e n , x = 1 , y = 2.$

Subst.ing in the GS, we get, $\frac{\left(1\right) \left(2\right)}{\left(1 + 1\right) \left(2 + 1\right)} = c = \frac{1}{3.}$

This gives us the complete soln. of the eqn. :

$\frac{x y}{\left(x + 1\right) \left(y + 1\right)} = \frac{1}{3} , \mathmr{and} , 2 x y = x + y + 1.$

2

## Answer the differential equation (dx/x^2+x) + (dy/y^2+y)=0 ; y(2)=1?

Steve
Featured 2 weeks ago

$y = \frac{x + 1}{2 x - 1}$

#### Explanation:

We have an differential equation equation in the form of differentials:

$\frac{\mathrm{dx}}{{x}^{2} + x} + \frac{\mathrm{dy}}{{y}^{2} + y} = 0$

We can write this in "separated variable" form as follows and integrate both sides

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{{y}^{2} + y} = - \frac{\mathrm{dx}}{{x}^{2} + x}$

$\int \setminus \frac{1}{{y}^{2} + y} \setminus \mathrm{dy} = - \int \setminus \frac{1}{{x}^{2} + x} \setminus \mathrm{dx}$

Now et us find the partial fraction decomposition of $\frac{1}{{u}^{2} + u}$ which we can use on both integrals:

$\frac{1}{{u}^{2} + u} \equiv \frac{1}{u \left(u + 1\right)}$
$\text{ } = \frac{A}{u} + \frac{B}{u + 1}$
$\text{ } = \frac{A \left(u + 1\right) + B u}{u \left(u + 1\right)}$

$1 \equiv A \left(u + 1\right) + B u$

We can find the constant coefficients $A$ and 'B' vis substitution (effectively the "cover-up method")

Put $u = 0 \setminus \setminus \setminus \setminus \setminus \implies 1 = A$
Put $u = - 1 \implies 1 = - B$

Thus:

$\frac{1}{{u}^{2} + u} \equiv \frac{1}{u} - \frac{1}{u + 1}$

Using this in the above we get:

$\int \setminus \frac{1}{y} - \frac{1}{y + 1} \setminus \mathrm{dy} = - \int \setminus \frac{1}{x} - \frac{1}{x + 1} \setminus \mathrm{dx}$

We can now evaluate the integrals (not forgetting the constant of integration) to get:

$\ln | y | - \ln | y + 1 | = - \left\{\ln | x | - \ln | x + 1 |\right\} + c$

Then rearranging and using the properties of logarithms we have:

$\ln | y | - \ln | y + 1 | + \ln | x | - \ln | x + 1 | = c$
$\therefore \ln \left(\frac{| x | | y |}{| x + 1 | | y + 1 |}\right) = c$
$\therefore \ln \left(| \frac{x y}{\left(x + 1\right) \left(y + 1\right)} |\right) = c$
$\therefore | \frac{x y}{\left(x + 1\right) \left(y + 1\right)} | = {e}^{c}$

Now ${e}^{c} > 0 \forall c \in \mathbb{R}$, thus we can remove the modulus operator, giving the General Solution:

$\frac{x y}{\left(x + 1\right) \left(y + 1\right)} = A$, say, where $A > 0$.

We are also given that $y \left(2\right) = 1$, this tells us that:

$\frac{2 \cdot 1}{\left(2 + 1\right) \left(1 + 1\right)} = A \implies A = \frac{2}{3}$

Thus the required solution is:

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

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

3

## How to find formula for nth derivative of f(x)=e^(2x)(x^2-3x+2)?

Ratnaker Mehta
Featured 2 weeks ago

${2}^{n - 2} {e}^{2 x} \left\{4 {x}^{2} + 4 \left(n - 3\right) x + \left({n}^{2} - 7 n + 8\right)\right\} .$

#### Explanation:

We will solve this Problem, using the following Leibnitz Theorem

(LT) for ${n}^{t h}$ Derivative of Product of two functions $u \mathmr{and} v .$

Denoting the ${n}^{t h}$ der. of fun. $u$ by ${u}_{n} ,$ we have,

${\left(u v\right)}_{n} = {u}_{n} v + {\text{_nC_1u_(n-1)v_1+}}_{n} {C}_{2} {u}_{n - 2} {v}_{2} + \ldots + u {v}_{n} .$

We take, $u = {e}^{2 x} \Rightarrow {u}_{n} = {2}^{n} \cdot {e}^{2 x} , \mathmr{and} , v = {x}^{2} - 3 x + 2.$

:. v_1=2x-3, v_2=2, &, v_3=v_4=...=v_n=0, n>=3.

$\therefore \frac{{d}^{n}}{\mathrm{dx}} ^ n \left\{\left({x}^{2} - 3 x + 2\right) {e}^{2 x}\right\} = \left({2}^{n} \cdot {e}^{2 x}\right) \left({x}^{2} - 3 x + 2\right)$
$+ n \cdot \left({2}^{n - 1} {e}^{2 x}\right) \left(2 x - 3\right) + \frac{n}{2} \cdot \left(n - 1\right) \cdot \left({2}^{n - 2} {e}^{2 x}\right) \cdot 2 ,$

$= {2}^{n} \cdot {e}^{2 x} \left({x}^{2} - 3 x + 2\right) + {2}^{n - 1} {e}^{2 x} \left(2 n x - 3 n\right) + {2}^{n - 2} {e}^{2 x} \left({n}^{2} - n\right) ,$

$= {2}^{n - 2} {e}^{2 x} \left\{4 \left({x}^{2} - 3 x + 2\right) + 2 \left(2 n x - 3 n\right) + \left({n}^{2} - n\right)\right\} ,$

$= {2}^{n - 2} {e}^{2 x} \left\{4 {x}^{2} + 4 \left(n - 3\right) x + \left({n}^{2} - 7 n + 8\right)\right\} .$

Enjoy Maths.!

1

## Determine the general solution of given differential equation y''' - 3y' - 2y =0 ?

Truong-Son N.
Featured 2 weeks ago

$y = {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}$

Well, this is a linear third-order ordinary differential equation. We can assume a solution of

$y = {e}^{r x}$

so that

${r}^{3} {e}^{r x} - 3 r {e}^{r x} - 2 {e}^{r x} = 0$

And since ${e}^{r x} \ne 0$, we obtain the auxiliary equation:

${r}^{3} - 3 r - 2 = 0$

Now we need to figure out how this can be factored. By synthetic division, we have a guess for $r + 1$ as one factor:

$\underline{- 1} | \text{ "1" "" "0" "-3" } - 2$
$+ \text{ "" "" "" "-1" "" "1" "" } 2$
$\text{----------------------------------------}$
$\text{ "" "" "1" "-1" "-2" "" } 0$

And we happen to be correct. Thus, we obtain a factorization of:

$\left(r + 1\right) \left({r}^{2} - r - 2\right) = 0$

${\left(r + 1\right)}^{2} \left(r - 2\right) = 0$

So, the roots are $r = - 1$ (multiplicity $2$) and $r = 2$. This gives a general solution as a linear combination of these three roots (i.e. not all the coefficients ${c}_{i}$ are zero).

$\textcolor{b l u e}{y = {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}}$

Note that the multiplicity-2 root has two unique constants in its part of the linear combination, but the second root instance is distinguished by another $x$ term in front.

And we should check whether we are correct or not... Do we still have that

$y ' ' ' - 3 y ' - 2 y = 0$?

Firstly, we have a straightforward differentiation using the product rule for the first term:

$y ' ' ' = - {c}_{1} {e}^{- x} + {c}_{2} \frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left[- x {e}^{- x} + {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \frac{d}{\mathrm{dx}} \left[- \left[- x {e}^{- x} + {e}^{- x}\right] - {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \frac{d}{\mathrm{dx}} \left[x {e}^{- x} - 2 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \left[\left[- x {e}^{- x} + {e}^{- x}\right] + 2 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - 3 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

And for the second term:

$y ' = - {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - {e}^{- x}\right] + 2 {c}_{3} {e}^{2 x}$

So, we shall verify whether our solution works.

y''' - 3y' - 2y stackrel(?" ")(=) 0

$= \left\{- {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - 3 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}\right\} - 3 \left\{- {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - {e}^{- x}\right] + 2 {c}_{3} {e}^{2 x}\right\} - 2 \left\{{c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}\right\}$

$= - {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} - 3 {c}_{2} {e}^{- x} + 8 {c}_{3} {e}^{2 x} + 3 {c}_{1} {e}^{- x} + 3 {c}_{2} x {e}^{- x} - 3 {c}_{2} {e}^{- x} - 6 {c}_{3} {e}^{2 x} - 2 {c}_{1} {e}^{- x} - 2 {c}_{2} x {e}^{- x} - 2 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} - {c}_{2} x {e}^{- x} + \cancel{3 {c}_{2} {e}^{- x}} + \cancel{8 {c}_{3} {e}^{2 x}} + 3 {c}_{1} {e}^{- x} + 3 {c}_{2} x {e}^{- x} - \cancel{3 {c}_{2} {e}^{- x}} - \cancel{6 {c}_{3} {e}^{2 x}} - 2 {c}_{1} {e}^{- x} - 2 {c}_{2} x {e}^{- x} - \cancel{2 {c}_{3} {e}^{2 x}}$

$= - {c}_{1} {e}^{- x} - \cancel{{c}_{2} x {e}^{- x}} + 3 {c}_{1} {e}^{- x} + \cancel{3 {c}_{2} x {e}^{- x}} - 2 {c}_{1} {e}^{- x} - \cancel{2 {c}_{2} x {e}^{- x}}$

$= - \cancel{{c}_{1} {e}^{- x}} + \cancel{3 {c}_{1} {e}^{- x}} - \cancel{2 {c}_{1} {e}^{- x}}$

$= 0$ color(blue)(sqrt"")

So indeed, we do have the general solution.

1

## What is a x b?

Truong-Son N.
Featured 6 days ago

I'm guessing the question refers to the $x$ coordinates where $f$ changes direction, and that $a \times b = - \frac{1}{7}$:

$a = - \sqrt{\frac{1}{7}}$, local minimum

b = sqrt((1/7), local maximum

$f \left(x\right) = \frac{6 x}{1 + 7 {x}^{2}}$:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}

Well, first of all, I think the question has a funky typo. It should say, "changes from decreasing to increasing at the point $a$ . . . "

Whenever a function changes from increasing to decreasing or vice versa, it exhibits either a local maximum or minimum.

For example:

• $y = {x}^{2}$ has one local minimum at $x = 0$, since the graph has only an upwards concavity.

graph{x^2 [-2.464, 2.464, -2, 2]}

• $y = \sin x$ from $0$ to $2 \pi$ changes from increasing to decreasing at $x = \frac{\pi}{2}$, and changes from decreasing to increasing at $x = \frac{3 \pi}{2}$.

graph{sinx [-0.05, 6.25, -1.2, 1.01]}

For your graph, the first derivative with respect to $x$, denoted $\frac{\mathrm{df}}{\mathrm{dx}}$, shows you whether the function increases, decreases, or neither at a certain $x$.

Here, we have (using the product rule):

$\frac{\mathrm{df}}{\mathrm{dx}} = 6 x \cdot - \frac{1}{1 + 7 {x}^{2}} ^ 2 \cdot 14 x + \frac{1}{1 + 7 {x}^{2}} \cdot 6$

$= \frac{- 6 \left(14 {x}^{2}\right)}{1 + 7 {x}^{2}} ^ 2 - \frac{- 6 \left(1 + 7 {x}^{2}\right)}{1 + 7 {x}^{2}} ^ 2$

$= \frac{- 6 \left(14 {x}^{2} - \left(1 + 7 {x}^{2}\right)\right)}{1 + 7 {x}^{2}} ^ 2$

$= - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$

So, you took the derivative correctly. Now, to find the points $a$ and $b$ where the function has a local maximum or minimum, find the zeroes of this derivative (i.e. find when $\frac{\mathrm{df}}{\mathrm{dx}} = 0$, the function changes direction).

$0 = - \frac{{\cancel{6}}^{\ne 0} \left(7 {x}^{2} - 1\right)}{\cancel{{\left(1 + 7 {x}^{2}\right)}^{2}}} ^ \left(\ne 0\right)$

The denominator can never equal $0$ because it's an upwards-shifted quadratic, so we don't have to worry about it.

$\implies 7 {x}^{2} = 1$

So, your critical x values (for your local maxima and/or minima) are at

$\implies \textcolor{g r e e n}{x} = \pm \sqrt{\frac{1}{7}} \approx \textcolor{g r e e n}{\pm 0.3780}$

Now that you know the $x$ coordinates, you'll need the $y$ coordinates, so plug in your $x$ values back into the ORIGINAL function:

$\textcolor{g r e e n}{f \left(\pm \sqrt{\frac{1}{7}}\right)} = \frac{6 \left(\pm \sqrt{\frac{1}{7}}\right)}{1 + 7 {\left(\pm \sqrt{\frac{1}{7}}\right)}^{2}}$

$= \frac{\pm \frac{6}{\sqrt{7}}}{1 + 7 \cdot \frac{1}{7}} = \pm \frac{3}{\sqrt{7}}$

$\approx \textcolor{g r e e n}{\pm 1.134}$

Right now, all you know is where the points $a$ and $b$ are, but not which is which.

The second derivative tells you the concavity; up, down, or an inflection point. By the question wording, one is concave up and the other is concave down.

This derivative would be somewhat ugly

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - \frac{d}{\mathrm{dx}} \left[\frac{42 {x}^{2} - 6}{1 + 7 {x}^{2}} ^ 2\right]$,

but you should get:

$= \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$

At the same critical $x$ values you got earlier, evaluate this second derivative.

• If it is positive, the function is concave up and is at a minimum.
• If it is negative, the function is concave down and is at a maximum.
• Otherwise, the function is at an inflection point.

At $\underline{x = - \sqrt{\frac{1}{7}}}$:

$\underline{\frac{{d}^{2} f}{{\mathrm{dx}}^{2}}} = \frac{84 \left(- \sqrt{\frac{1}{7}}\right) \left(7 {\left(- \sqrt{\frac{1}{7}}\right)}^{2} - 3\right)}{1 + 7 {\left(- \sqrt{\frac{1}{7}}\right)}^{2}} ^ 3$

$= \frac{\left(- \frac{84}{\sqrt{7}}\right) \left(- 2\right)}{2} ^ 3 \text{ } \underline{> 0}$

At $\underline{x = + \sqrt{\frac{1}{7}}}$, we similarly get:

$\underline{\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} < 0}$

So, the left-hand critical point is a minimum and the right-hand critical point is a maximum.

$\textcolor{b l u e}{\left(- \sqrt{\frac{1}{7}} , - \frac{3}{\sqrt{7}}\right) \approx \left(- 0.3780 , - 1.134\right)}$ (minimum)

$\textcolor{b l u e}{\left(\sqrt{\frac{1}{7}} , \frac{3}{\sqrt{7}}\right) \approx \left(0.3780 , 1.134\right)}$ (maximum)

And we can check by graphing. Here is the function, followed by its first derivative, and then its second derivative.

$f \left(x\right) = \frac{6 x}{1 + 7 {x}^{2}}$:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]