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2

## How do you derive taylor polynomial for x/(1+x)?

Truong-Son N.
Featured 2 weeks ago

The Taylor polynomial is just another name for the full Taylor series truncated at a finite $n$. In other words, it is a partial Taylor series (i.e. one we could write down in a reasonable amount of time).

Some common errors are:

• Letting $x = a$ within the ${\left(x - a\right)}^{n}$ term.
• Taking the derivatives at the same time as writing out the series, which is not necessarily wrong, but in my opinion, it allows more room for silly mistakes.
• Not plugging $a$ in for the $n$th derivative, or plugging in $0$.

The formula to write out the series was:

sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n

So, we would have to take some derivatives of $f$.

${f}^{\left(0\right)} \left(x\right) = f \left(x\right) = \frac{x}{1 + x}$

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

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

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

$= \frac{1}{1 + x} ^ 2$

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

$f ' ' ' \left(x\right) = \frac{6}{1 + x} ^ 4$

$f ' ' ' ' \left(x\right) = - \frac{24}{1 + x} ^ 5$

etc.

So plugging things in gives (truncated at $n = 4$):

=> (f^((0))(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + . . .

$= \frac{a}{1 + a} + \frac{1}{1 + a} ^ 2 \left(x - a\right) + \frac{- \frac{2}{1 + a} ^ 3}{2} {\left(x - a\right)}^{2} + \frac{\frac{6}{1 + a} ^ 4}{6} {\left(x - a\right)}^{3} + \frac{- \frac{24}{1 + a} ^ 5}{24} {\left(x - a\right)}^{4} + . . .$

$= \textcolor{b l u e}{\frac{a}{1 + a} + \frac{1}{1 + a} ^ 2 \left(x - a\right) - \frac{1}{1 + a} ^ 3 {\left(x - a\right)}^{2} + \frac{1}{1 + a} ^ 4 {\left(x - a\right)}^{3} - \frac{1}{1 + a} ^ 5 {\left(x - a\right)}^{4} + . . .}$

3

## How do I use implicit differentiation to find the derivative of x^2 + y^2 = (5x^2 + 4y^2 -x)^2 ? And then use that derivative to find the tangent line to the curve at point (0, 0.25)

Steve
Featured 4 weeks ago

The derivative is given (implicitly) by;

$x + y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

The equation of the tangent at $\left(0 , 0.25\right)$ is:

$y = x + \frac{1}{4}$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

${x}^{2} + {y}^{2} = {\left(5 {x}^{2} + 4 {y}^{2} - x\right)}^{2}$

Differentiate wrt $x$ (applying chain rule):

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

$\therefore x + y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

Whilst we could spend time and effort to find an explicit equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we have no requirement to, so lets just find $\frac{\mathrm{dy}}{\mathrm{dx}}$ at the point $\left(0 , 0.25\right)$;

$\implies 0 + \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(0 + 4 \left(\frac{1}{16}\right) - 0\right) \left(0 + \frac{8}{4} \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{4}\right) \left(2 \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{4}$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4}$

$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

So the tangent passes through $\left(0 , 0.25\right)$ and has gradient $1$ so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek is;

$\setminus \setminus \setminus \setminus \setminus y - \frac{1}{4} = 1 \left(x - 0\right)$
$\therefore y - \frac{1}{4} = x$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus y = x + \frac{1}{4}$

We can verify this solution graphically;

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = {x}^{2} + {y}^{2} - {\left(5 {x}^{2} + 4 {y}^{2} - x\right)}^{2}$; Then;

$\frac{\partial F}{\partial x} = 2 x - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)$

$\frac{\partial F}{\partial y} = 2 y - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(8 y\right)$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)}{2 y - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(8 y\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{x - \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)}{y - 8 y \left(5 {x}^{2} + 4 {y}^{2} - x\right)}$

Here we get an immediate implicit function for the derivative, so again at $\left(0 , 0.25\right)$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{0 - \left(0 + \frac{4}{16} - 0\right) \left(0 - 1\right)}{\frac{1}{4} - \frac{8}{4} \left(0 + \frac{4}{16} - 0\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{- \left(\frac{1}{4}\right) \left(- 1\right)}{\frac{1}{4} - 2 \left(\frac{1}{4}\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{\frac{1}{4}}{- \frac{1}{4}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 1 \setminus \setminus \setminus \setminus$, as before

1

## 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.

1

## 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.

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

$\textcolor{red}{\downarrow}$

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

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

1

## What is \lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - e ^ { x } } { x }?

Steve
Featured 3 weeks ago

${\lim}_{x \rightarrow 0} \frac{{e}^{2 x} - {e}^{x}}{x} = 1$

#### Explanation:

Method 1 : Graphically
graph{(e^(2x)-e^x)/(x) [-8.594, 9.18, -1.39, 7.494]}

Although far from conclusive, it appears that:

${\lim}_{x \rightarrow 0} \frac{{e}^{2 x} - {e}^{x}}{x} = 1$

Method 2 : L'Hôpital's rule

The limit:

${\lim}_{x \rightarrow 0} \frac{{e}^{2 x} - {e}^{x}}{x}$

is of an indeterminate form $\frac{0}{0}$, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then:

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

And so applying L'Hôpital's rule we get:

${\lim}_{x \rightarrow 0} \frac{{e}^{2 x} - {e}^{x}}{x} = {\lim}_{x \rightarrow 0} \frac{2 {e}^{2 x} - {e}^{x}}{1}$
$\text{ } = 2 - 1$
$\text{ } = 1$

Method 3 - Power Series

The power series for ${e}^{x}$ is as follows;

 e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...

And so we have:

 e^(2x) = 1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + (2x)^4/(4!) + ...

Therefore;

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

And so:

${\lim}_{x \rightarrow 0} \frac{{e}^{2 x} - {e}^{x}}{x} = {\lim}_{x \rightarrow 0} \left(1 + O \left(x\right)\right)$
$\text{ } = 1$

2

## Let R be the region in the first quadrant bounded by the x and y axis and the graphs of f(x) = 9/25 x +b and y = f^-1 (x). If the area of R is 49, then the value of b, is ? A) 18/5 B) 22/5 C) 28/5 D) none

mason m
Featured 2 weeks ago

C) $\frac{28}{5}$

#### Explanation:

Let's first imagine what the region would look like. Inverse functions are reflections of themselves over the line $x = y$, so the lines would resemble the following (ignore any numbers, this is just a speculation):

We see that the top line will be given by ${L}_{1} \left(x\right) = f \left(x\right) = \frac{9}{25} x + b$.

The inverse of this is given by:

$y = \frac{9}{25} x + b \text{ "=>" } x = \frac{9}{25} y + b$

$\textcolor{w h i t e}{y = \frac{9}{25} x + b} \text{ "=>" } y = \frac{25}{9} \left(x - b\right) = {f}^{-} 1 \left(x\right)$

Since this is the lower line we say that ${L}_{2} \left(x\right) = \frac{25}{9} \left(x - b\right)$.

The area of the region is given by two distinct parts: the first is the trapezoid that lies under ${L}_{1}$. This extends from $x = 0$ to the $x$ intercept of ${L}_{2}$.

The second part is a triangle that extends from the $x$ intercept of ${L}_{2}$ to the intersection point of ${L}_{1}$ and ${L}_{2}$. Its height is the value of ${L}_{1}$ at the $x$ intercept of ${L}_{2}$.

Let's find these important points first:

$m a t h b f \left({L}_{2}\right)$ $m a t h b f x$-intercept

${L}_{2} \left(x\right) = \frac{9}{25} \left(x - b\right) = 0 \text{ "=>" } x = b$

Intersection Point

${L}_{1} \left(x\right) = {L}_{2} \left(x\right) \text{ "=>" } \frac{9}{25} x + b = \frac{25}{9} \left(x - b\right)$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } \frac{81}{625} x + \frac{9}{25} b = x - b$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } \frac{34}{25} b = \frac{544}{625} x$

$\textcolor{w h i t e}{{L}_{1} \left(x\right) = {L}_{2} \left(x\right)} \text{ "=>" } x = \frac{25}{16} b$

We can now, without any integration necessary (since we're dealing with lines) find these areas.

Trapezoid Area

One base is $b$, which is the $y$ intercept of ${L}_{1}$. The other base is ${L}_{1} \left(b\right) = \frac{34}{25} b$. The height is $b$ as well, so the trapezoid's area is

${A}_{1} = \frac{1}{2} b \left(b + \frac{34}{25} b\right) = \frac{59}{50} {b}^{2}$

Triangle Area

The base is again ${L}_{1} \left(b\right) = \frac{34}{25} b$. The height (horizontal) is $\frac{25}{16} b - b = \frac{9}{16} b$. The area of the triangle is then

${A}_{2} = \frac{1}{2} \left(\frac{34}{25} b\right) \left(\frac{9}{16} b\right) = \frac{153}{400} {b}^{2}$

The total region $R$ has area $49$.

$R = {A}_{1} + {A}_{2} = \frac{59}{50} {b}^{2} + \frac{153}{400} {b}^{2} = 49$

$\frac{25}{16} {b}^{2} = 49$

$b = \sqrt{\frac{49 \left(16\right)}{25}} = \frac{28}{5}$

2

## (4) Find all points on the curve x^2y^2 + xy = 2 where the slope of the tangent line is −1?

Shwetank Mauria
Featured 2 weeks ago

At points $\left(1 , 1\right)$ and -1.-1)

#### Explanation:

Slope of a tangent at a point on the curve ${x}^{2} {y}^{2} + x y = 2$ will be given by the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at that point. So let us find its derivative, which is given by,

$2 x \times {y}^{2} + {x}^{2} \times 2 y \times \frac{\mathrm{dy}}{\mathrm{dx}} + 1 \times y + x \times \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 {x}^{2} y + x\right] = - 2 x {y}^{2} - y$ and

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

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

and as we have to identify, where slope of tangent is $- 1$, we have solution $y = x$

and as ${x}^{2} {y}^{2} + x y = 2$, this is at

${x}^{4} + {x}^{2} - 2 = 0$

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

or ${x}^{2} - 1 = 0$

i.e. $x = \pm 1$ and as $y = x$

we have slope of tangent as $- 1$ at $\left(1 , 1\right)$ and $\left(- 1 , - 1\right)$
graph{(x^2y^2+xy-2)(x-y)=0 [-10, 10, -5, 5]}

2

## Let p(x)=x^3-9x^2+10 Find the absolute maximum and minimum value of p on the interval [-2,8]?

HSBC244
Featured 2 weeks ago

Absolute maximum: $\left(0 , 10\right)$
Absolute minimum: $\left(6 , - 98\right)$

#### Explanation:

We start by differentiating.

$p ' \left(x\right) = 3 {x}^{2} - 18 x$

We now find the critical numbers, which occur when the derivative is $0$ or is undefined. Since polynomials are continuous, we will only have critical numbers when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

$0 = 3 x \left(x - 6\right)$

$x = 0 \mathmr{and} 6$

We now test around the points to check whether the derivative is increasing or decreasing.

Test Point: $x = - 1$

$p \left(- 1\right) = 3 {\left(- 1\right)}^{2} - 18 \left(- 1\right) = 3 + 18 = 21$

This means that $p \left(x\right)$ is increasing on $\left(- \infty , 0\right)$. Before confirming that $x = 0$ is an absolute maximum though we must check the y-values of the function at the end points.

$p \left(- 2\right) = {\left(- 2\right)}^{3} - 9 {\left(- 2\right)}^{2} + 10$

$p \left(- 2\right) = - 8 - 9 \left(4\right) + 10$

$p \left(- 2\right) = - 34$

This works!

$p \left(8\right) = {8}^{3} - 9 {\left(8\right)}^{2} + 10$

$p \left(8\right) = 512 - 576 + 10$

$p \left(8\right) = - 54$

This works!

Therefore, $\left(0 , 10\right)$ is an absolute maximum on $\left[- 2 , 8\right]$ (always specify, because on $\left(- \infty , \infty\right)$, this function has no absolute maximum).

Finally, we must ensure that $x = 6$ is an absolute minimum.

Test Point $x = 5$

$p \left(5\right) = {5}^{3} - 9 {\left(5\right)}^{2} + 10$

$p \left(5\right) = - 90$

We can now see that $p \left(x\right)$ is decreasing on $\left(0 , 6\right)$. This will mean that we have a minimum of some sort at $x = 6$. We know for sure that it will be either an absolute or local minimum because this critical point has a derivative equal to $0$, so the function changes vertical direction, or the derivative goes from decreasing to increasing.

$p \left(6\right) = {6}^{3} - 9 {\left(6\right)}^{2} + 10$

$p \left(6\right) = - 98$

Since this is smaller than both $p \left(- 2\right)$ and $p \left(8\right)$, we can see that this is an absolute minimum.

Hopefully this helps!

2

## Find the area inside r=1+cos2theta and outside circle r=1 ?

Steve
Featured 5 days ago

$2 + \frac{\pi}{4}$

#### Explanation:

Here is the graph of the two curves. The shaded area, $A$, is the area of interest:

It is a symmetrical problems so we only need find the shaded area of the RHS of Quadrant $1$ and multiply by $4$.

We could find the angle $\theta$ in $Q 1$ for the point of intrestion by solving the simultaneous equations:

$r = 1 + \cos 2 \theta$
$r = 1$

However, intuition is faster, and it looks like angle of intersection in $Q 1$ is $\frac{\pi}{4}$, we can verify this by a quick evaluation:

$\theta = \frac{\pi}{4} \implies r = 1 + \cos 2 \theta = 1 + \cos \left(\frac{\pi}{2}\right) = 1$

Confirming our intuition. So now we have the following:

Where the shaded area repents $\frac{1}{4}$ of the total area sought:

We can now start to set up a double integral to calculate this area

$r$ sweeps out a ray from $1$ to $1 + \cos \left(2 \theta\right)$
$\theta$ varies for $0$ to $\frac{\pi}{4}$

So then:

$\frac{1}{4} A = {\int}_{0}^{\frac{\pi}{4}} {\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} \setminus d \theta$

If we evaluate the inner integral first then we get:

${\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} = {\left[\frac{1}{2} {r}^{2}\right]}_{1}^{1 + \cos 2 \theta}$
$\text{ } = \frac{1}{2} \left({\left(1 + \cos 2 \theta\right)}^{2} - {1}^{2}\right)$
$\text{ } = \frac{1}{2} \left(1 + 2 \cos 2 \theta + {\cos}^{2} 2 \theta - 1\right)$
$\text{ } = \frac{1}{2} \left(2 \cos 2 \theta + {\cos}^{2} 2 \theta\right)$
$\text{ } = \frac{1}{2} \left(2 \cos 2 \theta + \frac{1}{2} \left(\cos 4 \theta + 1\right)\right)$
$\text{ } = \cos 2 \theta + \frac{1}{4} \cos 4 \theta + \frac{1}{4}$

And so our double integral becomes:

$\frac{1}{4} A \setminus = {\int}_{0}^{\frac{\pi}{4}} {\int}_{1}^{1 + \cos 2 \theta} r \setminus \mathrm{dr} \setminus d \theta$
$\text{ } = {\int}_{0}^{\frac{\pi}{4}} \cos 2 \theta + \frac{1}{4} \cos 4 \theta + \frac{1}{4} \setminus d \theta$
$\text{ } = {\left[\frac{1}{2} \sin 2 \theta + \frac{1}{16} \sin 4 \theta + \frac{1}{4} \theta\right]}_{0}^{\frac{\pi}{4}}$
$\text{ } = \left(\frac{1}{2} \sin \left(\frac{\pi}{2}\right) + \frac{1}{16} \sin \pi + \frac{1}{4} \frac{\pi}{4}\right) - \left(0\right)$
$\text{ } = \frac{1}{2} + \frac{\pi}{16}$
$\therefore A = 2 + \frac{\pi}{4}$



METHOD 2

If you are not happy with double integrals, then we can evaluate the area using the polar area formula $A = \int \setminus \frac{1}{2} {r}^{2} \setminus d \theta$ and calculating the two shaded sections separately:

Here this shaded area is given by:

${A}_{1} = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus {\left(1 + \cos 2 \theta\right)}^{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + {\cos}^{2} 2 \theta \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + \frac{1}{2} \left(\cos 4 \theta + 1\right) \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus 1 + 2 \cos 2 \theta + \frac{1}{2} \cos 4 \theta + \frac{1}{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} {\left[\frac{3}{2} \theta + \sin 2 \theta + \frac{1}{8} \sin 4 \theta\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \left(\frac{3 \pi}{8} + 1 + 0\right)$
$\setminus \setminus \setminus \setminus = \frac{3 \pi}{16} + \frac{1}{2}$

is given by:

${A}_{2} = \frac{1}{2} \setminus {\int}_{0}^{\frac{\pi}{4}} \setminus {\left(1\right)}^{2} \setminus d \theta$
$\setminus \setminus \setminus \setminus = \frac{1}{2} {\left[\theta\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus \setminus = \frac{1}{2} \left(\frac{\pi}{4} - 0\right)$
$\setminus \setminus \setminus \setminus = \frac{\pi}{8}$

And so the total sought area is $4$ times the difference between these results:

$\frac{1}{4} A \setminus = {A}_{1} - {A}_{2}$
$\text{ } = \frac{3 \pi}{16} + \frac{1}{2} - \frac{\pi}{8}$
$\text{ } = \frac{\pi}{16} + \frac{1}{2}$
$\therefore A = \frac{\pi}{4} + 2$

3

## Is there a certain formula for derivatives?

HSBC244
Featured 4 days ago

Yes, but not just one formula. There are many.

The derivative (of a differentiable function), $y = f \left(x\right)$, at the point $x = a$ is defined by the following limit

$f ' \left(a\right) = {\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$

With a slight change of notation we can write:

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

In some older texts the notation may involve $\delta x$ or $\Delta x$, instead of $h$ giving an identical result:

$f ' \left(x\right) = {\lim}_{\delta x \rightarrow 0} \frac{f \left(x + \delta x\right) - f \left(x\right)}{\delta x} = {\lim}_{\Delta x \rightarrow 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

It represents both the rate of change of the function, and the gradient of the tangent line at any particular point. If the limit does not exists then the function is not differentiable.

In practice we do not derive the derivative from first principles using the limit definition, but instead we use various rules that can be proved to be true;

Power Rule: $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

Example: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $y = 2 {x}^{7}$.

We have $n = 7$, so the derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = 14 {x}^{7 - 1} = 14 {x}^{6}$

Chain Rule: d/dx(f(g(x)) = f'(g(x)) * g'(x), in other words the derivative of the composition $f \left(g \left(x\right)\right)$ is the inner function times the outer function.

Example: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $y = {\left(x + 2\right)}^{7}$

We let $u = x + 2$ and $y = {u}^{7}$. Then, by the power rule, $\frac{\mathrm{dy}}{\mathrm{du}} = 7 {u}^{6}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 1$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot 7 {u}^{6}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 7 {\left(x + 2\right)}^{6}$

Product Rule: $\frac{d}{\mathrm{dx}} \left(f \left(x\right) \times g \left(x\right)\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Example: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $y = \left(x + 2\right) \left({x}^{2} - 1\right)$

We let $f \left(x\right) = x + 2$ and $g \left(x\right) = {x}^{2} - 1$. Then $f ' \left(x\right) = 1$ and $g ' \left(x\right) = 2 x$.

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} - 1 + 2 {x}^{2} + 4 x$

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

Quotient rule: $\frac{d}{\mathrm{dx}} \left(\frac{f \left(x\right)}{g \left(x\right)}\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Example: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $y = \frac{x}{x + 1}$

We let $f \left(x\right) = x$ and $g \left(x\right) = x + 1$. Then $f ' \left(x\right) = 1$ and $g ' \left(x\right) = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 \left(x + 1\right) - 1 \left(x\right)}{x + 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x + 1} ^ 2$

Here are a few other useful derivative formulas I think you should know.

$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

$\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$

Once you get good at the basic differentiation rules, you may be asked to solve problems that combine the differentiation rules in interesting ways.

Example: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ for $y = \ln \left(\sec x\right)$

First of all, note that $\sec x = \frac{1}{\cos} x$. The function becomes

$y = \ln \left(\frac{1}{\cos} x\right)$

There are a few ways of differentiating this.

a) you could use the quotient rule to differentiate $\frac{1}{\cos} x$ and then the chain rule to differentiate $y$.

b) you could rewrite $\frac{1}{\cos} x$ as ${\left(\cos x\right)}^{-} 1$ and then differentiate using the chain rule twice.

c) you could use the laws of logarithms to simplify and then differentiate. I'll use this method.

By the rule $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$, we have:

$y = \ln \left(\frac{1}{\cos} x\right) = \ln 1 - \ln \cos x = 0 - \ln \left(\cos x\right) = - \ln \left(\cos x\right)$

By the chain rule, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} x \cdot - \sin x = \sin \frac{x}{\cos} x = \tan x$

Hopefully you now get a good idea what differentiation is about!