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# If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of

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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of  [#permalink]

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20 Nov 2017, 12:07
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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?

A. 1/A < 1/B < 1/C

B. 1/A < 1/C < 1/B

C. 1/C < 1/A < 1/B

D. 1/C < 1/B < 1/A

E. 1/B < 1/C < 1/A

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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of  [#permalink]

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20 Nov 2017, 12:15
Bunuel wrote:
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?

A. 1/A < 1/B < 1/C

B. 1/A < 1/C < 1/B

C. 1/C < 1/A < 1/B

D. 1/C < 1/B < 1/A

E. 1/B < 1/C < 1/A

A < 0, 10 < B < 30, and 50 < C < 80

Let A = -5
Let B = 20
Let C = 70

$$\frac{1}{A}=-\frac{1}{5}$$

$$\frac{1}{B}=\frac{1}{20}$$

$$\frac{1}{C}=\frac{1}{70}$$

$$-\frac{1}{5} < \frac{1}{70} < \frac{1}{20}$$

$$\frac{1}{A} < \frac{1}{C} < \frac{1}{B}$$

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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of  [#permalink]

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20 Nov 2017, 12:19
Bunuel wrote:
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?

A. 1/A < 1/B < 1/C

B. 1/A < 1/C < 1/B

C. 1/C < 1/A < 1/B

D. 1/C < 1/B < 1/A

E. 1/B < 1/C < 1/A

B.$$\frac{1}{A} < \frac{1}{C} < \frac{1}{B}$$

$$A$$ is negative, then would be the smallest, reciprocal or not. $$C > B$$, the reciprocal would have opposite relative order.
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of &nbs [#permalink] 20 Nov 2017, 12:19
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# Area between two circles

1. Jun 25, 2013

### skybox

1. The problem statement, all variables and given/known data

2. Relevant equations
The area of a circle:
$$A_c = \pi r^{2}$$

3. The attempt at a solution
I know that the diameter of the oval shape is 10m since the problem says that it touches the circumference of the center of each circle. I am not sure how to approach the problem now. Any hints would be great.

Thanks

2. Jun 25, 2013

Well we have two equations to work with
$$x^2 + y^2 = 10^2$$
$$(x-10)^2 + y^2 = 10^2$$

Theses give the equations of the two circles if we place the origin at the center of the circle on the left. Next I would advise finding the intersection points. Hopefully you can continue to work on it some more and determine the area using these equations.

Last edited: Jun 25, 2013
3. Jun 25, 2013

If you get stuck again, I imagine it will be finding the area once you get those two points. I would highly recommend reading this page on wikipedia http://en.wikipedia.org/wiki/Spherical_cap. You can obtain the equation for spherical caps. I'm not sure if this is the way your instructor wants to do it or not though. It is possible to derive it yourself using the equations, but this will make it much less painless =)

4. Jun 25, 2013

### verty

This is very much like an olympiad question, so much so that I think we should refrain from explaining how to solve it and leave it as a challenge for each reader to solve.

As a hint, there is an easier way to find the points of intersection that uses only geometry, and to solve it, only geometry and some very elementary trig knowledge is needed.

5. Jun 25, 2013

I find this to be easier than whatever geometry you're using

$$x^2 - 20x + 100 + y^2 = x^2 + y^2$$
$$x = 100/20 = 5, y = \pm 5\sqrt{3}$$

haha, but its always nice to have multiple ways to solve things. Of course, the next part of the problem is really the hard part =p

6. Jun 25, 2013

### verty

In case it is needed, here is a second hint: Divide the area down the middle, find half, then double it.

This way I know Skybox has ample hints to solve it. So I hope we can now leave it unanswered, thank you.

7. Jun 25, 2013

### skybox

Thanks all for the replies. I think I have an idea on how to solve the problem with all the hints given. I will not post the solution once I do come to a solution. Thanks again!

format: plain


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# Dual of a normed vector space

Let $(E;\Vert \cdot \Vert )$ be a normed vector space. Show that given $x, y \in E$, $x \neq y$, there exists $f \in E'$ ($E'$ is the bounded dual space of $E$) such that $f(x) \neq f(y)$.

I started trying to show that there always is an injective map on the dual, which would imply this result, but it happens to be case that some bounded dual spaces do not have any injective maps.

Any help would be greatly appreciated. Thanks

• Hahn–Banach Theorem – Yiorgos S. Smyrlis Oct 29 '16 at 12:39

Let $Z$ be the subspace $\{\lambda(x-y)\in E:\lambda\in \mathbb{R}\}$ of $E$, generated by the vector $x-y\neq 0_E$ and consider the functional $f:Z\rightarrow \mathbb{R}:~f(\lambda(x-y))=\lambda\|x-y\|$. Then $f$ is clearly linear and bounded ($Z$ is finite dimensional), so there is a linear bounded extension $\tilde{f}: E\rightarrow \mathbb{R}$ of $f$, according to the Hahn - Banach theorem. Therefore $\tilde{f}\in E'$ and $$\tilde{f}(x)-\tilde{f}(y)=\tilde{f}(x-y)=f(x-y)=\|x-y\|\neq 0.$$


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Updated date:

# Constant Multiple Rule for Derivatives (With Proof and Examples)

Author:

Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.

## What Is the Constant Multiple Rule?

The derivative of constant times a function is the constant times the derivative of the function. If c is a constant and f is a differentiable function, then,

(d/dx) [c f(x)] = c (d/dx) [f(x)]

### Constant Multiple Rule Proof

When new functions are formed from old functions by multiplication by a constant or any other operations, their derivatives can be calculated using derivatives of the old functions. In particular, the Constant Multiple Rule states that the derivative of a constant multiplied by a function is the constant multiplied by the function's derivative.

Here is ample proof of Constant Multiple Rule using limits. Let g(x) = c f(x)

g’(x) = limh→o [g(x+h)-g(x)] / h

g’(x) = limh→o [cf(x+h)-cf(x)] / h

g’(x) = limh→o c [(f(x+h)-f(x)) / h]

g’(x) = c limh→o [(f(x+h)-f(x)) / h]

g’(x) = cf’(x)

Below is the geometric interpretation of the Constant Multiple Rule. Multiplying by c=2 stretches the graph vertically by a factor of 2. All the rises have been doubled, but the runs stay the same. So the slopes are doubled, too.

## Example 1: Derivative of a Function to the Fourth Power

Find the derivative of the function (d/dx) 3x4 using the Constant Multiple Rule.

Solution

Apply the Constant Multiple Rule by taking the derivative of the power function first and then multiply with the coefficient 3.

(d/dx) 3x4 = 3 (d/dx) x4

Apply the Power Rule in differentiating the power function.

(d/dx) 3x4 = 3 (d/dx) 4x4-1

(d/dx) 3x4 = 3 (d/dx) 4x3

Simplify the algebraic expression.

(d/dx) 3x4 = 3 (d/dx) 4x3

(d/dx) 3x4 = 3(4x3)

(d/dx) 3x4 = 12x3

The derivative of the function (d/dx) 3x4 is 12x3.

## Example 2: Derivative of a Negative Function

What is the derivative of the function (d/dx) -x?

Solution

Apply the Constant Multiple Rule by taking the derivative of the power function first and then multiply with the coefficient -1.

(d/dx) -x = (d/dx) [(-1) x]

Apply the Power Rule in differentiating the power function.

(d/dx) -x = (-1) (d/dx) x

Recall that the derivative of x is 1. Simplify further the algebraic expression.

(d/dx) -x = -1(1)

(d/dx) -x = -1

The derivative of the function -x is -1.

## Example 3: Derivative of a Fraction

What is the derivative of the function y= (1/2) x?

Solution

Apply the Constant Multiple Rule by taking the derivative of the power function first and then multiply with the coefficient \frac {1} {2}. Apply the Power Rule in differentiating the power function.

(d/dx) (1/2)x = (1/2) (d/dx) x

Recall that the derivative of x is 1. Simplify further the algebraic expression.

(d/dx) (1/2) x = (1/2) (1)

(d/dx) (1/2) x = ½

The derivative of the function y = (1/2) x is ½.

## Example 4: Derivative of a Function of X

Find the derivative of the constant multiple function f(x)=6x.

Solution

Apply the Constant Multiple Rule by taking the derivative of the power function first and then multiply with the coefficient 6. Apply the Power Rule in differentiating the power function.

(d/dx) 6x = 6 (d/dx) x

Recall that the derivative of x is 1. Simplify further the algebraic expression.

(d/dx) 6x = 6(1)

(d/dx) 6x = 6

The derivative of the function f(x)=6x is 6.

## Example 5: Derivative of a Complex Cube Root Function

What is the derivative of the constant multiple function y = (3√8) x3?

Solution

Apply the Constant Multiple Rule by taking the derivative of the power function first and then multiply with the coefficient 3√8. Apply the Power Rule in differentiating the power function.

(d/dx) (3√8) x3 = (3√8) (d/dx) x3

Recall the Power Rule and solve for the derivative of the power function x3.

(d/dx) (3√8) x3 = 3√8 (3x2)

(d/dx) (3√8) x3 = 2(3x2)

(d/dx) (3√8) x3 = 6x2

The derivative of the function y = (3√8) x3 is 6x2.

## Example 6: Derivative of a Function to the Seventh Power

Find the derivative of the function y = 3x7 using the Constant Multiple Rule.

Solution

First, separate the constant value of 3 from the whole function. Then, take the derivative of the power function x7 using the Power Rule.

(d/dx) 3x7 = 3 (d/dx) x7

Apply the Power Rule then multiply the result to the constant 3, as stated from the Constant Multiple Rule.

(d/dx) 3x7 = 3 (d/dx) x7 = 3(7x7-1)

(d/dx) 3x7 = 3(7x6)

(d/dx) 3x7 = 21x6

The derivative of the function y = 3x7 is 21x6.

## Example 7: Derivative of a Negative Function

What is the derivative of the constant multiple function (d/dx) (-5x)?

Solution

From the given function, segregate the negative value constant from the x variable.

(d/dx) -5x= -5 (d/dx) x

Apply the Power Rule, then multiply the outcome with a negative five.

(d/dx) -5x= -5 (d/dx) x

(d/dx) -5x= -5 (d/dx) x1-1

(d/dx) -5x= -5 (d/dx) x0

(d/dx) -5x= -5 (d/dx) (1)

(d/dx) -5x= -5

The derivative of the function y = -5x is -5.

## Example 8: Derivative of a Pi Function

Find the derivative of the function y = 3πt.

Solution

The given function is a radian function of variable t. Recall that pi is a constant value of 3.14. It means that the part with 3π will be the constant of the pi function. Separate the constant value 3π from the variable t and differentiate t alone.

(d/dt) 3πt= 3π (d/dt) t

Apply the Power Rule and the Constant Multiple Rule to the equation.

(d/dt) 3πt= 3π (d/dt) t1-1

(d/dt) 3πt= 3π (d/dt) t0

(d/dt) 3πt= 3π

The derivative of the function y = 3πt is 3π.

## Example 9: Derivative of a Function to the Fourth Power

Solve for the derivative of the function f(x) = 4x4.

Solution

In getting the derivative of the function 4x4, the Constant Multiple Rule applies. Like the usual method, separate the constant four from x4 and then solve the power function's derivative.

(d/dx) 4x4= 4 (d/dx) x4

Apply the Power Rule and the Constant Multiple Rule to the equation.

(d/dx) 4x4= 4 (d/dx) x4

(d/dx) 4x4= 4 (d/dx) 4x4-1

(d/dx) 4x4= 4 (d/dx) 4x3

(d/dx) 4x4= 16x3

The derivative of the function 4x4 is 16x3.

## Example 10: Derivative of a Sum of Power Functions

Find the derivative of the function f(x) = 6x3 + 9x2 + 2x + 8.

Solution

The given equation is a run of power functions. To solve, differentiate the terms individually. Start with the 6x3 and apply the Constant Multiple Rule.

(d/dx) 6x3= 6 (d/dx) x3

(d/dx) 6x3= 6 (3x3-1)

(d/dx) 6x3= 6 (3x2)

(d/dx) 6x3= 18x2

(d/dx) 9x2= 9 (d/dx) x2

(d/dx) 9x2= 9 (2x2-1)

(d/dx) 9x2= 9 (2x1)

(d/dx) 9x2= 18x

(d/dx) 2x= 2 (d/dx) x

(d/dx) 2x = 2 (x1-1)

(d/dx) 2x = 2 (x0)

(d/dx) 2x = 2 (1)

(d/dx) 2x = 2


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# Derivative/chain rule

#### daigo

##### Member
$$f(x) = \frac{1}{x\sqrt{5-2x}}$$
$$f'(x) = (x\sqrt{5-2x})^{-1} \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\ = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \$/extract_tex] Have I done anything wrong so far? #### daigo ##### Member After I simplify, I get this: $$= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\ \\ \\ = \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}$$ But the solution says I'm wrong...what have I done? #### Sudharaka ##### Well-known member MHB Math Helper $$f'(x) = (x\sqrt{5-2x})^{-1} \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\ = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \\$$ Have I done anything wrong so far? Hi diago, Up to this point you are correct. After I simplify, I get this: $$= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\ \\ \\ = \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}$$ But the solution says I'm wrong...what have I done? Note that, you have missed the minus sign before the fraction. Kind Regards, Sudharaka. #### daigo ##### Member I can't see where I've missed any negative sign #### daigo ##### Member Oh wait, I see it now...the very first negative sign at the very beginning...I completely missed that this entire time and I couldn't see it...wow #### CaptainBlack ##### Well-known member $$f(x) = \frac{1}{x\sqrt{5-2x}}$$ You are making a meal of this. Rewrite as: \[ f(x) = x^{-1}(5-2x)^{-1/2}$
then using the product rule:
\begin{aligned}f'(x)&=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)\\ \\ &=-x^{-1}f(x)+f(x)(5-2x)\\ \\&=f(x)\left[-\frac{1}{x}+\frac{1}{5-2x} \right]\\ \\ &=f(x)\left[ \frac{3x-5}{x(5-2x)}\right] \\ \\ &= \frac{3x-5} {x^2(5-2x)^{3/2}}\end{aligned}

#### DeusAbscondus

##### Active member
CB: strategy at line 3; please unpick this for me

You are making a meal of this. Rewrite as:

$f(x) = x^{-1}(5-2x)^{-1/2}$
then using the product rule:
\begin{aligned}f'(x)&=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)\\ \\ &=-x^{-1}f(x)+f(x)(5-2x)\\ \\&=f(x)\left[-\frac{1}{x}+\frac{1}{5-2x} \right]\\ \\ &=f(x)\left[ \frac{3x-5}{x(5-2x)}\right] \\ \\ &= \frac{3x-5} {x^2(5-2x)^{3/2}}\end{aligned}
Hello CB,
I note with interest the 2nd line of your calculations after: "using the product rule", which is
$$=x^{-1}(f(x)+f(x)(5-2x)$$
Would you kindly:
1) explain what is behind this move?
It looks like you are factoring out f(x) from both sides of the addition. Is that right? If so, could you enlighten me as to the arithmetic/logic involved?
(Looks like whatever it is, they didn't show us in class the other day when introducing the product rule, and looks like this would make these problems a hell of a lot easier to deal with)

(Edited 1 hour later) I have just surprised myself to see how easy, pleasant and fun it was to apply basic algebra and advance myself towards an answer to question (1) above:
(Edited 1 hour 30 mins from original post) {On further testing my hypothesis, set out below, I realize it is wrong, making what follows redundant, but perhaps not entirely: I will let it stand as an example of "going off half-cocked" with the result: shot in the pants!}
It goes like this (plse comment, emend, correct, contradict as and where appropriate )

Let $x^{-1}=u$
and $(5-2x)^{-1/2}=v$
Since $f'(x)=(u'v)+(uv')$ then, your equation $f'(x)=\left(x^-1(f(x))+(f(x)\times(5-2x)\right)$ implies that $f'(x)=u^2v+v^2u$ which can then be factored as:
$(uv)(u+v)$ (a very interesting result, if my reasoning is correct, which no-one told me about at school )

Now,since $f(x)=(uv)$ this explains why you were able to write $f'(x)$ as the sum of $u(uv)$ and $v(vu)$ at line 2 of your calculations.

Am I right?
If so, it is then simply a matter of factoring $f(x)$ outside square brackets in the next line. And, if I am right, would this be the way to go as a point of "best practice" when working on all such problems?

Finally, if I have, if fact, anwered question (1) please go ahead and answer the next 2 typographical questions, or direct me to a convenient and usable (searchable) source of such short-cuts:

2) explain the short-cut I see in your latex "&="
3) tell me how to make a "hotkey" for things like "\begin{aligned}" so as to avoid having to type it out in full every time I want to use it

(Edited 1 hour 30 mins from original post) {On further testing my hypothesis, set out above, I realize it is wrong, making it largely redundant, but perhaps not entirely: I will let it stand as an example of "going off half-cocked" with the result: shot in the pants!}

Regs,
DeusAbs

Last edited:

#### Ackbach

Staff member
$$f'(x) = (x\sqrt{5-2x})^{-1} \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\ = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\ = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \$/extract_tex] Have I done anything wrong so far? At the risk of adding too many cooks, I would say that the quotient rule is not a bad way to go here. The nice thing about the quotient rule is that it sometimes saves algebra after taking all your derivatives, because you don't have to get a common denominator. #### Jameson ##### Administrator Staff member DeusAbcondus, I think it's really great you're diving into math this way of trying out new things and not memorizing "the book" to get by. You're already way ahead of most who encounter math in some way. This line, $$\displaystyle f'(x)=u^2v+v^2u$$ is not generally true for $$\displaystyle f(x)=u(x) \cdot v(x)$$ (note that u and v are both themselves functions of x). You can in theory always factor out f(x) from something, but many times it won't do you any good. CB simply noticed that in this problem it would save time by doing so. For example, if you have $$\displaystyle f(x)=x^3+2$$ then $$\displaystyle f'(x)=3x^2$$. You could write this as $$\displaystyle f(x) \cdot \frac{3x^2}{x^3+2}$$ ,which is "factoring out" f(x) in a sense, but that would be silly. So factoring f(x) out for simplification purposes seems to be a situational trick to apply. I hope this addresses what you were thinking of somewhat. Please let me know if I'm not getting your comments properly. #### DeusAbscondus ##### Active member DeusAbcondus, I think it's really great you're diving into math this way of trying out new things and not memorizing "the book" to get by. You're already way ahead of most who encounter math in some way. This line, $$\displaystyle f'(x)=u^2v+v^2u$$ is not generally true for $$\displaystyle f(x)=u(x) \cdot v(x)$$ (note that u and v are both themselves functions of x). You can in theory always factor out f(x) from something, but many times it won't do you any good. CB simply noticed that in this problem it would save time by doing so. For example, if you have $$\displaystyle f(x)=x^3+2$$ then $$\displaystyle f'(x)=3x^2$$. You could write this as $$\displaystyle f(x) \cdot \frac{3x^2}{x^3+2}$$ ,which is "factoring out" f(x) in a sense, but that would be silly. So factoring f(x) out for simplification purposes seems to be a situational trick to apply. I hope this addresses what you were thinking of somewhat. Please let me know if I'm not getting your comments properly. Thanks Jameson, not least for your encouraging (and very generous) appraisal of my faltering steps. I was momentarily stunned and delighted by that move by CB, which I'd not encountered before (early days for me) so I leapt upon it like a treasure-seeker, thinking it was going to prove to be a key to a treasure-trove! However, that still leaves me in a quandary as to just what he did do by that step: I still can't see it, as if it is a conjurer's trick with the pea under one of three cups, or none at all! Having been so good to read this far, could you put me out of my misery and explain a bit further that particular line of his reasoning? Thanks ever so much, D'abs #### Jameson ##### Administrator Staff member I'll do my best In all honesty I'm not really one of the "math guys" here. I handle all of the administrative ins and outs to keep the site running and promote it. Of course I have a certain passion for math, otherwise I wouldn't be here, but my level isn't as high as many others. Ok, so let's start at the beginning. CB rewrote $$\displaystyle f(x) = \frac{1}{x\sqrt{5-2x}} \text{ as } f(x) = x^{-1}(5-2x)^{-1/2}$$. I think that should be clear. Then he took the derivative using the product rule. If y = uv then y' = uv'+u'v. It looks like the way he wrote it though he just reversed the order and did u'v+uv' but that's the same result naturally. That's this line. $$\displaystyle f'(x)=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)$$ Now comes the factoring part. There are two main terms in the derivative separated by an addition sign. Let's look at the first. $$\displaystyle (-1)x^{-2}(5-2x)^{-1/2}$$ <- this term. Remember that $$\displaystyle f(x)=x^{-1}(5-2x)^{-1/2}$$ When looking at the term you'll notice many similarities. The $$\displaystyle (5-2x)^{-1/2}$$ is exactly the same and the $$\displaystyle -x^{-2}$$ is not that different from $$\displaystyle x^{-1}$$ which is in f(x). Here is where the factoring begins. This uses a very common and powerful rule of combining exponents which are multiplied together that says $$\displaystyle x^a \cdot x^b=x^{a+b}$$ Using that in our case we note that $$\displaystyle x^{-2}=x^{-1} \cdot x^{-1}$$. So finally we start with $$\displaystyle (-1)x^{-2}(5-2x)^{-1/2}=(-1) \cdot \left( x^{-1} \cdot x^{-1} \right) \cdot (5-2x)^{-1/2}$$ Now we just regroup the parentheses, getting $$\displaystyle \left( (-1) \cdot x^{-1} \right) \cdot \left( x^{-1} \cdot (5-2x)^{-1/2} \right)= \left( (-1) \cdot x^{-1} \right) \cdot f(x)=-x^{-1} \cdot f(x)$$ Is this along the lines of what you were asking? #### DeusAbscondus ##### Active member I'll do my best This uses a very common and powerful rule of combining exponents which are multiplied together that says $$\displaystyle x^a \cdot x^b=x^{a+b}$$ Using that in our case we note that $$\displaystyle x^{-2}=x^{-1} \cdot x^{-1}$$. So finally we start with $$\displaystyle (-1)x^{-2}(5-2x)^{-1/2}=(-1) \cdot \left( x^{-1} \cdot x^{-1} \right) \cdot (5-2x)^{-1/2}$$ Now we just regroup the parentheses, getting $$\displaystyle \left( (-1) \cdot x^{-1} \right) \cdot \left( x^{-1} \cdot (5-2x)^{-1/2} \right)= \left( (-1) \cdot x^{-1} \right) \cdot f(x)=-x^{-1} \cdot f(x)$$ Is this along the lines of what you were asking? Jameson, Not only does it hit the nail squarely and sweetly on the spot, it does so with concision and elegance. I love maths and I love this site that puts me in touch with excellent minds and good-hearted people, and sometimes, with these qualities mixed in the same person, such as in your case. Big thanks and big respect! DeusAbs #### CaptainBlack ##### Well-known member At the risk of adding too many cooks, I would say that the quotient rule is not a bad way to go here. The nice thing about the quotient rule is that it sometimes saves algebra after taking all your derivatives, because you don't have to get a common denominator. There is an excellent reason why I did not even consider using the quotient rule; that is that I do not know it. Why do I not know it you ask, well when we did it when I was 16 or so I realised that it was redundant, every thing that can be done with the quotient rule can be done with the product rule since: \[\frac{d}{dx}\left( \frac{u(x)}{v(x)}\right)=\frac{d}{dx}\left( u(x) \times (v(x))^{-1}\right)=u'v^{-1}-u v^{-2}v'$

by the product rule, which saves valuable wet storage.

CB


text:
## Intrinsic volumes of inscribed random polytopes in smooth convex bodies

Bárány, Imre
Fodor, Ferenc
Vígh, Viktor
##### Description
Let $K$ be a $d$ dimensional convex body with a twice continuously differentiable boundary and everywhere positive Gauss-Kronecker curvature. Denote by $K_n$ the convex hull of $n$ points chosen randomly and independently from $K$ according to the uniform distribution. Matching lower and upper bounds are obtained for the orders of magnitude of the variances of the $s$-th intrinsic volumes $V_s(K_n)$ of $K_n$ for $s\in\{1, ..., d\}$. Furthermore, strong laws of large numbers are proved for the intrinsic volumes of $K_n$. The essential tools are the Economic Cap Covering Theorem of B\'ar\'any and Larman, and the Efron-Stein jackknife inequality.
##### Keywords
Mathematics - Metric Geometry, Mathematics - Probability, 52A22